https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Dubna+ismycity&feedformat=atom AoPS Wiki - User contributions [en] 2021-11-27T17:58:35Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_8&diff=99190 1994 AIME Problems/Problem 8 2018-11-29T21:18:46Z <p>Dubna ismycity: </p> <hr /> <div>== Problem ==<br /> The points &lt;math&gt;(0,0)\,&lt;/math&gt;, &lt;math&gt;(a,11)\,&lt;/math&gt;, and &lt;math&gt;(b,37)\,&lt;/math&gt; are the vertices of an equilateral triangle. Find the value of &lt;math&gt;ab\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Consider the points on the [[complex plane]]. The point &lt;math&gt;b+37i&lt;/math&gt; is then a rotation of &lt;math&gt;60&lt;/math&gt; degrees of &lt;math&gt;a+11i&lt;/math&gt; about the origin, so:<br /> <br /> &lt;cmath&gt;(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.&lt;/cmath&gt;<br /> <br /> Equating the real and imaginary parts, we have:<br /> <br /> &lt;cmath&gt;\begin{align*}b&amp;=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&amp;=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}&lt;/cmath&gt;<br /> <br /> Solving this system, we find that &lt;math&gt;a=21\sqrt{3}, b=5\sqrt{3}&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> '''Note''': There is another solution where the point &lt;math&gt;b+37i&lt;/math&gt; is a rotation of &lt;math&gt;-60&lt;/math&gt; degrees of &lt;math&gt;a+11i&lt;/math&gt;; however, this triangle is just a reflection of the first triangle by the &lt;math&gt;y&lt;/math&gt;-axis, and the signs of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are flipped. However, the product &lt;math&gt;ab&lt;/math&gt; is unchanged.<br /> <br /> == Solution Two ==<br /> Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? &lt;math&gt;\sqrt{3}&lt;/math&gt; and perpendiculars inspires this solution:<br /> <br /> First, drop a perpendicular from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. Call this midpoint of &lt;math&gt;AB M&lt;/math&gt;. Thus, &lt;math&gt;M=(\frac{a+b}{2}, 24)&lt;/math&gt;. The vector from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;[\frac{a+b}{2}, 24]&lt;/math&gt;. Meanwhile from point &lt;math&gt;M&lt;/math&gt; we can use a vector with &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; the distance; we have to switch the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; and our displacement is &lt;math&gt;[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]&lt;/math&gt;. (Do you see why we switched &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; due to the rotation of 90 degrees?)<br /> <br /> <br /> We see this displacement from &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;[\frac{a-b}{2}, 13]&lt;/math&gt; as well. Equating the two vectors, we get &lt;math&gt;a+b=26\sqrt{3}&lt;/math&gt; and &lt;math&gt;a-b=16\sqrt{3}&lt;/math&gt;. Therefore, &lt;math&gt;a=21\sqrt{3}&lt;/math&gt; and &lt;math&gt;b=5\sqrt{3}&lt;/math&gt;. And the answer is &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> '''Note''': This solution was also present in Titu Andreescu and Zuming Feng's &quot;103 Trigonometry Problems&quot;.<br /> <br /> == Solution Three ==<br /> Plot this equilateral triangle on the complex plane.<br /> Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives &lt;math&gt;(\frac{a+b}{3}, 16i)&lt;/math&gt;. The new coordinates of the equilateral triangle are &lt;math&gt;(-\frac{a+b}{3}-16i), (a-\frac{a+b}{3}-5i), (b-\frac{a+b}{3}+21i)&lt;/math&gt;. These three vertices are solutions of a cubic polynomial of form &lt;math&gt;x^3 + C&lt;/math&gt;. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots r1, r2, and r3, r1r2 + r2r3 + r3r1 = 0.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation &lt;math&gt;5a = 21b.