https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Cocohearts&feedformat=atom AoPS Wiki - User contributions [en] 2020-11-29T11:14:36Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125558 2011 USAJMO Problems/Problem 3 2020-06-15T23:42:36Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\Im(w_n^3)}{\Re(w_n^3)}<br /> &amp;= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\<br /> &amp;= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=-\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ~ cocohearts<br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125557 2011 USAJMO Problems/Problem 3 2020-06-15T23:19:32Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\Im(w_n^3)}{\Re(w_n^3)}<br /> &amp;= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\<br /> &amp;= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ~ cocohearts<br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125556 2011 USAJMO Problems/Problem 3 2020-06-15T23:07:59Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\mathb{Im}(w_n^3)}{\mathb{Re}(w_n^3)}<br /> &amp;= \frac{\mathb{Im}((1+2a_ni)^3)}{\mathb{Re}((1+2a_ni)^3)}\\<br /> &amp;= \frac{\mathb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ~ cocohearts<br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125555 2011 USAJMO Problems/Problem 3 2020-06-15T23:04:04Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\mathbb{Im}(w_n^3)}{\mathbb{Re}(w_n^3)}<br /> &amp;= \frac{\mathbb{Im}((1+2a_ni)^3)}{\mathbb{Re}((1+2a_ni)^3)}\\<br /> &amp;= \frac{\mathbb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathbb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ~ cocohearts<br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125552 2011 USAJMO Problems/Problem 3 2020-06-15T23:00:48Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> \begin{align*}<br /> \frac{\mathbb{Im}(w_n^3)}{\mathbb{Re}(w_n^3)}<br /> &amp;= \frac{\mathbb{Im}((1+2a_ni)^3)}{\mathbb{Re}((1+2a_ni)^3)}\\<br /> &amp;= \frac{\mathbb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathbb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.\\<br /> \end{align*}<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=125551 2011 USAJMO Problems/Problem 3 2020-06-15T22:58:34Z <p>Cocohearts: /* Solutions */</p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; Define the three complex numbers &lt;math&gt;w_n = 1+2a_ni&lt;/math&gt; for &lt;math&gt;n=1,2,3&lt;/math&gt;. Then note that the slope - that is, the imaginary part divided by the real part - of all &lt;math&gt;w_n^3&lt;/math&gt; is constant, say it is &lt;math&gt;k&lt;/math&gt;. Then for &lt;math&gt;n=1,2,3&lt;/math&gt;,<br /> <br /> \begin{align*}<br /> \frac{\mathbb{Im}(w_n^3)}{\mathbb{Re}(w_n^3)}<br /> &amp;= \frac{\mathbb{Im}((1+2a_ni)^3)}{\mathbb{Re}((1+2a_ni)^3)}\\<br /> &amp;= \frac{\mathbb{Im}(1+6a_ni-12a_n^2-8a_n^3i)}{\mathbb{Re}(1+6a_ni-12a_n^2-8a_n^3i)}\\<br /> &amp;= \frac{6a_n-8a_n^3}{1-12a_n^2}\\<br /> &amp;= k.<br /> \end{align*}<br /> <br /> Rearranging, we get that&lt;cmath&gt;8a_n^3 -12ka_n^2-6a_n+k=0,&lt;/cmath&gt;or&lt;cmath&gt;a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.&lt;/cmath&gt;Note that this is a cubic, and the roots are &lt;math&gt;a_1,a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; which are all distinct, and so there are no other roots. Using Vieta's, we get that<br /> &lt;cmath&gt;a_1+a_2+a_3=\frac{3k}2,&lt;/cmath&gt;and&lt;cmath&gt;a_1a_2+a_2a_3+a_3a_1=\frac34.&lt;/cmath&gt;<br /> Obviously all values of &lt;math&gt;k&lt;/math&gt; are possible, and so our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> <br /> ===Solution 2===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 3===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> <br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Nesbitt%27s_Inequality&diff=118706 Nesbitt's Inequality 2020-03-03T21:41:19Z <p>Cocohearts: /* Proofs */</p> <hr /> <div>'''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs. It states that for positive &lt;math&gt;a, b, c &lt;/math&gt;,<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{3}{2}<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> with equality when all the variables are equal.<br /> <br /> All of the proofs below generalize to proof the following more general inequality.<br /> <br /> If &lt;math&gt; a_1, \ldots a_n &lt;/math&gt; are positive and &lt;math&gt; \sum_{i=1}^{n}a_i = s &lt;/math&gt;, then <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{i=1}^{n}\frac{a_i}{s-a_i} \geq \frac{n}{n-1}<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> or equivalently<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{i=1}^{n}\frac{s}{s-a_i} \geq \frac{n^2}{n-1}<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> with equality when all the &lt;math&gt;a_i &lt;/math&gt; are equal.<br /> <br /> == Proofs ==<br /> <br /> === By Smoothing ===<br /> <br /> We may normalize so that &lt;math&gt;a+b+c=1&lt;/math&gt;. Then we wish to show&lt;cmath&gt;\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq\frac32.&lt;/cmath&gt;The equality case is when &lt;math&gt;a=b=c&lt;/math&gt; so we can use smoothing. We wish to show that if &lt;math&gt;a\neq b&lt;/math&gt; then&lt;cmath&gt;\frac{a}{1-a}+\frac{b}{1-b}&gt;2\cdot\frac{\frac{a+b}2}{1-\frac{a+b}2}.&lt;/cmath&gt;If this is true, then if the minimum occurs at a point where &lt;math&gt;a\neq b&lt;/math&gt; then we can replace &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; with their average and then the expression decreases, so that cannot be the minimum. Thus the minimum occurs where they are all equal, and we would be done.<br /> <br /> Expanding both sides of the inequality and clearing denominators we get &lt;cmath&gt;a^2+b^2&gt;2ab,&lt;/cmath&gt;which is true because &lt;math&gt;a\neq b&lt;/math&gt;.<br /> <br /> === By Rearrangement ===<br /> <br /> Note that &lt;math&gt;a,b,c &lt;/math&gt; and &lt;math&gt; \frac{1}{b+c} = \frac{1}{a+b+c -a}&lt;/math&gt;, &lt;math&gt; \frac{1}{c+a} = \frac{1}{a+b+c -b} &lt;/math&gt;, &lt;math&gt; \frac{1}{a+b} = \frac{1}{a+b+c -c} &lt;/math&gt; are sorted in the same order. Then by the [[rearrangement inequality]],<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> 2 \left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right) \ge \frac{b}{b+c} + \frac{c}{b+c} + \frac{c}{c+a} + \frac{a}{c+a} + \frac{a}{a+b} + \frac{b}{a+b} = 3<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> For equality to occur, since we changed &lt;math&gt;{} a \cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} &lt;/math&gt; to &lt;math&gt; b \cdot \frac{1}{b+c} + a \cdot \frac{1}{c+a} &lt;/math&gt;, we must have &lt;math&gt;a=b &lt;/math&gt;, so by symmetry, all the variables must be equal.<br /> <br /> === By Cauchy ===<br /> <br /> By the [[Cauchy-Schwarz Inequality]], we have<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> [(b+c) + (c+a) + (a+b)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 9<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> or<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> 2\left( \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{a+b} \right) \geq 9<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> as desired. Equality occurs when &lt;math&gt;(b+c)^2 = (c+a)^2 = (a+b)^2 &lt;/math&gt;, i.e., when &lt;math&gt;a=b=c &lt;/math&gt;.<br /> <br /> We also present three closely related variations of this proof, which illustrate how [[AM-HM]] is related to [[AM-GM]] and Cauchy.<br /> <br /> ==== By AM-GM ====<br /> <br /> By applying [[AM-GM]] twice, we have<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> [(b+c) + (c+a) + (a+b)] \left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 3 [(b+c)(c+a)(a+b)]^{\frac{1}{3}} \cdot \left(\frac{1}{(b+c)(c+a)(a+b)}\right)^{\frac{1}{3}} = 9<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> which yields the desired inequality.<br /> <br /> ==== By Expansion and AM-GM ====<br /> <br /> We consider the equivalent inequality<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> [(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 9<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> Setting &lt;math&gt;x = b+c, y= c+a, z= a+b &lt;/math&gt;, we expand the left side to obtain<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> 3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \geq 9<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> which follows from &lt;math&gt; \frac{x}{y} + \frac{y}{x} \geq 2 &lt;/math&gt;, etc., by [[AM-GM]], with equality when &lt;math&gt;x=y=z &lt;/math&gt;.<br /> <br /> ==== By AM-HM ====<br /> <br /> The [[AM-HM]] inequality for three variables,<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> is equivalent to<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> (x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 9<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> Setting &lt;math&gt;x=b+c, y=c+a, z=a+b &lt;/math&gt; yields the desired inequality.<br /> <br /> === By Substitution ===<br /> <br /> The numbers &lt;math&gt;x = \frac{a}{b+c}, y = \frac{b}{c+a}, z = \frac{c}{a+b} &lt;/math&gt; satisfy the condition &lt;math&gt;xy + yz + zx + 2xyz = 1 &lt;/math&gt;. Thus it is sufficient to prove that if any numbers &lt;math&gt;x,y,z &lt;/math&gt; satisfy &lt;math&gt;xy + yz + zx + 2xyz = 1 &lt;/math&gt;, then &lt;math&gt; x+y+z \geq \frac{3}{2} &lt;/math&gt;.<br /> <br /> Suppose, on the contrary, that &lt;math&gt; x+y+z &lt; \frac{3}{2} &lt;/math&gt;. We then have &lt;math&gt;xy + yz + zx \leq \frac{(x+y+z)^2}{3} &lt; \frac{3}{4} &lt;/math&gt;, and &lt;math&gt; 2xyz \leq 2 \left( \frac{x+y+z}{3} \right)^3 &lt; \frac{1}{4} &lt;/math&gt;. Adding these inequalities yields &lt;math&gt;xy + yz + zx + 2xyz &lt; 1 &lt;/math&gt;, a contradiction.