&lt;/math&gt;<br /> Now use the equation with only real parts. This should give you a quadratic &lt;math&gt;a^2 - ab + b^2 = 1083&lt;/math&gt;. Use your previously obtained equation to plug in for a and solve for b, which should yield &lt;math&gt;5\sqrt{3}&lt;/math&gt;. a is then &lt;math&gt;21/5\sqrt{3}&lt;/math&gt;. Multiplying a and b yields &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> {{AIME box|year=1994|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=97125 2011 AIME I Problems/Problem 4 2018-08-10T02:48:03Z <p>Dubna ismycity: /* Solution 1 */</p> <hr /> <div>== Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=125&lt;/math&gt;, &lt;math&gt;AC=117&lt;/math&gt; and &lt;math&gt;BC=120&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at point &lt;math&gt;L&lt;/math&gt;, and the angle bisector of angle &lt;math&gt;B&lt;/math&gt; intersects &lt;math&gt; \overline{AC} &lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt; \overline{BK}&lt;/math&gt; and &lt;math&gt; \overline{AL}&lt;/math&gt;, respectively. Find &lt;math&gt;MN&lt;/math&gt;.<br /> <br /> <br /> == Solution 1 == <br /> Extend &lt;math&gt;{CM}&lt;/math&gt; and &lt;math&gt;{CN}&lt;/math&gt; such that they intersect line &lt;math&gt;{AB}&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. <br /> Since &lt;math&gt;{BM}&lt;/math&gt; is the angle bisector of angle B, and &lt;math&gt;{CM}&lt;/math&gt; is perpendicular to &lt;math&gt;{BM}&lt;/math&gt;, so &lt;math&gt;BP=BC=120&lt;/math&gt;, M is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;,N is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. But &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{56}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> [There seem to be some mislabeled points going on here but the idea is sound.]<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;ABC&lt;/math&gt;. Now, since &lt;math&gt;AM \perp LC&lt;/math&gt; and &lt;math&gt;AN \perp KB&lt;/math&gt;, we have &lt;math&gt;AMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = AI&lt;/math&gt;. Since &lt;math&gt;\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK&lt;/math&gt;, we have that &lt;math&gt;MN = AI \cdot \cos \angle IAK&lt;/math&gt;. Letting &lt;math&gt;X&lt;/math&gt; be the point of contact of the incircle of &lt;math&gt;ABC&lt;/math&gt; with side &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;AX=MN&lt;/math&gt; thus &lt;math&gt;MN=\frac{117+120-125}{2}=\boxed{56}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) == <br /> Project &lt;math&gt;I&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; as &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;ID&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; are both in-radii of &lt;math&gt;\triangle ABC&lt;/math&gt; so we get right triangles with legs &lt;math&gt;r&lt;/math&gt; (the in-radius length) and &lt;math&gt;s - c = 56&lt;/math&gt;. Since &lt;math&gt;IC&lt;/math&gt; is the hypotenuse for the 4 triangles (&lt;math&gt;\triangle INC, \triangle IMC, \triangle IDC,&lt;/math&gt; and &lt;math&gt;\triangle IEC&lt;/math&gt;), &lt;math&gt;C, D, M, I, N, E&lt;/math&gt; are con-cyclic on a circle we shall denote as &lt;math&gt;\omega&lt;/math&gt; which is also the circumcircle of &lt;math&gt;\triangle CMN&lt;/math&gt; and &lt;math&gt;\triangle CDE&lt;/math&gt;. To find &lt;math&gt;MN&lt;/math&gt;, we can use the Law of Cosines on &lt;math&gt;\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;. Now, the circumradius &lt;math&gt;R&lt;/math&gt; can be found with Pythagorean Theorem with &lt;math&gt;\triangle CDI&lt;/math&gt; or &lt;math&gt;\triangle CEI&lt;/math&gt;: &lt;math&gt;r^2 + 56^2 = (2R)^2&lt;/math&gt;. To find &lt;math&gt;r&lt;/math&gt;, we can use the formula &lt;math&gt;rs = [ABC]&lt;/math&gt; and by Heron's, &lt;math&gt;[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}&lt;/math&gt;. To find &lt;math&gt;\angle MCN&lt;/math&gt;, we can find &lt;math&gt;\angle MIN&lt;/math&gt; since &lt;math&gt;\angle MCN = 180 - \angle MIN&lt;/math&gt;. &lt;math&gt;\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}&lt;/math&gt;. Thus, &lt;math&gt;\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}&lt;/math&gt; and