<br /> <br /> === By Normalization and AM-HM ===<br /> <br /> We may normalize so that &lt;math&gt;a+b+c = 1 &lt;/math&gt;. It is then sufficient to prove<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \geq \frac{9}{2}<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> which follows from [[AM-HM]].<br /> <br /> === By Weighted AM-HM ===<br /> <br /> We may normalize so that &lt;math&gt;a+b+c =1 &lt;/math&gt;.<br /> <br /> We first note that by the [[rearrangement inequality]] or the fact that &lt;math&gt;(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0&lt;/math&gt;,<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> 3 (ab + bc + ca) \leq a^2 + b^2 + c^2 + 2(ab + bc + ca) <br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> so<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{1}{a(b+c) + b(c+a) + c(a+b)} \geq \frac{1}{\frac{2}{3}(a+b+c)^2} = \frac{3}{2}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Since &lt;math&gt;a+b+c = 1 &lt;/math&gt;, weighted AM-HM gives us<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> a\cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} + c \cdot \frac{1}{a+b} \geq \frac{1}{a(b+c) + b(c+a) + c(a+b)} \geq \frac{3}{2}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> ===By Muirhead's and Cauchy===<br /> <br /> By Cauchy, &lt;cmath&gt;\sum_{\text{cyc}}\frac{a^2}{ab + ac} \geq \frac{(a + b + c)^2}{2(ab + ac + bc)} = \frac{a^2 + b^2 + c^2 + 2(ab + ac + bc)}{2(ab + ac + bc)}&lt;/cmath&gt; &lt;cmath&gt; = 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \geq \frac{3}{2}&lt;/cmath&gt; since &lt;math&gt;a^2 + b^2 + c^2 \geq ab + ac + bc&lt;/math&gt; by Muirhead as &lt;math&gt;[1, 1, 0]\prec [2, 0, 0]&lt;/math&gt;<br /> <br /> ===Another Interesting Method===<br /> <br /> Let &lt;cmath&gt;S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&lt;/cmath&gt;<br /> And &lt;cmath&gt;M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}&lt;/cmath&gt;<br /> And &lt;cmath&gt;N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}&lt;/cmath&gt;<br /> Now, we get &lt;cmath&gt;M+N=3&lt;/cmath&gt;<br /> Also by AM-GM; &lt;cmath&gt;M+S\geq 3&lt;/cmath&gt; and &lt;cmath&gt;N+S\geq 3&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies M+N+2S\geq 6&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 2S\geq 3&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies S\geq \frac{3}{2}&lt;/cmath&gt;<br /> <br /> ===By Muirhead's and expansion===<br /> <br /> Let &lt;math&gt;[x,y,z]=\sum_{sym} a^xb^yc^z&lt;/math&gt;. Expanding out we get that our inequality is equivalent to &lt;cmath&gt;\sum_{cyc} a^3+\sum_{sym} a^2b+\sum_{cyc} abc \geq \frac{3(a+b)(b+c)(c+a)}{2}&lt;/cmath&gt; This means &lt;cmath&gt;[3,0,0]/2+[2,1,0]+[1,1,1]/2\geq \frac{3}{2}(a+b)(b+c)(a+c)&lt;/cmath&gt; So it follows that we must prove &lt;cmath&gt;[3,0,0]+2[2,1,0]+[1,1,1]\geq 3([2,1,0]+[1,1,1]/3)&lt;/cmath&gt; So it follows that we must prove &lt;math&gt;[3,0,0]\geq [2,1,0]&lt;/math&gt; which immediately follows from Muirhead's.<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Medial_triangle&diff=114313 Medial triangle 2020-01-05T00:54:30Z <p>Cocohearts: </p> <hr /> <div>The medial triangle is created by joining the midpoints of a triangle's sides. It is similar to the original triangle by a spiral similarity at G the centroid with a scale factor of 2 and a rotation of 180 degrees.<br /> <br /> {{stub}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106222 2007 AIME I Problems/Problem 10 2019-06-10T01:13:06Z <p>Cocohearts: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start by showing that every group of &lt;math&gt;6&lt;/math&gt; rows can be grouped into &lt;math&gt;3&lt;/math&gt; complementary pairs.<br /> We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; shaded. Note how if there is no complement to this, then all the other five rows must have at least one square in the first two columns shaded. That means that in total, the first two rows have &lt;math&gt;2+5=7&lt;/math&gt; squares shaded in- that is false since it should only be &lt;math&gt;6&lt;/math&gt;.<br /> Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of &lt;math&gt;6&lt;/math&gt; rows can be grouped into &lt;math&gt;3&lt;/math&gt; complementary pairs.<br /> <br /> Now we proceed with three cases.<br /> <br /> *There are &lt;math&gt;\frac{\binom42}2=3&lt;/math&gt; pairs of complementary pairs. The first case is that the three pairs are all different, meaning that every single possible pair of shaded squares is used once. This gives us &lt;math&gt;6!=720.&lt;/math&gt;<br /> <br /> *Our second case is that two of the pairs are the same, and the third is different. We have &lt;math&gt;3&lt;/math&gt; to choose the pair that shows up twice and &lt;math&gt;2&lt;/math&gt; for the other, giving us &lt;math&gt;3\cdot2\cdot\binom62\binom42\binom21=6\cdot15\cdot6\cdot2=1080.&lt;/math&gt;<br /> <br /> *Our final case is that all three pairs are the same. This is just &lt;math&gt;3\cdot\binom63=60.&lt;/math&gt;<br /> <br /> Our answer is thus &lt;math&gt;720+1080+60=1860,&lt;/math&gt; leaving us with a final answer of &lt;math&gt;\boxed{860}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106221 2007 AIME I Problems/Problem 10 2019-06-10T01:09:01Z <p>Cocohearts: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start by showing that every group of &lt;math&gt;6&lt;/math&gt; rows can be grouped into &lt;math&gt;3&lt;/math&gt; complementary pairs.<br /> We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; shaded. Note how if there is no complement to this, then all the other five rows must have at least one square in the first two columns shaded. That means that in total, the first two rows have &lt;math&gt;2+5=7&lt;/math&gt; squares shaded in- that is false since it should only be &lt;math&gt;6&lt;/math&gt;.<br /> Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of &lt;math&gt;6&lt;/math&gt; rows can be grouped into &lt;math&gt;3&lt;/math&gt; complementary pairs.<br /> <br /> Now we proceed with three cases.<br /> <br /> &lt;math&gt;1.&lt;/math&gt; There are &lt;math&gt;\frac{\binom42}2=3&lt;/math&gt; pairs of complementary pairs. The first case is that the three pairs are all different, meaning that every single possible pair of shaded squares is used once. This gives us &lt;math&gt;6!=720.&lt;/math&gt;<br /> <br /> &lt;math&gt;2.&lt;/math&gt; Our second case is that two of the pairs are the same, and the third is different. We have &lt;math&gt;3&lt;/math&gt; to choose the pair that shows up twice and &lt;math&gt;2&lt;/math&gt; for the other, giving us &lt;math&gt;3\cdot2\cdot\binom62\binom42\binom21=6\cdot15\cdot6\cdot2=1080.&lt;/math&gt;<br /> <br /> &lt;math&gt;3.&lt;/math&gt; Our final case is that all three pairs are the same. This is just &lt;math&gt;3\cdot\binom63=60.&lt;/math&gt;<br /> <br /> Our answer is thus &lt;math&gt;720+1080+60=1860,&lt;/math&gt; leaving us with a final answer of &lt;math&gt;\boxed{860}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106220 2007 AIME I Problems/Problem 10 2019-06-10T00:55:38Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106219 2007 AIME I Problems/Problem 10 2019-06-10T00:55:05Z <p>Cocohearts: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106218 2007 AIME I Problems/Problem 10 2019-06-10T00:54:52Z <p>Cocohearts: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106217 2007 AIME I Problems/Problem 10 2019-06-10T00:54:41Z <p>Cocohearts: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106216 2007 AIME I Problems/Problem 10 2019-06-10T00:54:17Z <p>Cocohearts: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106215 2007 AIME I Problems/Problem 10 2019-06-10T00:54:03Z <p>Cocohearts: /* Solution 7 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 8===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106214 2007 AIME I Problems/Problem 10 2019-06-10T00:53:52Z <p>Cocohearts: /* Solution 6 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 7===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 7===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106213 2007 AIME I Problems/Problem 10 2019-06-10T00:53:40Z <p>Cocohearts: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 6 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 6===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 7===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106212 2007 AIME I Problems/Problem 10 2019-06-10T00:53:30Z <p>Cocohearts: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> === Solution 5 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 6===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 7===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_10&diff=106211 2007 AIME I Problems/Problem 10 2019-06-10T00:53:12Z <p>Cocohearts: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let &lt;math&gt;N&lt;/math&gt; be the number of shadings with this property. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> {|<br /> |-<br /> |__TOC__<br /> |&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;<br /> |[[Image:AIME I 2007-10.png]]<br /> |}<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Consider the first column. There are &lt;math&gt;{6\choose3} = 20&lt;/math&gt; ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.<br /> <br /> Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:<br /> <br /> *All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.<br /> <br /> *Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have &lt;math&gt;3 \cdot 2 \cdot 3 \cdot 3 = 54&lt;/math&gt;.<br /> <br /> *All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in. But there are 6 ways to do this. So in all there are 36 ways.<br /> <br /> So there are &lt;math&gt;20(3+54+36) = 1860&lt;/math&gt; different shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{6\choose3}&lt;/math&gt; to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:<br /> <br /> [[Image:AIME I 2007-10b.PNG|thumbnail|left|200px|One example of each case for the first two columns]]<br /> <br /> *If column 1 and column 2 do not share any two filled squares on the same row, then there are &lt;math&gt;{6\choose3}&lt;/math&gt; combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives &lt;math&gt;{6\choose3}^2 = 400&lt;/math&gt; arrangements.<br /> *If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row (&lt;math&gt;6 - 1 = 5&lt;/math&gt; places), share another empty square on a row, and have 2 squares each on different rows. This gives &lt;math&gt;6 \cdot 5 \cdot {4\choose2} = 180&lt;/math&gt;. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get &lt;math&gt;{4\choose2} = 6&lt;/math&gt;. We get &lt;math&gt;180 \cdot 6 = 1080&lt;/math&gt;.<br /> * If column 1 and column 2 share 2 filled squares on the same row (&lt;math&gt;{6\choose2} = 15&lt;/math&gt; places), they must also share 2 empty squares on the same row (&lt;math&gt;{4\choose2} = 6&lt;/math&gt;). The last two squares can be arranged in &lt;math&gt;{2\choose1} = 2&lt;/math&gt; positions; this totals to &lt;math&gt;15 \cdot 6 \cdot 2 = 180&lt;/math&gt;. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have &lt;math&gt;{2\choose1} = 2&lt;/math&gt; places, giving &lt;math&gt;180 \cdot 2 = 360&lt;/math&gt;.<br /> * If column 1 and column 2 share 3 filled squares on the same row (&lt;math&gt;{6\choose3} = 20&lt;/math&gt; places), then the squares on columns 3 and 4 are fixed.<br /> <br /> Thus, there are &lt;math&gt;400 + 1080 + 360 + 20 = 1860&lt;/math&gt; number of shadings, and the solution is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We draw a [[bijection]] between walking from &lt;math&gt;(0,0,0,0)&lt;/math&gt; to &lt;math&gt;(3,3,3,3)&lt;/math&gt; as follows: if in the &lt;math&gt;i&lt;/math&gt;th row, the &lt;math&gt;j&lt;/math&gt;th and &lt;math&gt;k&lt;/math&gt;th columns are shaded, then the &lt;math&gt;(2i-1)&lt;/math&gt;st step is in the direction corresponding to &lt;math&gt;j&lt;/math&gt;, and the &lt;math&gt;(2i)&lt;/math&gt;th step is in the direction corresponding to &lt;math&gt;k&lt;/math&gt; (&lt;math&gt;j &lt; k&lt;/math&gt;) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that &lt;math&gt;j!=k&lt;/math&gt; to solve the problem:<br /> <br /> &lt;math&gt;\frac{1}{2^6}\left(\frac{12!}{3!^4}-4\cdot{6 \choose 1}\frac{10!}{3!^3}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^2}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}4!\right) = 1860&lt;/math&gt;<br /> <br /> So that the answer is &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> === Solution 4 ===<br /> <br /> Consider all possible shadings for a single row. There are &lt;math&gt;{4 \choose 2}=6&lt;/math&gt; ways to do so, and denote these as &lt;math&gt;a=1+2&lt;/math&gt;, &lt;math&gt;b=3+4&lt;/math&gt;, &lt;math&gt;c=1+4&lt;/math&gt;, &lt;math&gt;d=2+3&lt;/math&gt;, &lt;math&gt;e=1+3&lt;/math&gt;, and &lt;math&gt;f=2+4&lt;/math&gt; where &lt;math&gt;x+y&lt;/math&gt; indicates that columns &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are shaded. From our condition on the columns, we have &lt;math&gt;a+c+e=a+d+f=b+d+e=b+c+f=3&lt;/math&gt; Summing the first two and the last two equations, we have &lt;math&gt;2a+c+d+e+f=6=2b+c+d+e+f&lt;/math&gt;, from which we have &lt;math&gt;a=b&lt;/math&gt;. Likewise, &lt;math&gt;c=d&lt;/math&gt; and &lt;math&gt;e=f&lt;/math&gt; since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for &lt;math&gt;c&lt;/math&gt;/&lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;/&lt;math&gt;f&lt;/math&gt;, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives &lt;math&gt;3{6 \choose 3}=60&lt;/math&gt; solutions; the second gives &lt;math&gt;6\cdot 6\cdot5\cdot{4 \choose 2}=1080&lt;/math&gt; solutions, and the final case gives &lt;math&gt;6!=720&lt;/math&gt; solutions. In all, we have 1860 solutions, for an answer of &lt;math&gt;860&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Each shading can be brought, via row swapping operations, to a state with a &lt;math&gt;3\times2&lt;/math&gt; shaded &lt;math&gt;L&lt;/math&gt; in the lower left hand corner. The number of such arrangements multiplied by &lt;math&gt;{5 \choose 2}{3\choose 2}&lt;/math&gt; will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form &lt;math&gt;\{3,0,0\},\{0,1,2\},\{1,1,1\}&lt;/math&gt;.<br /> Form 1: The entire lower left &lt;math&gt;3\times2&lt;/math&gt; rectangle is shaded, forcing the opposite &lt;math&gt;3\times2&lt;/math&gt; rectangle to also be shaded; thus 1 arrangement<br /> Form 2: There is a column with nothing shaded in the bottom right &lt;math&gt;3\times2&lt;/math&gt;, so it must be completely shaded in the upper right &lt;math&gt;3\times2&lt;/math&gt;. Now consider the upper right half column that will have &lt;math&gt;1&lt;/math&gt; shade. There are &lt;math&gt;3&lt;/math&gt; ways of choosing this shade, and all else is determined from here; thus 3 arrangements<br /> Form 3: The upper right &lt;math&gt;3\times3&lt;/math&gt; will have exactly &lt;math&gt;2&lt;/math&gt; shades per column and row. This is equivalent to the number of terms in a &lt;math&gt;3\times3&lt;/math&gt; determinant, or &lt;math&gt;6&lt;/math&gt; arrangements<br /> <br /> Of the &lt;math&gt;3^2&lt;/math&gt; ways of choosing to complete the bottom half of the &lt;math&gt;4\times6&lt;/math&gt;, form 1 is achieved in exactly 1 way; form 2 is achieved in &lt;math&gt;3\times2&lt;/math&gt; ways; and form &lt;math&gt;3&lt;/math&gt; in the remaining &lt;math&gt;2&lt;/math&gt; ways.<br /> Thus, the weighted total is &lt;math&gt;1+6\times3+2\times6=31&lt;/math&gt;.<br /> Complete: &lt;math&gt;31\times60=860\mod{1000}&lt;/math&gt;<br /> <br /> ===Solution 6===<br /> Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.<br /> <br /> There are &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways to choose which rows have 1 shaded square (which we'll call a &quot;1-row&quot;) within the first 3 columns and which rows have 2 (we'll call these &quot;2-rows&quot;) within the first 3 columns. Next, we do some casework:<br /> <br /> *If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only &lt;math&gt;{3 \choose 3}{3 \choose 0} \times 1= 1&lt;/math&gt; valid shading in this case.<br /> <br /> *If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are &lt;math&gt;{3 \choose 2}{3 \choose 1}\times 2=18&lt;/math&gt; valid shadings in this case.<br /> <br /> *If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are &lt;math&gt;{3 \choose 1}{3 \choose 2}\times {4 \choose 2}=54&lt;/math&gt; valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).<br /> <br /> *If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are &lt;math&gt;{3 \choose 0}{3 \choose 3}\times{6 \choose 3} = 20&lt;/math&gt; valid shadings in this case.<br /> <br /> In total, we have &lt;math&gt;{6\choose3}(1+18+54+20)=20*93=1860&lt;/math&gt;. Thus our answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> ===Solution 7===<br /> <br /> We can use generating functions. Suppose that the variables &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function &lt;math&gt;ab+ac+ad+bc+bd+cd&lt;/math&gt;, which we can write as &lt;math&gt;P(a,b,c,d)=(ab+cd)+(a+b)(c+d)&lt;/math&gt;. Therefore, &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt; represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in &lt;math&gt;P(a,b,c,d)^6&lt;/math&gt;.<br /> <br /> By the Binomial Theorem,<br /> &lt;cmath&gt;P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}&lt;/cmath&gt;<br /> If we expand &lt;math&gt;(ab+cd)^k&lt;/math&gt;, then the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are always equal. Therefore, to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt;, the powers of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;(a+b)^{6-k}&lt;/math&gt; must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that &lt;math&gt;k&lt;/math&gt; must be even. We can use the same logic for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;. Therefore, the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in the following expression is the same as the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (1).<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}&lt;/cmath&gt;<br /> Now we notice that the only way to obtain terms of the form &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; is if we take the central term in the binomial expansion of &lt;math&gt;(ab+cd)^{2k}&lt;/math&gt;. Therefore, the terms that contribute to the coefficient of &lt;math&gt;a^3b^3c^3d^3&lt;/math&gt; in (2) are<br /> &lt;cmath&gt;\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.&lt;/cmath&gt;<br /> This sum is &lt;math&gt;400+1080+360+20=1860&lt;/math&gt; so the answer is &lt;math&gt;\boxed{860}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=104245 2011 AIME I Problems/Problem 14 2019-03-11T00:40:34Z <p>Cocohearts: /* Diagram */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 4===<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> O2=(A+E)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> label(&quot;$O$&quot;,O2,dir(152.5));<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> draw(O2--(W+X)/2,red);<br /> draw(O2--N,red);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_2$&quot;,(W+X)/2,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=104244 2011 AIME I Problems/Problem 14 2019-03-11T00:39:31Z <p>Cocohearts: /* Diagram */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 4===<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> O2=(A+E)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> label(&quot;$O$&quot;,O2,NE);<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> draw(O2--(W+X)/2,red);<br /> draw(O2--N);<br /> dot(O2);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_2$&quot;,(W+X)/2,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=104242 2011 AIME I Problems/Problem 14 2019-03-11T00:38:48Z <p>Cocohearts: /* Diagram */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 4===<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> O2=(A+E)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> label(&quot;$O$&quot;,O2,NE);<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> draw(O2--(W+X)/2,red);<br /> dot(O2);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_2$&quot;,(W+X)/2,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=104233 2011 AIME I Problems/Problem 14 2019-03-11T00:30:37Z <p>Cocohearts: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 4===<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=104232 2011 AIME I Problems/Problem 14 2019-03-11T00:30:13Z <p>Cocohearts: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 4===<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angleOM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{\sqrt2/4}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Chicken_McNugget_Theorem&diff=104077 Chicken McNugget Theorem 2019-03-06T05:33:04Z <p>Cocohearts: /* Proof 1 */</p> <hr /> <div>The '''Chicken McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s &lt;math&gt;m,n&lt;/math&gt;, the greatest integer that cannot be written in the form &lt;math&gt;am + bn&lt;/math&gt; for [[nonnegative]] integers &lt;math&gt;a, b&lt;/math&gt; is &lt;math&gt;mn-m-n&lt;/math&gt;.