since &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt;, we have &lt;math&gt;\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}&lt;/math&gt;. Plugging this into our Law of Cosines formula gives &lt;math&gt;MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})&lt;/math&gt;. To find &lt;math&gt;\cos{\angle C}&lt;/math&gt;, we use LoC on &lt;math&gt;\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}&lt;/math&gt;. Our formula now becomes &lt;math&gt;MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}&lt;/math&gt;. After simplifying, we get &lt;math&gt;MN^2 = 3136 \implies MN = \boxed{056}&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> --lucasxia01<br /> <br /> == Solution 4==<br /> <br /> Because &lt;math&gt;\angle CMI = \angle CNI = 90&lt;/math&gt;, &lt;math&gt;CMIN&lt;/math&gt; is cyclic. <br /> <br /> Ptolemy on CMIN:<br /> <br /> &lt;math&gt;CN*MI+CM*IN=CI*MN&lt;/math&gt;<br /> <br /> &lt;math&gt;CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN&lt;/math&gt;<br /> <br /> &lt;math&gt;MN = CI \sin \angle MCN&lt;/math&gt; by angle addition formula.<br /> <br /> &lt;math&gt;\angle MCN = 180 - \angle MIN = 90 - \angle BCI&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be where the incircle touches &lt;math&gt;BC&lt;/math&gt;, then &lt;math&gt;CI \cos \angle BCI = CH = \frac{a+b-c}{2}&lt;/math&gt;.<br /> &lt;math&gt;a=120, b=117, c=125&lt;/math&gt;, for a final answer of &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=97124 2011 AIME I Problems/Problem 4 2018-08-10T02:47:18Z <p>Dubna ismycity: /* Solution 1 */</p> <hr /> <div>== Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=125&lt;/math&gt;, &lt;math&gt;AC=117&lt;/math&gt; and &lt;math&gt;BC=120&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at point &lt;math&gt;L&lt;/math&gt;, and the angle bisector of angle &lt;math&gt;B&lt;/math&gt; intersects &lt;math&gt; \overline{AC} &lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt; \overline{BK}&lt;/math&gt; and &lt;math&gt; \overline{AL}&lt;/math&gt;, respectively. Find &lt;math&gt;MN&lt;/math&gt;.<br /> <br /> <br /> == Solution 1 == <br /> Extend &lt;math&gt;{CM}&lt;/math&gt; and &lt;math&gt;{CN}&lt;/math&gt; such that they intersect lines &lt;math&gt;{AB}&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. <br /> Since &lt;math&gt;{BM}&lt;/math&gt; is the angle bisector of angle B,and &lt;math&gt;{CM}&lt;/math&gt; is perpendicular to &lt;math&gt;{BM}&lt;/math&gt; ,so , &lt;math&gt;BP=BC=120&lt;/math&gt;, M is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;,N is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. But &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;,so &lt;math&gt;MN=\boxed{56}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> [There seem to be some mislabeled points going on here but the idea is sound.]<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;ABC&lt;/math&gt;. Now, since &lt;math&gt;AM \perp LC&lt;/math&gt; and &lt;math&gt;AN \perp KB&lt;/math&gt;, we have &lt;math&gt;AMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = AI&lt;/math&gt;. Since &lt;math&gt;\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK&lt;/math&gt;, we have that &lt;math&gt;MN = AI \cdot \cos \angle IAK&lt;/math&gt;. Letting &lt;math&gt;X&lt;/math&gt; be the point of contact of the incircle of &lt;math&gt;ABC&lt;/math&gt; with side &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;AX=MN&lt;/math&gt; thus &lt;math&gt;MN=\frac{117+120-125}{2}=\boxed{56}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) == <br /> Project &lt;math&gt;I&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; as &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;ID&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; are both in-radii of &lt;math&gt;\triangle ABC&lt;/math&gt; so we get right triangles with