<br /> <br /> A consequence of the theorem is that there are exactly &lt;math&gt;\frac{(m - 1)(n - 1)}{2}&lt;/math&gt; positive integers which cannot be expressed in the form &lt;math&gt;am + bn&lt;/math&gt;. The proof is based on the fact that in each pair of the form &lt;math&gt;(k, (m - 1)(n - 1) - k+1)&lt;/math&gt;, exactly one element is expressible.<br /> <br /> == Origins ==<br /> There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.<br /> <br /> <br /> <br /> <br /> <br /> ==Proof 1==<br /> &lt;b&gt;Definition&lt;/b&gt;. An integer &lt;math&gt;N \in \mathbb{Z}&lt;/math&gt; will be called &lt;i&gt;purchasable&lt;/i&gt; if there exist nonnegative integers &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;am+bn = N&lt;/math&gt;.<br /> <br /> We would like to prove that &lt;math&gt;mn-m-n&lt;/math&gt; is the largest non-purchasable integer. We are required to show that (1) &lt;math&gt;mn-m-n&lt;/math&gt; is non-purchasable, and (2) every &lt;math&gt;N &gt; mn-m-n&lt;/math&gt; is purchasable. <br /> Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. Let &lt;math&gt;A_{N} \subset \mathbb{Z} \times \mathbb{Z}&lt;/math&gt; be the set of solutions &lt;math&gt;(x,y)&lt;/math&gt; to &lt;math&gt;xm+yn = N&lt;/math&gt;. Then &lt;math&gt;A_{N} = \{(x+kn,y-km) \;:\; k \in \mathbb{Z}\}&lt;/math&gt; for any &lt;math&gt;(x,y) \in A_{N}&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By [[Bezout's Lemma]], there exist integers &lt;math&gt;x',y'&lt;/math&gt; such that &lt;math&gt;x'm+y'n = 1&lt;/math&gt;. Then &lt;math&gt;(Nx')m+(Ny')n = N&lt;/math&gt;. Hence &lt;math&gt;A_{N}&lt;/math&gt; is nonempty. It is easy to check that &lt;math&gt;(Nx'+kn,Ny'-km) \in A_{N}&lt;/math&gt; for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. We now prove that there are no others. Suppose &lt;math&gt;(x_{1},y_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2})&lt;/math&gt; are solutions to &lt;math&gt;xm+yn=N&lt;/math&gt;. Then &lt;math&gt;x_{1}m+y_{1}n = x_{2}m+y_{2}n&lt;/math&gt; implies &lt;math&gt;m(x_{1}-x_{2}) = n(y_{2}-y_{1})&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime and &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;n(y_{2}-y_{1})&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;y_{2}-y_{1}&lt;/math&gt; and &lt;math&gt;y_{2} \equiv y_{1} \pmod{m}&lt;/math&gt;. Similarly &lt;math&gt;x_{2} \equiv x_{1} \pmod{n}&lt;/math&gt;. Let &lt;math&gt;k_{1},k_{2}&lt;/math&gt; be integers such that &lt;math&gt;x_{2}-x_{1} = k_{1}n&lt;/math&gt; and &lt;math&gt;y_{2}-y_{1} = k_{2}m&lt;/math&gt;. Then &lt;math&gt;m(-k_{1}n) = n(k_{2}m)&lt;/math&gt; implies &lt;math&gt;k_{1} = -k_{2}.&lt;/math&gt; We have the desired result. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. For any integer &lt;math&gt;N&lt;/math&gt;, there exists unique &lt;math&gt;(a_{N},b_{N}) \in \mathbb{Z} \times \{0,1,\ldots,m-1\}&lt;/math&gt; such that &lt;math&gt;a_{N}m + b_{N}n = N&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By the division algorithm, there exists one and only one &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;0 \le y-km \le m-1&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. &lt;math&gt;N&lt;/math&gt; is purchasable if and only if &lt;math&gt;a_{N} \ge 0&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: If &lt;math&gt;a_{N} \ge 0&lt;/math&gt;, then we may simply pick &lt;math&gt;(a,b) = (a_{N},b_{N})&lt;/math&gt; so &lt;math&gt;N&lt;/math&gt; is purchasable. If &lt;math&gt;a_{N} &lt; 0&lt;/math&gt;, then &lt;math&gt;a_{N}+kn &lt; 0&lt;/math&gt; if &lt;math&gt;k \le 0&lt;/math&gt; and &lt;math&gt;b_{N}-km &lt; 0&lt;/math&gt; if &lt;math&gt;k &gt; 0&lt;/math&gt;, hence at least one coordinate of &lt;math&gt;(a_{N}+kn,b_{N}-km)&lt;/math&gt; is negative for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. Thus &lt;math&gt;N&lt;/math&gt; is not purchasable. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Thus the set of non-purchasable integers is &lt;math&gt;\{xm+yn \;:\; x&lt;0,0 \le y \le m-1\}&lt;/math&gt;. We would like to find the maximum of this set. <br /> Since both &lt;math&gt;m,n&lt;/math&gt; are positive, the maximum is achieved when &lt;math&gt;x = -1&lt;/math&gt; and &lt;math&gt;y = m-1&lt;/math&gt; so that &lt;math&gt;xm+yn = (-1)m+(m-1)n = mn-m-n&lt;/math&gt;.<br /> <br /> ==Proof 2==<br /> We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]:<br /> <br /> &quot;Let &lt;math&gt;S = \{1,2,3,\cdots, p-1\}&lt;/math&gt;. Then, we claim that the set &lt;math&gt;a \cdot S&lt;/math&gt;, consisting of the product of the elements of &lt;math&gt;S&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;, taken modulo &lt;math&gt;p&lt;/math&gt;, is simply a permutation of &lt;math&gt;S&lt;/math&gt;. In other words, <br /> <br /> &lt;center&gt;&lt;cmath&gt;S \equiv \{1a, 2a, \cdots, (p-1)a\} \pmod{p}.&lt;/cmath&gt;&lt;/center&gt;&lt;br&gt;<br /> <br /> Clearly none of the &lt;math&gt;ia&lt;/math&gt; for &lt;math&gt;1 \le i \le p-1&lt;/math&gt; are divisible by &lt;math&gt;p&lt;/math&gt;, so it suffices to show that all of the elements in &lt;math&gt;a \cdot S&lt;/math&gt; are distinct. Suppose that &lt;math&gt;ai \equiv aj \pmod{p}&lt;/math&gt; for &lt;math&gt;i \neq j&lt;/math&gt;. Since &lt;math&gt;\text{gcd}\, (a,p) = 1&lt;/math&gt;, by the cancellation rule, that reduces to &lt;math&gt;i \equiv j \pmod{p}&lt;/math&gt;, which is a contradiction.&quot;<br /> <br /> Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, we know that multiplying the residues of &lt;math&gt;m&lt;/math&gt; by &lt;math&gt;n&lt;/math&gt; simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form &lt;math&gt;am+bn&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is the original residue. We now prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;: For any nonnegative integer &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;cn&lt;/math&gt; is the least purchasable number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Any number that is less than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; can be represented in the form &lt;math&gt;cn-dm&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is a positive integer. If this is purchasable, we can say &lt;math&gt;cn-dm=am+bn&lt;/math&gt; for some nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. This can be rearranged into &lt;math&gt;(a+d)m=(c-b)n&lt;/math&gt;, which implies that &lt;math&gt;(a+d)&lt;/math&gt; is a multiple of &lt;math&gt;n&lt;/math&gt; (since &lt;math&gt;\gcd(m, n)=1&lt;/math&gt;). We can say that &lt;math&gt;(a+d)=gn&lt;/math&gt; for some positive integer &lt;math&gt;g&lt;/math&gt;, and substitute to get &lt;math&gt;gmn=(c-b)n&lt;/math&gt;. Because &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;(c-b)n &lt; mn&lt;/math&gt;, and &lt;math&gt;gmn &lt; mn&lt;/math&gt;. We divide by &lt;math&gt;mn&lt;/math&gt; to get &lt;math&gt;g&lt;1&lt;/math&gt;. However, we defined &lt;math&gt;g&lt;/math&gt; to be a positive integer, and all positive integers are greater than or equal to &lt;math&gt;1&lt;/math&gt;. Therefore, we have a contradiction, and &lt;math&gt;cn&lt;/math&gt; is the least purchasable number congruent to &lt;math&gt;cn \bmod m&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> This means that because &lt;math&gt;cn&lt;/math&gt; is purchasable, every number that is greater than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; is also purchasable (because these numbers are in the form &lt;math&gt;am+bn&lt;/math&gt; where &lt;math&gt;b=c&lt;/math&gt;). Another result of this Lemma is that &lt;math&gt;cn-m&lt;/math&gt; is the greatest number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt; that is not purchasable. &lt;math&gt;c \leq m-1&lt;/math&gt;, so &lt;math&gt;cn-m \leq (m-1)n-m=mn-m-n&lt;/math&gt;, which shows that &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number in the form &lt;math&gt;cn-m&lt;/math&gt;. Any number greater than this and congruent to some &lt;math&gt;cn \bmod m&lt;/math&gt; is purchasable, because that number is greater than &lt;math&gt;cn&lt;/math&gt;. All numbers are congruent to some &lt;math&gt;cn&lt;/math&gt;, and thus all numbers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable.<br /> <br /> Putting it all together, we can say that for any coprime &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number not representable in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Corollary==<br /> This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt; For any integer &lt;math&gt;k&lt;/math&gt;, exactly one of the integers &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Because every number is congruent to some residue of &lt;math&gt;m&lt;/math&gt; permuted by &lt;math&gt;n&lt;/math&gt;, we can set &lt;math&gt;k \equiv cn \bmod m&lt;/math&gt; for some &lt;math&gt;c&lt;/math&gt;. We can break this into two cases.<br /> <br /> &lt;i&gt;Case 1&lt;/i&gt;: &lt;math&gt;k \leq cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is not purchasable, and that &lt;math&gt;mn-m-n-k \geq mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. &lt;math&gt;n(m-1-c)&lt;/math&gt; is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable. Therefore, &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k \geq n(m-1-c)&lt;/math&gt;, so &lt;math&gt;mn-m-n-k&lt;/math&gt; is purchasable.<br /> <br /> &lt;i&gt;Case 2&lt;/i&gt;: &lt;math&gt;k &gt; cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is purchasable, and that &lt;math&gt;mn-m-n-k &lt; mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. Again, because &lt;math&gt;n(m-1-c)&lt;/math&gt; is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable, and because &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k &lt; n(m-1-c)&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> We now limit the values of &lt;math&gt;k&lt;/math&gt; to all integers &lt;math&gt;0 \leq k \leq \frac{mn-m-n}{2}&lt;/math&gt;, which limits the values of &lt;math&gt;mn-m-n-k&lt;/math&gt; to &lt;math&gt;mn-m-n \geq mn-m-n-k \geq \frac{mn-m-n}{2}&lt;/math&gt;. Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, only one of them can be a multiple of &lt;math&gt;2&lt;/math&gt;. Therefore, &lt;math&gt;mn-m-n \equiv (0)(1)-0-1 \equiv -1 \equiv 1 \bmod 2&lt;/math&gt;, showing that &lt;math&gt;\frac{mn-m-n}{2}&lt;/math&gt; is not an integer and that &lt;math&gt;\frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2}&lt;/math&gt; are integers. We can now set limits that are equivalent to the previous on the values of &lt;math&gt;k&lt;/math&gt; and &lt;math&gt;mn-m-n-k&lt;/math&gt; so that they cover all integers form &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; without overlap: &lt;math&gt;0 \leq k \leq \frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2} \leq mn-m-n-k \leq mn-m-n&lt;/math&gt;. There are &lt;math&gt;\frac{mn-m-n-1}{2}+1=\frac{(m-1)(n-1)}{2}&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt;, and each is paired with a value of &lt;math&gt;mn-m-n-k&lt;/math&gt;, so we can make &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different ordered pairs of the form &lt;math&gt;(k, mn-m-n-k)&lt;/math&gt;. The coordinates of these ordered pairs cover all integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; inclusive, and each contains exactly one not-purchasable integer, so that means that there are &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different not-purchasable integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt;. All integers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable, so that means there are a total of &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; integers &lt;math&gt;\geq 0&lt;/math&gt; that are not purchasable.<br /> <br /> In other words, for every pair of coprime integers &lt;math&gt;m, n&lt;/math&gt;, there are exactly &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; nonnegative integers that cannot be represented in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Generalization==<br /> If &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are not relatively prime, then we can simply rearrange &lt;math&gt;am+bn&lt;/math&gt; into the form<br /> &lt;cmath&gt;\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)&lt;/cmath&gt;<br /> &lt;math&gt;\frac{m}{\gcd(m,n)}&lt;/math&gt; and &lt;math&gt;\frac{n}{\gcd(m,n)}&lt;/math&gt; are relatively prime, so we apply Chicken McNugget to find a bound<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)^{2}}-\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}&lt;/cmath&gt;<br /> We can simply multiply &lt;math&gt;\gcd(m,n)&lt;/math&gt; back into the bound to get<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)}-m-n=\textrm{lcm}(m, n)-m-n&lt;/cmath&gt;<br /> Therefore, all multiples of &lt;math&gt;\gcd(m, n)&lt;/math&gt; greater than &lt;math&gt;\textrm{lcm}(m, n)-m-n&lt;/math&gt; are representable in the form &lt;math&gt;am+bn&lt;/math&gt; for some positive integers &lt;math&gt;a, b&lt;/math&gt;.<br /> <br /> =Problems=<br /> <br /> ===Simple===<br /> *Marcy buys paint jars in containers of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;. What's the largest number of paint jars that Marcy can't obtain? <br /> <br /> Answer: &lt;math&gt;5&lt;/math&gt; containers<br /> <br /> *Bay Area Rapid food sells chicken nuggets. You can buy packages of &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;7&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; such that there is no way to buy exactly &lt;math&gt;n&lt;/math&gt; nuggets? Can you Generalize ?(ACOPS) <br /> <br /> Answer: &lt;math&gt;n=59&lt;/math&gt; <br /> <br /> *If a game of American Football has only scores of field goals (3 points) and touchdowns with the extra point (7 points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?<br /> <br /> Answer: &lt;math&gt;11&lt;/math&gt; points<br /> <br /> ===Intermediate===<br /> *Ninety-four bricks, each measuring &lt;math&gt;4''\times10''\times19'',&lt;/math&gt; are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes &lt;math&gt;4''\,&lt;/math&gt; or &lt;math&gt;10''\,&lt;/math&gt; or &lt;math&gt;19''\,&lt;/math&gt; to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? [[1994 AIME Problems/Problem 11|AIME]]<br /> <br /> ===Olympiad===<br /> *On the real number line, paint red all points that correspond to integers of the form &lt;math&gt;81x+100y&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are positive integers. Paint the remaining integer points blue. Find a point &lt;math&gt;P&lt;/math&gt; on the line such that, for every integer point &lt;math&gt;T&lt;/math&gt;, the reflection of &lt;math&gt;T&lt;/math&gt; with respect to &lt;math&gt;P&lt;/math&gt; is an integer point of a different colour than &lt;math&gt;T&lt;/math&gt;. (India TST)<br /> <br /> *Let &lt;math&gt;S&lt;/math&gt; be a set of integers (not necessarily positive) such that<br /> <br /> (a) there exist &lt;math&gt;a,b \in S&lt;/math&gt; with &lt;math&gt;\gcd(a,b)=\gcd(a-2,b-2)=1&lt;/math&gt;;<br /> <br /> (b) if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt; (possibly equal), then &lt;math&gt;x^2-y&lt;/math&gt; also belongs to &lt;math&gt;S&lt;/math&gt;. <br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; is the set of all integers. (USAMO)<br /> <br /> ==See Also==<br /> *[[Theorem]]<br /> *[[Prime]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Number theory]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_17&diff=103442 2019 AMC 12A Problems/Problem 17 2019-02-19T05:24:18Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;s_k&lt;/math&gt; denote the sum of the &lt;math&gt;\textit{k}&lt;/math&gt;th powers of the roots of the polynomial &lt;math&gt;x^3-5x^2+8x-13&lt;/math&gt;. In particular, &lt;math&gt;s_0=3&lt;/math&gt;, &lt;math&gt;s_1=5&lt;/math&gt;, and &lt;math&gt;s_2=9&lt;/math&gt;. Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be real numbers such that &lt;math&gt;s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}&lt;/math&gt; for &lt;math&gt;k = 2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;....&lt;/math&gt; What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Applying Newton's Sums (see [https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums this link]), we have&lt;cmath&gt;s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,&lt;/cmath&gt;so&lt;cmath&gt;s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},&lt;/cmath&gt;we get the answer as &lt;math&gt;5+(-8)+13=10&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;p, q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the roots of the polynomial. Then,<br /> <br /> &lt;math&gt;p^3 - 5p^2 + 8p - 13 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;q^3 - 5q^2 + 8q - 13 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;r^3 - 5r^2 + 8r - 13 = 0&lt;/math&gt;<br /> <br /> Adding these three equations, we get<br /> <br /> &lt;math&gt;(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;s_3 - 5s_2 + 8s_1 = 39&lt;/math&gt;<br /> <br /> &lt;math&gt;39&lt;/math&gt; can be written as &lt;math&gt;13s_0&lt;/math&gt;, giving<br /> <br /> &lt;math&gt;s_3 = 5s_2 - 8s_1 + 13s_0&lt;/math&gt;<br /> <br /> We are given that &lt;math&gt;s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}&lt;/math&gt; is satisfied for &lt;math&gt;k = 2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;....&lt;/math&gt;, meaning it must be satisfied when &lt;math&gt;k = 2&lt;/math&gt;, giving us &lt;math&gt;s_3 = a \, s_2 + b \, s_1 + c \, s_0&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;a = 5, b = -8&lt;/math&gt;, and &lt;math&gt;c = 13&lt;/math&gt; by matching coefficients.<br /> <br /> &lt;math&gt;5 - 8 + 13 = \boxed{\textbf{(D) } 10}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;p, q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the roots of the polynomial. By Vieta's Formulae, we have<br /> <br /> &lt;math&gt;p+q+r = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;pq+qr+rp = 8&lt;/math&gt;<br /> <br /> &lt;math&gt;pqr=13&lt;/math&gt;.<br /> <br /> We know &lt;math&gt;s_k = p^k + q^k + r^k&lt;/math&gt;. Consider &lt;math&gt;(p+q+r)(s_k) =5s_k&lt;/math&gt;.<br /> <br /> &lt;math&gt;5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k&lt;/math&gt;<br /> <br /> Using &lt;math&gt;pqr = 13&lt;/math&gt; and &lt;math&gt;s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}&lt;/math&gt;, we see<br /> &lt;math&gt;13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}&lt;/math&gt;.