legs &lt;math&gt;r&lt;/math&gt; (the in-radius length) and &lt;math&gt;s - c = 56&lt;/math&gt;. Since &lt;math&gt;IC&lt;/math&gt; is the hypotenuse for the 4 triangles (&lt;math&gt;\triangle INC, \triangle IMC, \triangle IDC,&lt;/math&gt; and &lt;math&gt;\triangle IEC&lt;/math&gt;), &lt;math&gt;C, D, M, I, N, E&lt;/math&gt; are con-cyclic on a circle we shall denote as &lt;math&gt;\omega&lt;/math&gt; which is also the circumcircle of &lt;math&gt;\triangle CMN&lt;/math&gt; and &lt;math&gt;\triangle CDE&lt;/math&gt;. To find &lt;math&gt;MN&lt;/math&gt;, we can use the Law of Cosines on &lt;math&gt;\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;. Now, the circumradius &lt;math&gt;R&lt;/math&gt; can be found with Pythagorean Theorem with &lt;math&gt;\triangle CDI&lt;/math&gt; or &lt;math&gt;\triangle CEI&lt;/math&gt;: &lt;math&gt;r^2 + 56^2 = (2R)^2&lt;/math&gt;. To find &lt;math&gt;r&lt;/math&gt;, we can use the formula &lt;math&gt;rs = [ABC]&lt;/math&gt; and by Heron's, &lt;math&gt;[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}&lt;/math&gt;. To find &lt;math&gt;\angle MCN&lt;/math&gt;, we can find &lt;math&gt;\angle MIN&lt;/math&gt; since &lt;math&gt;\angle MCN = 180 - \angle MIN&lt;/math&gt;. &lt;math&gt;\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}&lt;/math&gt;. Thus, &lt;math&gt;\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}&lt;/math&gt; and since &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt;, we have &lt;math&gt;\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}&lt;/math&gt;. Plugging this into our Law of Cosines formula gives &lt;math&gt;MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})&lt;/math&gt;. To find &lt;math&gt;\cos{\angle C}&lt;/math&gt;, we use LoC on &lt;math&gt;\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}&lt;/math&gt;. Our formula now becomes &lt;math&gt;MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}&lt;/math&gt;. After simplifying, we get &lt;math&gt;MN^2 = 3136 \implies MN = \boxed{056}&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> --lucasxia01<br /> <br /> == Solution 4==<br /> <br /> Because &lt;math&gt;\angle CMI = \angle CNI = 90&lt;/math&gt;, &lt;math&gt;CMIN&lt;/math&gt; is cyclic. <br /> <br /> Ptolemy on CMIN:<br /> <br /> &lt;math&gt;CN*MI+CM*IN=CI*MN&lt;/math&gt;<br /> <br /> &lt;math&gt;CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN&lt;/math&gt;<br /> <br /> &lt;math&gt;MN = CI \sin \angle MCN&lt;/math&gt; by angle addition formula.<br /> <br /> &lt;math&gt;\angle MCN = 180 - \angle MIN = 90 - \angle BCI&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be where the incircle touches &lt;math&gt;BC&lt;/math&gt;, then &lt;math&gt;CI \cos \angle BCI = CH = \frac{a+b-c}{2}&lt;/math&gt;.<br /> &lt;math&gt;a=120, b=117, c=125&lt;/math&gt;, for a final answer of &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&diff=96847 2005 AIME II Problems/Problem 5 2018-08-08T00:05:05Z <p>Dubna ismycity: /* Solution I */</p> <hr /> <div>== Problem ==<br /> Determine the number of [[ordered pair]]s &lt;math&gt; (a,b) &lt;/math&gt; of [[integer]]s such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> == Solution I ==<br /> The equation can be rewritten as &lt;math&gt; \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 &lt;/math&gt; Multiplying through by &lt;math&gt;\log a \log b &lt;/math&gt; and factoring yields &lt;math&gt;(\log b - 3\log a)(\log b - 2\log a)=0 &lt;/math&gt;. Therefore, &lt;math&gt;\log b=3\log a &lt;/math&gt; or &lt;math&gt;\log b=2\log a &lt;/math&gt;, so either &lt;math&gt; b=a^3 &lt;/math&gt; or &lt;math&gt; b=a^2 &lt;/math&gt;. <br /> *For the case &lt;math&gt; b=a^2 &lt;/math&gt;, note that &lt;math&gt; 44^2=1936 &lt;/math&gt; and &lt;math&gt; 45^2=2025 &lt;/math&gt;. Thus, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; will work. <br /> *For the case &lt;math&gt; b=a^3 &lt;/math&gt;, note that &lt;math&gt; 12^3=1728 &lt;/math&gt; while &lt;math&gt; 13^3=2197 &lt;/math&gt;. Therefore, for this case, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt; work. <br /> There are &lt;math&gt; 44-2+1=43 &lt;/math&gt; possibilities for the square case and &lt;math&gt; 12-2+1=11 &lt;/math&gt; possibilities for the cube case. Thus, the answer is &lt;math&gt; 43+11= \boxed{054}&lt;/math&gt;.