<br /> <br /> We have &lt;cmath&gt;\begin{split} 5s_k + 13s_{k-2} &amp;= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\<br /> &amp;= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\<br /> &amp;= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\<br /> &amp;= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}&lt;/cmath&gt;<br /> <br /> Rearrange to get<br /> &lt;math&gt;s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2019|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=101919 Mass points 2019-02-12T18:23:03Z <p>Cocohearts: /* Examples */</p> <hr /> <div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br /> <br /> Mass point geometry was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems. Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br /> <br /> The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Any line passing this central point will balance the figure. If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. If two points are balanced, the point on the balancing line used to balance them has a mass of the sum of the masses of the two points.<br /> <br /> == Examples ==<br /> Consider a triangle &lt;math&gt;ABC&lt;/math&gt; with its three [[Median_(geometry)|median]]s drawn, with the intersection points being &lt;math&gt;D, E, F,&lt;/math&gt; corresponding to &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively. Thus, if we label point &lt;math&gt;A&lt;/math&gt; with a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; must also have a weight of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are equidistant from &lt;math&gt;D&lt;/math&gt;. By the same process, we find &lt;math&gt;C&lt;/math&gt; must also have a weight of 1. Now, since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; both have a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; must have a weight of &lt;math&gt;2&lt;/math&gt; (as is true for &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;). Thus, if we label the centroid &lt;math&gt;P&lt;/math&gt;, we can deduce that &lt;math&gt;DP:PC&lt;/math&gt; is &lt;math&gt;1:2&lt;/math&gt; - the inverse ratio of their weights.<br /> <br /> &lt;math&gt;\triangle ABC&lt;/math&gt; has point &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;AB&lt;/math&gt;, point &lt;math&gt;E&lt;/math&gt; on &lt;math&gt;BC&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. &lt;math&gt;AE&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt;, and &lt;math&gt;BF&lt;/math&gt; intersect at point &lt;math&gt;G&lt;/math&gt;. The ratio &lt;math&gt;AD:DB&lt;/math&gt; is &lt;math&gt;3:5&lt;/math&gt; and the ratio &lt;math&gt;CE:EB&lt;/math&gt; is &lt;math&gt;8:3&lt;/math&gt;. Find the ratio of &lt;math&gt;FG:GB&lt;/math&gt;<br /> <br /> ==Problems==<br /> [[2013 AMC 10B Problems/Problem 16]]<br /> <br /> [[2004 AMC 10B Problems/Problem 20]]<br /> <br /> [[2009 AIME I Problems/Problem 5]]<br /> <br /> [[2009 AIME I Problems/Problem 4]]<br /> <br /> [[2001 AIME I Problems/Problem 7 ]]<br /> <br /> [[2011 AIME II Problems/Problem 4]]<br /> <br /> [[ 1992 AIME Problems/Problem 14 ]]<br /> <br /> [[ 1988 AIME Problems/Problem 12]]<br /> <br /> [[1989 AIME Problems/Problem 15]]<br /> <br /> [[1985 AIME Problems/Problem 6]]<br /> &lt;!-- Previous external links led to errors, removed--&gt;<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_9&diff=93059 2018 AIME I Problems/Problem 9 2018-03-09T03:04:06Z <p>Cocohearts: /* Solution 1 */</p> <hr /> <div>Find the number of four-element subsets of &lt;math&gt;\{1,2,3,4,\dots, 20\}&lt;/math&gt; with the property that two distinct elements of a subset have a sum of &lt;math&gt;16&lt;/math&gt;, and two distinct elements of a subset have a sum of &lt;math&gt;24&lt;/math&gt;. For example, &lt;math&gt;\{3,5,13,19\}&lt;/math&gt; and &lt;math&gt;\{6,10,20,18\}&lt;/math&gt; are two such subsets.<br /> <br /> ==Solutions==<br /> <br /> ==Solution 1==<br /> This problem is tricky because it is the capital of a few &quot;bashy&quot; calculations. Nevertheless, the process is straightforward. Call the set &lt;math&gt;\{a, b, c, d\}&lt;/math&gt;.<br /> <br /> Note that there are only two cases: 1 where &lt;math&gt;a + b = 16&lt;/math&gt; and &lt;math&gt;c + d = 24&lt;/math&gt; or 2 where &lt;math&gt;a + b = 16&lt;/math&gt; and &lt;math&gt;a + c = 24&lt;/math&gt;. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you &lt;math&gt;a=d&lt;/math&gt;, which cannot be true.<br /> <br /> Case 1.<br /> This is probably the simplest: just make a list of possible combinations for &lt;math&gt;\{a, b\}&lt;/math&gt; and &lt;math&gt;\{c, d\}&lt;/math&gt;. We get &lt;math&gt;\{1, 15\}\dots\{7, 9\}&lt;/math&gt; for the first and &lt;math&gt;\{4, 20\}\dots\{11, 13\}&lt;/math&gt; for the second. That appears to give us &lt;math&gt;7*8=56&lt;/math&gt; solutions, right? NO. Because elements can't repeat, take out the supposed sets<br /> &lt;cmath&gt;\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 16, 4, 20\}, \{5, 11, 5, 19\},&lt;/cmath&gt;&lt;cmath&gt;\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}&lt;/cmath&gt; That's ten cases gone. So &lt;math&gt;46&lt;/math&gt; for Case 1.<br /> <br /> Case 2.<br /> We can look for solutions by listing possible &lt;math&gt;a&lt;/math&gt; values and filling in the blanks. Start with &lt;math&gt;a=4&lt;/math&gt;, as that is the minimum. We find &lt;math&gt;\{4, 12, 20, ?\}&lt;/math&gt;, and likewise up to &lt;math&gt;a=15&lt;/math&gt;. But we can't have &lt;math&gt;a=8&lt;/math&gt; or &lt;math&gt;a=12&lt;/math&gt; because &lt;math&gt;a=b&lt;/math&gt; or &lt;math&gt;a=c&lt;/math&gt;, respectively! Now, it would seem like there are &lt;math&gt;10&lt;/math&gt; values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;17&lt;/math&gt; unique values for each &lt;math&gt;?&lt;/math&gt;, giving a total of &lt;math&gt;170&lt;/math&gt;, but that is once again not true because there are some repeated values! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. There are 3 pairs about &lt;math&gt;a=8&lt;/math&gt; and 3 pairs about &lt;math&gt;a=12&lt;/math&gt;, meaning we lose &lt;math&gt;6&lt;/math&gt;. That's &lt;math&gt;164&lt;/math&gt; for Case 2.<br /> <br /> Total gives &lt;math&gt;\boxed{210}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_6&diff=93058 2018 AIME I Problems/Problem 6 2018-03-09T03:01:01Z <p>Cocohearts: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of complex numbers &lt;math&gt;z&lt;/math&gt; with the properties that &lt;math&gt;|z|=1&lt;/math&gt; and &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is a real number. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a=z^{120}&lt;/math&gt;. This simplifies the problem constraint to &lt;math&gt;a^6-a \in \mathbb{R}&lt;/math&gt;. This is true if &lt;math&gt;Im(a^6)=Im(a)&lt;/math&gt;. Let &lt;math&gt;\theta&lt;/math&gt; be the angle &lt;math&gt;a&lt;/math&gt; makes with the positive x-axis. Note that there is exactly one &lt;math&gt;a&lt;/math&gt; for each angle &lt;math&gt;0\le\theta&lt;2\pi&lt;/math&gt;. This must be true for &lt;math&gt;12&lt;/math&gt; values of &lt;math&gt;a&lt;/math&gt; (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time &lt;math&gt;\sin\theta=\sin{6\theta}&lt;/math&gt;). For each of these solutions for &lt;math&gt;a&lt;/math&gt;, there are necessarily &lt;math&gt;120&lt;/math&gt; solutions for &lt;math&gt;z&lt;/math&gt;. Thus, there are &lt;math&gt;12*120=1440&lt;/math&gt; solutions for &lt;math&gt;z&lt;/math&gt;, yielding an answer of &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2018|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_25&diff=90697 2018 AMC 10A Problems/Problem 25 2018-02-09T00:54:26Z <p>Cocohearts: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For a positive integer &lt;math&gt;n&lt;/math&gt; and nonzero digits &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, let &lt;math&gt;A_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;-digit integer each of whose digits is equal to &lt;math&gt;a&lt;/math&gt;; let &lt;math&gt;B_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;-digit integer each of whose digits is equal to &lt;math&gt;b&lt;/math&gt;, and let &lt;math&gt;C_n&lt;/math&gt; be the &lt;math&gt;2n&lt;/math&gt;-digit (not &lt;math&gt;n&lt;/math&gt;-digit) integer each of whose digits is equal to &lt;math&gt;c&lt;/math&gt;. What is the greatest possible value of &lt;math&gt;a + b + c&lt;/math&gt; for which there are at least two values of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;C_n - B_n = A_n^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> <br /> Observe &lt;math&gt;A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}&lt;/math&gt;; similarly &lt;math&gt;B_n = b \cdot \tfrac{10^n - 1}{9}&lt;/math&gt; and &lt;math&gt;C_n = c \cdot \tfrac{10^{2n} - 1}{9}&lt;/math&gt;. The relation &lt;math&gt;C_n - B_n = A_n^2&lt;/math&gt; rewrites as<br /> &lt;cmath&gt;c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.&lt;/cmath&gt;Since &lt;math&gt;n &gt; 0&lt;/math&gt;, &lt;math&gt;10^n &gt; 1&lt;/math&gt; and we may cancel out a factor of &lt;math&gt;\tfrac{10^n - 1}{9}&lt;/math&gt; to obtain<br /> &lt;cmath&gt;c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.&lt;/cmath&gt;This is a linear equation in &lt;math&gt;10^n&lt;/math&gt;. Thus, if two distinct values of &lt;math&gt;n&lt;/math&gt; satisfy it, then all values of &lt;math&gt;n&lt;/math&gt; will. Matching coefficients, we need<br /> &lt;cmath&gt;c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.