<br /> <br /> Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs &lt;math&gt;(a,b)&lt;/math&gt;, and not for the number of possible values of &lt;math&gt;b&lt;/math&gt;. Were the problem to ask for the number of possible values of &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;b^6&lt;/math&gt; under &lt;math&gt;2005&lt;/math&gt; would have to be subtracted, which would just be &lt;math&gt;2&lt;/math&gt; values: &lt;math&gt;2^6&lt;/math&gt; and &lt;math&gt;3^6&lt;/math&gt;. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)<br /> <br /> ==Solution II ==<br /> Let &lt;math&gt;k=\log_a b&lt;/math&gt;. Then our equation becomes &lt;math&gt;k+\frac{6}{k}=5&lt;/math&gt;. Multiplying through by &lt;math&gt;k&lt;/math&gt; and solving the quadratic gives us &lt;math&gt;k=2&lt;/math&gt; or &lt;math&gt;k=3&lt;/math&gt;. Hence &lt;math&gt;a^2=b&lt;/math&gt; or &lt;math&gt;a^3=b&lt;/math&gt;. <br /> <br /> For the first case &lt;math&gt;a^2=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 44, a total of 43 values.<br /> For the second case &lt;math&gt;a^3=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 12, a total of 11 values.<br /> <br /> <br /> Thus the total number of possible values is &lt;math&gt;43+11=\boxed{54}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_2&diff=96842 2005 AIME II Problems/Problem 2 2018-08-07T22:57:08Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the [[probability]] each guest got one roll of each type is &lt;math&gt; \frac mn, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are [[relatively prime]] [[integer]]s, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Use [[construction]]. We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined. <br /> <br /> *Person 1: &lt;math&gt;\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}&lt;/math&gt;<br /> *Person 2: &lt;math&gt;\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25&lt;/math&gt;<br /> *Person 3: One roll of each type is left, so the probability here is &lt;math&gt;1&lt;/math&gt;. <br /> <br /> Our answer is thus &lt;math&gt;\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}&lt;/math&gt;, and &lt;math&gt;m + n = \boxed{79}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Call the three different type of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, etc. This can occur in &lt;math&gt;\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216&lt;/math&gt; different manners. The total number of possible strings is &lt;math&gt;\frac{9!}{3!3!3!} = 1680&lt;/math&gt;. The solution is therefore &lt;math&gt;\frac{216}{1680} = \frac{9}{70}&lt;/math&gt;, and &lt;math&gt;m + n = \boxed{79}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to &lt;math&gt;{9 \choose 3}{6 \choose3}&lt;/math&gt; as the amount of ways to select three rolls out of 9 to give to the first person is &lt;math&gt;{9 \choose 3}&lt;/math&gt;, and three rolls out of 6 is &lt;math&gt;{6 \choose3}&lt;/math&gt;. After that, the three remaining rolls have no more configurations. <br /> <br /> The numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored. <br /> <br /> xxx, yyy, zzz<br /> <br /> So you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving &lt;math&gt;3^3&lt;/math&gt; combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently &lt;math&gt;2^2&lt;/math&gt; since there are two of each letter left. <br /> <br /> &lt;math&gt;(27*8)/{9 \choose 3}{6 \choose3}&lt;/math&gt; simplifies down to our fraction m/n, which is &lt;math&gt;9/70&lt;/math&gt;. Adding them up gives &lt;math&gt;9 + 70 = \boxed{79}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_5&diff=96804 2011 AIME II Problems/Problem 5 2018-08-06T03:57:35Z <p>Dubna ismycity: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of the first 2011 terms of a [[geometric sequence]] is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.<br /> <br /> ==Solution==<br /> Since the sum of the first &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;200&lt;/math&gt;, and the sum of the fist &lt;math&gt;4022&lt;/math&gt; terms is &lt;math&gt;380&lt;/math&gt;, the sum of the second &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;180&lt;/math&gt;.<br /> This is decreasing from the first 2011, so the common ratio is less than one.<br /> <br /> Because it is a geometric sequence and the sum of the first 2011 terms is &lt;math&gt;200&lt;/math&gt;, second &lt;math&gt;2011&lt;/math&gt; is &lt;math&gt;180&lt;/math&gt;, the ratio of the second &lt;math&gt;2011&lt;/math&gt; terms to the first &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;\frac{9}{10}&lt;/math&gt;. Following the same pattern, the sum of the third &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;\frac{9}{10}*180 = 162&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;200+180+162=542&lt;/math&gt;, so the sum of the first &lt;math&gt;6033&lt;/math&gt; terms is &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Solution by e_power_pi_times_i<br /> <br /> The sum of the first &lt;math&gt;2011&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{2011})}{1-k}&lt;/math&gt;, and the first &lt;math&gt;4022&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{4022})}{1-k}&lt;/math&gt;. Dividing these equations, we get &lt;math&gt;\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}&lt;/math&gt;. Noticing that &lt;math&gt;k^{4022}&lt;/math&gt; is just the square of &lt;math&gt;k^{2011}&lt;/math&gt;, we substitute &lt;math&gt;x = k^{2011}&lt;/math&gt;, so &lt;math&gt;\dfrac{1}{x+1} = \dfrac{10}{19}&lt;/math&gt;. That means that &lt;math&gt;k^{2011} = \dfrac{9}{10}&lt;/math&gt;. Since the sum of the first &lt;math&gt;6033&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{6033})}{1-k}&lt;/math&gt;, dividing gives &lt;math&gt;\dfrac{1-k^{2011}}{1-k^{6033}}&lt;/math&gt;. Since &lt;math&gt;k^{6033} = \dfrac{729}{1000}&lt;/math&gt;, plugging all the values in gives &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> The sum of the first 2011 terms of the sequence is expressible as &lt;math&gt;a_1 + a_1r + a_1r^2 + a_1r^3&lt;/math&gt; .... until &lt;math&gt;a_1r^{2010}&lt;/math&gt;. The sum of the 2011 terms following the first 2011 is expressible as &lt;math&gt;a_1r^{2011} + a_1r^{2012} + a_1r^{2013}&lt;/math&gt; .... until &lt;math&gt;a_1r^{4021}&lt;/math&gt;. Notice that the latter sum of terms can be expressed as &lt;math&gt;(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})&lt;/math&gt;. We also know that the latter sum of terms can be obtained by subtracting 200 from 180, which then means that &lt;math&gt;r^{2011} = 9/10&lt;/math&gt;. The terms from 4023 to 6033 can be expressed as &lt;math&gt;(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})&lt;/math&gt;, which is equivalent to &lt;math&gt;((9/10)^2)(200) = 162&lt;/math&gt;. Adding 380 and 162 gives the answer of &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_2&diff=96621 2008 AIME I Problems/Problem 2 2018-07-31T03:34:33Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> Square &lt;math&gt;AIME&lt;/math&gt; has sides of length &lt;math&gt;10&lt;/math&gt; units. Isosceles triangle &lt;math&gt;GEM&lt;/math&gt; has base &lt;math&gt;EM&lt;/math&gt;, and the area common to triangle &lt;math&gt;GEM&lt;/math&gt; and square &lt;math&gt;AIME&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt; square units. Find the length of the altitude to &lt;math&gt;EM&lt;/math&gt; in &lt;math&gt;\triangle GEM&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> Note that if the altitude of the triangle is at most &lt;math&gt;10&lt;/math&gt;, then the maximum area of the intersection of the triangle and the square is &lt;math&gt;5\cdot10=50&lt;/math&gt;.