&lt;/cmath&gt;To maximize &lt;math&gt;a + b + c = a + \tfrac{a^2}{3}&lt;/math&gt;, we need to maximize &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be integers, &lt;math&gt;a&lt;/math&gt; must be a multiple of 3. If &lt;math&gt;a = 9&lt;/math&gt; then &lt;math&gt;b&lt;/math&gt; exceeds 9. However, if &lt;math&gt;a = 6&lt;/math&gt; then &lt;math&gt;b = 8&lt;/math&gt; and &lt;math&gt;c = 4&lt;/math&gt; for an answer of &lt;math&gt;\boxed{\textbf{(D)} \text{ 18}}&lt;/math&gt;. (CantonMathGuy)<br /> <br /> == Solution 2==<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2018|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_25&diff=89632 2015 AMC 10A Problems/Problem 25 2018-01-09T05:44:25Z <p>Cocohearts: /* Solution 2 */</p> <hr /> <div>==Problem 25==<br /> Let &lt;math&gt;S&lt;/math&gt; be a square of side length &lt;math&gt;1&lt;/math&gt;. Two points are chosen independently at random on the sides of &lt;math&gt;S&lt;/math&gt;. The probability that the straight-line distance between the points is at least &lt;math&gt;\dfrac{1}{2}&lt;/math&gt; is &lt;math&gt;\dfrac{a-b\pi}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers with &lt;math&gt;\gcd(a,b,c)=1&lt;/math&gt;. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Divide the boundary of the square into halves, thereby forming &lt;math&gt;8&lt;/math&gt; segments. Without loss of generality, let the first point &lt;math&gt;A&lt;/math&gt; be in the bottom-left segment. Then, it is easy to see that any point in the &lt;math&gt;5&lt;/math&gt; segments not bordering the bottom-left segment will be distance at least &lt;math&gt;\dfrac{1}{2}&lt;/math&gt; apart from &lt;math&gt;A&lt;/math&gt;. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least &lt;math&gt;0.5&lt;/math&gt; apart from &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;\dfrac{0 + 1}{2} = \dfrac{1}{2}&lt;/math&gt; because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)<br /> <br /> If the second point &lt;math&gt;B&lt;/math&gt; is on the left-bottom segment, then if &lt;math&gt;A&lt;/math&gt; is distance &lt;math&gt;x&lt;/math&gt; away from the left-bottom vertex, then &lt;math&gt;B&lt;/math&gt; must be at least &lt;math&gt;\dfrac{1}{2} - \sqrt{0.25 - x^2}&lt;/math&gt; away from that same vertex. Thus, using an averaging argument we find that the probability in this case is<br /> &lt;cmath&gt;\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.&lt;/cmath&gt;<br /> <br /> (Alternatively, one can equate the problem to finding all valid &lt;math&gt;(x, y)&lt;/math&gt; with &lt;math&gt;0 &lt; x, y &lt; \dfrac{1}{2}&lt;/math&gt; such that &lt;math&gt;x^2 + y^2 \ge \dfrac{1}{4}&lt;/math&gt;, i.e. &lt;math&gt;(x, y)&lt;/math&gt; is outside the unit circle with radius &lt;math&gt;0.5.&lt;/math&gt;)<br /> <br /> Thus, averaging the probabilities gives<br /> &lt;cmath&gt;P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).&lt;/cmath&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{\textbf{(A) } 59}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, on an adjacent side is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, and on the opposite side is &lt;math&gt;\frac{1}{4}&lt;/math&gt;. We discuss these three cases. <br /> <br /> Case 1: Two points are on the same side. Let the first point be &lt;math&gt;a&lt;/math&gt; and the second point be &lt;math&gt;b&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-axis with &lt;math&gt;0\le a, b\le 1&lt;/math&gt;. Consider &lt;math&gt;(a, b)&lt;/math&gt; a point on the unit square &lt;math&gt;[0,1]\times [0,1]&lt;/math&gt; on the Cartesian plane. The region &lt;math&gt;\{(a,b): |b-a|&gt; \frac{1}{2}\}&lt;/math&gt; has the area of &lt;math&gt;(\frac{1}{2})^2&lt;/math&gt;. Therefore, the probability that &lt;math&gt;|b-a|&gt; \frac{1}{2}&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt;. <br /> <br /> Case 2: Two points are on two adjacent sides. Let the two sides be &lt;math&gt;[0,1]&lt;/math&gt; on the x-axis and &lt;math&gt;[0,1]&lt;/math&gt; on the y-axis and let one point be &lt;math&gt;(a, 0)&lt;/math&gt; and the other point be &lt;math&gt;(0, b)&lt;/math&gt;. Then &lt;math&gt;0\le a, b\le 1&lt;/math&gt; and the distance between the two points is &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt;. As in Case 1, &lt;math&gt;(a, b)&lt;/math&gt; is a point on the unit square &lt;math&gt;[0,1]\times [0,1]&lt;/math&gt;. The area of the region &lt;math&gt;\{(a,b): \sqrt{a^2+b^2} \le \frac{1}{2}, 0\le a, b\le 1\}&lt;/math&gt; is &lt;math&gt;\frac{\pi}{16}&lt;/math&gt; and the area of its complementary set inside the square (i.e. &lt;math&gt;\{(a,b): \sqrt{a^2+b^2} &gt; 1/2, 0\le a, b\le 1\}&lt;/math&gt; ) is &lt;math&gt;1-\frac{\pi}{16}&lt;/math&gt;. Therefore, the probability that the distance between &lt;math&gt;(a, 0)&lt;/math&gt; and &lt;math&gt;(0, b)&lt;/math&gt; is at least &lt;math&gt;\frac{1}{2}&lt;/math&gt; is &lt;math&gt;1-\frac{\pi}{16}&lt;/math&gt;. <br /> <br /> Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least &lt;math&gt;1/2&lt;/math&gt; is obviously &lt;math&gt;1&lt;/math&gt;. <br /> <br /> Thus the probability that the probability that the distance between the two points is at least &lt;math&gt;1/2&lt;/math&gt; is given by <br /> &lt;cmath&gt;<br /> \frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}.<br /> &lt;/cmath&gt;<br /> Therefore &lt;math&gt;a=26&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;, and &lt;math&gt;c=32&lt;/math&gt;. Thus, &lt;math&gt;a+b+c=59&lt;/math&gt; and the answer is &lt;math&gt;\textbf{(A).}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=88874 Nine point circle 2017-12-12T03:01:27Z <p>Cocohearts: /* Third Proof of Existence */</p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right| Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> <br /> It's also denoted Kimberling center &lt;math&gt;X_5&lt;/math&gt;.<br /> <br /> <br /> ==First Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, and &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> ==Second Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;{1}/{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle, and the vertices of the triangle to its [[Euler point]]s.<br /> Hence proved.<br /> ==See also==<br /> *[[Kimberling center]]<br /> *[[Center line]]<br /> *[[Evans point]]<br /> *[[Euler line]]<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=88873 Nine point circle 2017-12-12T03:00:35Z <p>Cocohearts: /* Second Proof of Existence */</p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right| Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> <br /> It's also denoted Kimberling center &lt;math&gt;X_5&lt;/math&gt;.<br /> <br /> <br /> ==First Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, and &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> ==Second Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;{1}/{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle, and the vertices of the triangle to its [[Euler point]]s.<br /> Hence proved.<br /> ==Third Proof of Existence==<br /> [asy]<br /> draw((0,0)--(2,4)--(5,0)--(0,0));<br /> [/asy]<br /> <br /> ==See also==<br /> *[[Kimberling center]]<br /> *[[Center line]]<br /> *[[Evans point]]<br /> *[[Euler line]]<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_14&diff=87546 2006 AMC 12A Problems/Problem 14 2017-09-17T19:36:08Z <p>Cocohearts: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}}<br /> <br /> == Problem ==<br /> <br /> Two farmers agree that pigs are worth &lt;math&gt;300&lt;/math&gt; dollars and that goats are worth &lt;math&gt;210&lt;/math&gt; dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with &quot;change&quot; received in the form of goats or pigs as necessary. (For example, a &lt;math&gt;390&lt;/math&gt; dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ } 210&lt;/math&gt;<br /> <br /> == Solutions ==<br /> <br /> ===Solution 1===<br /> <br /> The problem can be restated as an equation of the form &lt;math&gt;300p + 210g = x&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; is the number of pigs, &lt;math&gt;g&lt;/math&gt; is the number of goats, and &lt;math&gt;x&lt;/math&gt; is the positive debt. The problem asks us to find the lowest ''x'' possible. ''p'' and ''g'' must be [[integer]]s, which makes the equation a [[Diophantine equation]]. The [[Euclidean algorithm]] tells us that there are integer solutions to the Diophantine equation &lt;math&gt;am + bn = c&lt;/math&gt;, where &lt;math&gt;c&lt;/math&gt; is the [[greatest common divisor]] of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and no solutions for any smaller &lt;math&gt;c&lt;/math&gt;. Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, &lt;math&gt;\mathrm{(C) \ }&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Alternatively, note that &lt;math&gt;300p + 210g = 30(10p + 7g)&lt;/math&gt; is divisible by 30 no matter what &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us &lt;math&gt;630 - 600 = 30&lt;/math&gt; exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is &lt;math&gt;\mathrm{(C) \ }&lt;/math&gt;.<br /> debt that can be resolved.