<br /> This implies that vertex G must be located outside of square &lt;math&gt;AIME&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> pair E=(0,0), M=(10,0), I=(10,10), A=(0,10);<br /> draw(A--I--M--E--cycle);<br /> pair G=(5,25);<br /> draw(G--E--M--cycle);<br /> label(&quot;$$G$$&quot;,G,N);<br /> label(&quot;$$A$$&quot;,A,NW);<br /> label(&quot;$$I$$&quot;,I,NE);<br /> label(&quot;$$M$$&quot;,M,NE);<br /> label(&quot;$$E$$&quot;,E,NW);<br /> label(&quot;$$10$$&quot;,(M+E)/2,S);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Let &lt;math&gt;GE&lt;/math&gt; meet &lt;math&gt;AI&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and let &lt;math&gt;GM&lt;/math&gt; meet &lt;math&gt;AI&lt;/math&gt; at &lt;math&gt;Y&lt;/math&gt;. Clearly, &lt;math&gt;XY=6&lt;/math&gt; since the area of [[trapezoid]] &lt;math&gt;XYME&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt;. Also, &lt;math&gt;\triangle GXY \sim \triangle GEM&lt;/math&gt;.<br /> <br /> Let the height of &lt;math&gt;GXY&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By the similarity, &lt;math&gt;\dfrac{h}{6} = \dfrac{h + 10}{10}&lt;/math&gt;, we get &lt;math&gt;h = 15&lt;/math&gt;. Thus, the height of &lt;math&gt;GEM&lt;/math&gt; is &lt;math&gt;h + 10 = \boxed{025}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_4&diff=96599 2000 AIME I Problems/Problem 4 2018-07-30T05:16:05Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> The diagram shows a [[rectangle]] that has been dissected into nine non-overlapping [[square]]s. Given that the width and the height of the rectangle are relatively prime positive integers, find the [[perimeter]] of the rectangle.<br /> <br /> &lt;center&gt;&lt;asy&gt;draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));<br /> draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61));<br /> draw((34,36)--(34,45)--(25,45));<br /> draw((36,36)--(36,38)--(34,38));<br /> draw((36,38)--(41,38));<br /> draw((34,45)--(41,45));&lt;/asy&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Call the squares' side lengths from smallest to largest &lt;math&gt;a_1,\ldots,a_9&lt;/math&gt;, and let &lt;math&gt;l,w&lt;/math&gt; represent the dimensions of the rectangle.<br /> <br /> The picture shows that<br /> &lt;cmath&gt;\begin{align*}<br /> a_1+a_2 &amp;= a_3\\<br /> a_1 + a_3 &amp;= a_4\\<br /> a_3 + a_4 &amp;= a_5\\<br /> a_4 + a_5 &amp;= a_6\\<br /> a_2 + a_3 + a_5 &amp;= a_7\\<br /> a_2 + a_7 &amp;= a_8\\<br /> a_1 + a_4 + a_6 &amp;= a_9\\<br /> a_6 + a_9 &amp;= a_7 + a_8.\end{align*}&lt;/cmath&gt;<br /> <br /> Notice that &lt;math&gt;a_7 + a_9 = a_6 + a_8&lt;/math&gt;. Expressing all terms 3 to 9 in terms of &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_2&lt;/math&gt; and substituting their expanded forms into the previous equation will give the expression &lt;math&gt;5a_1 = 2a_2&lt;/math&gt;.<br /> <br /> We can guess that &lt;math&gt;a_1 = 2&lt;/math&gt;. (If we started with &lt;math&gt;a_1&lt;/math&gt; odd, the resulting sides would not be integers and we would need to scale up by a factor of &lt;math&gt;2&lt;/math&gt; to make them integers; if we started with &lt;math&gt;a_1 &gt; 2&lt;/math&gt; even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives &lt;math&gt;a_9 = 36&lt;/math&gt;, &lt;math&gt;a_6=25&lt;/math&gt;, &lt;math&gt;a_8 = 33&lt;/math&gt;, which gives us &lt;math&gt;l=61,w=69&lt;/math&gt;. These numbers are relatively prime, as desired. The perimeter is &lt;math&gt;2(61)+2(69)=\boxed{260}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96594 1986 AIME Problems/Problem 6 2018-07-30T03:55:55Z <p>Dubna ismycity: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> =Solution =<br /> <br /> ==Solution 1==<br /> <br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96593 1986 AIME Problems/Problem 6 2018-07-30T03:55:46Z <p>Dubna ismycity: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> =Solution =<br /> <br /> ==Solution 1==<br /> <br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> =Solution 2=<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96592 1986 AIME Problems/Problem 6 2018-07-30T03:55:37Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> =Solution =<br /> <br /> =Solution 1=<br /> <br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> =Solution 2=<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96591 1986 AIME Problems/Problem 6 2018-07-30T03:55:22Z <p>Dubna ismycity: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> == Solution ==<br /> <br /> =Solution 1=<br /> <br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> =Solution 2=<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96590 1986 AIME Problems/Problem 6 2018-07-30T03:54:55Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> == Solution ==<br /> <br /> =Solution 1=<br /> <br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> == Alternate Solution ==<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=96589 1986 AIME Problems/Problem 6 2018-07-30T03:54:38Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> == Solution ==<br /> <br /> Solution 1<br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> == Alternate Solution ==<br /> <br /> Use the same method as above where you represent the sum of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number &lt;math&gt;k&lt;/math&gt;. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Solving the two equations gives values that are respectively around &lt;math&gt;62.5&lt;/math&gt; and &lt;math&gt;61.5&lt;/math&gt; with the quadratic formula, and the only integer between the two is &lt;math&gt;62&lt;/math&gt;. <br /> <br /> This implies that we can plug in &lt;math&gt;62&lt;/math&gt; and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_6&diff=92602 1986 AIME Problems/Problem 6 2018-03-02T05:44:50Z <p>Dubna ismycity: /* Solution */</p> <hr /> <div>== Problem ==<br /> The pages of a book are numbered &lt;math&gt;1_{}^{}&lt;/math&gt; through &lt;math&gt;n_{}^{}&lt;/math&gt;. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of &lt;math&gt;1986_{}^{}&lt;/math&gt;. What was the number of the page that was added twice? <br /> <br /> == Solution ==<br /> Denote the page number as &lt;math&gt;x&lt;/math&gt;, with &lt;math&gt;x &lt; n&lt;/math&gt;. The sum formula shows that &lt;math&gt;\frac{n(n + 1)}{2} + x = 1986&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; cannot be very large, disregard it for now and solve &lt;math&gt;\frac{n(n+1)}{2} = 1986&lt;/math&gt;. The positive root for &lt;math&gt;n \approx \sqrt{3972} \approx 63&lt;/math&gt;. Quickly testing, we find that &lt;math&gt;63&lt;/math&gt; is too large, but if we plug in &lt;math&gt;62&lt;/math&gt; we find that our answer is &lt;math&gt;\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}&lt;/math&gt;.<br /> <br /> == Alternate Solution ==<br /> <br /> Use the same method as above where you represent the sum of integers from 1 to n expressed as &lt;math&gt;\frac{n(n + 1)}{2}&lt;/math&gt;, plus the additional page number k. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.) <br /> <br /> &lt;math&gt;\frac{n(n + 1)}{2} = 1986&lt;/math&gt; is the quadratic you must solve to obtain the upper bound of n and &lt;math&gt;\frac{n(n + 1)}{2} + n = 1986&lt;/math&gt; <br /> is the quadratic you must solve to obtain the lower bound of n. <br /> <br /> Solving the two give values that are respectively around 62.5 and 61.5 with the quadratic formula, where the only integer between the two is 62. <br /> <br /> This implies that we can plug in 62 and come to the same conclusion as the above solution where &lt;math&gt;x = \boxed{033}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Dubna ismycity