<br /> <br /> ===Solution 3===<br /> <br /> Let us simplify this problem. Dividing by &lt;math&gt;30&lt;/math&gt;, we get a pig to be: &lt;math&gt;\frac{300}{30} = 10&lt;/math&gt;, and a goat to be &lt;math&gt;\frac{210}{30}= 7&lt;/math&gt;. <br /> It becomes evident that if you exchange &lt;math&gt;5&lt;/math&gt; pigs for &lt;math&gt;7&lt;/math&gt; goats, we get the smallest positive difference - &lt;math&gt;5\cdot 10 - 7\cdot 7 = 50-49 = 1&lt;/math&gt;. <br /> Since we originally divided by &lt;math&gt;30&lt;/math&gt;, we need to multiply again, thus getting the answer: &lt;math&gt;1\cdot 30 = \mathrm{(C) 30}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}<br /> {{AMC10 box|year=2006|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=87364 2008 AMC 10A Problems/Problem 20 2017-09-03T23:38:06Z <p>Cocohearts: /* Solution */</p> <hr /> <div>==Problem==<br /> [[Trapezoid]] &lt;math&gt;ABCD&lt;/math&gt; has bases &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; and diagonals intersecting at &lt;math&gt;K&lt;/math&gt;. Suppose that &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;DC = 12&lt;/math&gt;, and the area of &lt;math&gt;\triangle AKD&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt;. What is the area of trapezoid &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;center&gt;&lt;asy&gt;<br /> pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br /> pen sm = fontsize(10); /* small font pen */<br /> pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br /> pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br /> D(MP(&quot;A&quot;,A,N)--MP(&quot;B&quot;,B,N)--MP(&quot;C&quot;,C)--MP(&quot;D&quot;,D)--A--C);D(B--D);D(A--MP(&quot;K&quot;,K)--D--cycle,linewidth(0.7));<br /> MP(&quot;9&quot;,(A+B)/2,N,sm);MP(&quot;12&quot;,(C+D)/2,sm);MP(&quot;24&quot;,(A+D)/2+(1,0),E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Since &lt;math&gt;\overline{AB} \parallel \overline{DC}&lt;/math&gt; it follows that &lt;math&gt;\triangle ABK \sim \triangle CDK&lt;/math&gt;. Thus &lt;math&gt;\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}&lt;/math&gt;. <br /> <br /> We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since &lt;math&gt;\triangle AKB, \triangle AKD&lt;/math&gt; share a common [[altitude]] to &lt;math&gt;\overline{BD}&lt;/math&gt;, it follows that (we let &lt;math&gt;[\triangle \ldots]&lt;/math&gt; denote the area of the triangle) &lt;math&gt;\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}&lt;/math&gt;, so &lt;math&gt;[\triangle AKB] = \frac{3}{4}(24) = 18&lt;/math&gt;. Similarly, we find &lt;math&gt;[\triangle DKC] = \frac{4}{3}(24) = 32&lt;/math&gt; and &lt;math&gt;[\triangle BKC] = 24&lt;/math&gt;.<br /> <br /> Therefore, the area of &lt;math&gt;ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_20&diff=87363 2008 AMC 10A Problems/Problem 20 2017-09-03T23:37:40Z <p>Cocohearts: /* Solution */</p> <hr /> <div>==Problem==<br /> [[Trapezoid]] &lt;math&gt;ABCD&lt;/math&gt; has bases &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; and diagonals intersecting at &lt;math&gt;K&lt;/math&gt;. Suppose that &lt;math&gt;AB = 9&lt;/math&gt;, &lt;math&gt;DC = 12&lt;/math&gt;, and the area of &lt;math&gt;\triangle AKD&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt;. What is the area of trapezoid &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize=5cm;<br /> pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */<br /> pen sm = fontsize(10); /* small font pen */<br /> pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */<br /> pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;<br /> D(MP(&quot;A&quot;,A,N)--MP(&quot;B&quot;,B,N)--MP(&quot;C&quot;,C)--MP(&quot;D&quot;,D)--A--C);D(B--D);D(A--MP(&quot;K&quot;,K)--D--cycle,linewidth(0.7));<br /> MP(&quot;9&quot;,(A+B)/2,N,sm);MP(&quot;12&quot;,(C+D)/2,sm);MP(&quot;24&quot;,(A+D)/2+(1,0),E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Since &lt;math&gt;\overline{AB} \parallel \overline{DC}&lt;/math&gt; it follows that &lt;math&gt;\triangle ABK \sim \triangle CDK&lt;/math&gt;. Thus &lt;math&gt;\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}&lt;/math&gt;. <br /> <br /> We now introduce the concept of [[area ratios]]: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since &lt;math&gt;\triangle AKB, \triangle AKD&lt;/math&gt; share a common [[altitude]] to &lt;math&gt;\overline{BD}&lt;/math&gt;, it follows that (we let &lt;math&gt;[\triangle \ldots]&lt;/math&gt; denote the area of the triangle) &lt;math&gt;\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}&lt;/math&gt;, so &lt;math&gt;[\triangle AKB] = \frac{3}{4}(24) = 18&lt;/math&gt;. Similarly, we find &lt;math&gt;[\triangle DKC] = \frac{4}{3}(24) = 32&lt;/math&gt; and &lt;math&gt;[\triangle BKC] = 24&lt;/math&gt;.<br /> <br /> Therefore, the area of &lt;math&gt;ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_11&diff=86744 1986 AHSME Problems/Problem 11 2017-08-01T23:58:33Z <p>Cocohearts: /* Problem */</p> <hr /> <div>==FATSO==<br /> <br /> In &lt;math&gt;\triangle ABC, AB = 13, BC = 14&lt;/math&gt; and &lt;math&gt;CA = 15&lt;/math&gt;. Also, &lt;math&gt;M&lt;/math&gt; is the midpoint of side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; is the foot of the altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;. <br /> The length of &lt;math&gt;HM&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A);<br /> draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C));<br /> label(&quot;A&quot;, A, NE);<br /> label(&quot;B&quot;, B, W);<br /> label(&quot;C&quot;, C, E);<br /> label(&quot;H&quot;, H, S);<br /> label(&quot;M&quot;, M, dir(M));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad<br /> \textbf{(B)}\ 6.5\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 7.5\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is &lt;math&gt;13&lt;/math&gt;, the median must be &lt;math&gt;6.5&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1986|num-b=10|num-a=12}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_11&diff=86743 1986 AHSME Problems/Problem 11 2017-08-01T23:57:59Z <p>Cocohearts: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC, AB = 13, BC = 14&lt;/math&gt; and &lt;math&gt;CA = 15&lt;/math&gt;. Also, &lt;math&gt;M&lt;/math&gt; is the midpoint of side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; is the foot of the altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;. <br /> The length of &lt;math&gt;HM&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A);<br /> draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C));<br /> label(&quot;A&quot;, A, NE);<br /> label(&quot;B&quot;, B, W);<br /> label(&quot;C&quot;, C, E);<br /> label(&quot;H&quot;, H, S);<br /> label(&quot;M&quot;, M, dir(M));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad<br /> \textbf{(B)}\ 6.5\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 7.5\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is &lt;math&gt;13&lt;/math&gt;, the median must be &lt;math&gt;6.5&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1986|num-b=10|num-a=12}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85929 2002 AMC 12A Problems/Problem 19 2017-06-02T01:10:13Z <p>Cocohearts: /* cocohearts is awesome */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85928 2002 AMC 12A Problems/Problem 19 2017-06-02T01:10:06Z <p>Cocohearts: /* cocohearts is awesome */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85927 2002 AMC 12A Problems/Problem 19 2017-06-02T01:09:57Z <p>Cocohearts: /* cocohearts is awesome */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85926 2002 AMC 12A Problems/Problem 19 2017-06-02T01:09:36Z <p>Cocohearts: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == cocohearts is awesome ==<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85925 2002 AMC 12A Problems/Problem 19 2017-06-02T01:09:26Z <p>Cocohearts: /* See Also */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85924 2002 AMC 12A Problems/Problem 19 2017-06-02T01:09:11Z <p>Cocohearts: /* See Also */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85923 2002 AMC 12A Problems/Problem 19 2017-06-02T01:08:48Z <p>Cocohearts: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85922 2002 AMC 12A Problems/Problem 19 2017-06-02T01:08:16Z <p>Cocohearts: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85921 2002 AMC 12A Problems/Problem 19 2017-06-02T01:06:41Z <p>Cocohearts: /* cocohearts is awesome */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85920 2002 AMC 12A Problems/Problem 19 2017-06-02T01:06:21Z <p>Cocohearts: /* See Also */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==</div> Cocohearts https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_19&diff=85919 2002 AMC 12A Problems/Problem 19 2017-06-02T01:05:40Z <p>Cocohearts: /* See Also */</p> <hr /> <div>== Problem ==<br /> <br /> The graph of the function &lt;math&gt;f&lt;/math&gt; is shown below. How many solutions does the equation &lt;math&gt;f(f(x))=6&lt;/math&gt; have? <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> draw(P1--P2--P3--P4--P5);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }2<br /> \qquad<br /> \text{(B) }4<br /> \qquad<br /> \text{(C) }5<br /> \qquad<br /> \text{(D) }6<br /> \qquad<br /> \text{(E) }7<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First of all, note that the equation &lt;math&gt;f(t)=6&lt;/math&gt; has two solutions: &lt;math&gt;t=-2&lt;/math&gt; and &lt;math&gt;t=1&lt;/math&gt;. <br /> <br /> Given an &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;f(x)=t&lt;/math&gt;. Obviously, to have &lt;math&gt;f(f(x))=6&lt;/math&gt;, we need to have &lt;math&gt;f(t)=6&lt;/math&gt;, and we already know when that happens. In other words, the solutions to &lt;math&gt;f(f(x))=6&lt;/math&gt; are precisely the solutions to (&lt;math&gt;f(x)=-2&lt;/math&gt; or &lt;math&gt;f(x)=1&lt;/math&gt;).<br /> <br /> Without actually computing the exact values, it is obvious from the graph that the equation &lt;math&gt;f(x)=-2&lt;/math&gt; has two and &lt;math&gt;f(x)=1&lt;/math&gt; has four different solutions, giving us a total of &lt;math&gt;2+4=\boxed{(D)6}&lt;/math&gt; solutions.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> <br /> pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);<br /> real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};<br /> <br /> path graph = P1--P2--P3--P4--P5;<br /> path line1 = (-7,1)--(6,1);<br /> path line2 = (-7,-2)--(6,-2);<br /> <br /> draw(graph);<br /> draw(line1, red);<br /> draw(line2, red);<br /> <br /> dot(&quot;(-7, -4)&quot;,P1);<br /> dot(&quot;(-2, 6)&quot;,P2,LeftSide);<br /> dot(&quot;(1, 6)&quot;,P4);<br /> dot(&quot;(5, -6)&quot;,P5);<br /> dot(intersectionpoints(graph,line1),red);<br /> dot(intersectionpoints(graph,line2),red);<br /> <br /> xaxis(&quot;$x$&quot;,-7.5,7,Ticks(xticks),EndArrow(6));<br /> yaxis(&quot;$y$&quot;,-6.5,7,Ticks(yticks),EndArrow(6));<br /> &lt;/asy&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> <br /> {{MAA Notice}}<br /> {{MAA Notice}}<br /> <br /> == cocohearts is awesome ==</div> Cocohearts