https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jerry122805&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-24T03:57:25Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Carnot%27s_Theorem&diff=147353 Carnot's Theorem 2021-02-18T05:45:05Z <p>Jerry122805: </p> <hr /> <div>'''Carnot's Theorem''' states that in a [[triangle]] &lt;math&gt;ABC&lt;/math&gt;, the signed sum of [[perpendicular]] distances from the [[circumcenter]] &lt;math&gt;O&lt;/math&gt; to the sides (i.e., signed lengths of the pedal lines from &lt;math&gt;O&lt;/math&gt;) is:<br /> <br /> &lt;math&gt;OO_A+OO_B+OO_C=R+r&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair a,b,c,O,i,d,f,g;<br /> a=(0,0);<br /> b=(4,0);<br /> c=(1,3);<br /> O=circumcenter(a,b,c);<br /> i=incenter(a,b,c);<br /> draw(a--b--c--cycle);<br /> draw(circumcircle(a,b,c));<br /> draw(incircle(a,b,c));<br /> dot(i);<br /> dot(O);<br /> label(&quot;$A$&quot;,a,W);<br /> label(&quot;$B$&quot;,b,E);<br /> label(&quot;$C$&quot;,c,N);<br /> label(&quot;$I$&quot;,i,N);<br /> label(&quot;$O$&quot;,O,N);<br /> d=foot(O,b,c);<br /> dot(d);<br /> draw(O--d);<br /> label(&quot;$O_A$&quot;,d,N);<br /> draw(rightanglemark(O,d,b));<br /> f=foot(O,a,b);<br /> dot(f);<br /> draw(O--f);<br /> draw(rightanglemark(O,f,a));<br /> label(&quot;$O_C$&quot;,f,S);<br /> g=foot(O,c,a);<br /> dot(g);<br /> draw(O--g);<br /> draw(rightanglemark(O,g,a));<br /> label(&quot;$O_B$&quot;,g,W);<br /> &lt;/asy&gt;<br /> <br /> where r is the [[inradius]] and R is the [[circumradius]]. The sign of the distance is chosen to be negative iff the entire segment &lt;math&gt;OO_i&lt;/math&gt; lies outside the triangle.<br /> Explicitly,<br /> <br /> &lt;math&gt;OO_A+OO_B+OO_C=\frac{abc(|\cos{A}|+|\cos{B}|+|\cos{C}|)}{4|\Delta|}&lt;/math&gt;<br /> <br /> where &lt;math&gt;\Delta&lt;/math&gt; is the area of triangle &lt;math&gt;\Delta ABC&lt;/math&gt;.<br /> <br /> <br /> Weisstein, Eric W. &quot;Carnot's Theorem.&quot; From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html<br /> <br /> <br /> =Carnot's Theorem=<br /> <br /> '''Carnot's Theorem''' states that in a [[triangle]] &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;A_1\in BC&lt;/math&gt;, &lt;math&gt;B_1\in AC&lt;/math&gt;, and &lt;math&gt;C_1\in AB&lt;/math&gt;, [[perpendicular]]s to the sides &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; are [[concurrent]] [[iff|if and only if]] &lt;math&gt;A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2&lt;/math&gt;.<br /> <br /> ====Proof====<br /> '''Only if:''' Assume that the given perpendiculars are concurrent at &lt;math&gt;M&lt;/math&gt;. Then, from the Pythagorean Theorem, &lt;math&gt;A_1B^2=BM^2-MA_1^2&lt;/math&gt;, &lt;math&gt;C_1A^2=AM^2-MC_1^2&lt;/math&gt;, &lt;math&gt;B_1C^2=CM^2-MB_1^2&lt;/math&gt;, &lt;math&gt;A_1C^2=MC^2-MA_1^2&lt;/math&gt;, &lt;math&gt;C_1B^2=MB^2-MC_1^2&lt;/math&gt;, and &lt;math&gt;B_1A^2=AM^2-MB_1^2&lt;/math&gt;. Substituting each and every one of these in and simplifying gives the desired result.<br /> <br /> <br /> ''' If:''' Consider the intersection of the perpendiculars from &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt;. Call this intersection point &lt;math&gt;N&lt;/math&gt;, and let &lt;math&gt;C_2&lt;/math&gt; be the perpendicular from &lt;math&gt;N&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. From the other direction of the desired result, we have that &lt;math&gt;A_1B^2+C_2A^2+B_1C^2=A_1C^2+C_2B^2+B_1A^2&lt;/math&gt;. We also have that &lt;math&gt;A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2&lt;/math&gt;, which implies that &lt;math&gt;C_1A^2-C_1B^2=C_2A^2-C_2B^2&lt;/math&gt;. This is a difference of squares, which we can easily factor into &lt;math&gt;(C_1A-C_1B)(C_1A+C_1B)=(C_2A-C_2B)(C_2A+C_2B)&lt;/math&gt;. Note that &lt;math&gt;C_1A+C_1=C_2A+C_2B=AB&lt;/math&gt;, so we have that &lt;math&gt;C_1A-C_1B=C_2A-C_2B&lt;/math&gt;. This implies that &lt;math&gt;C_1=C_2&lt;/math&gt;, which gives the desired result.<br /> <br /> =Carnot Extended=<br /> Let &lt;math&gt;P,Q,R&lt;/math&gt; be points in the plane of triangle &lt;math&gt;ABC&lt;/math&gt;. Then the perpendiculars from &lt;math&gt;P,Q,R&lt;/math&gt; to &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively are concurrent if and only if &lt;cmath&gt;PB^2-PC^2+QC^2-QA^2+RA^2-RB^2=0&lt;/cmath&gt;<br /> <br /> ====Proof====<br /> Let &lt;math&gt;X,Y,Z&lt;/math&gt; be the feet of perpendiculars from &lt;math&gt;P,Q,R&lt;/math&gt; to &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively. Note that &lt;math&gt;PB^2-PC^2=XB^2-XC^2&lt;/math&gt; from the application of pythogorean theorem to triangles &lt;math&gt;PXB,PXC&lt;/math&gt;. Now with similar relations for &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, Carnot's theorem finishes the job!<br /> <br /> <br /> =Problems=<br /> <br /> ===Olympiad===<br /> &lt;math&gt;\triangle ABC&lt;/math&gt; is a triangle. Take points &lt;math&gt;D, E, F&lt;/math&gt; on the perpendicular bisectors of &lt;math&gt;BC, CA, AB&lt;/math&gt; respectively. Show that the lines through &lt;math&gt;A, B, C&lt;/math&gt; perpendicular to &lt;math&gt;EF, FD, DE&lt;/math&gt; respectively are concurrent. ([[1997 USAMO Problems/Problem 2|Source]])<br /> <br /> ==See also==<br /> <br /> [[Carnot's Polygon Theorem]]<br /> [[Japanese Theorem]]<br /> [[Category:Geometry]]<br /> [[Category:Theorems]]</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=143392 2019 AMC 10B Problems/Problem 16 2021-01-27T04:53:28Z <p>Jerry122805: /* Problem */</p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with a right angle at &lt;math&gt;C&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;AC=CD,&lt;/math&gt; &lt;math&gt;DE=EB,&lt;/math&gt; and the ratio &lt;math&gt;AC:DE=4:3&lt;/math&gt;. What is the ratio &lt;math&gt;AD:DB?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2&lt;/math&gt;<br /> <br /> <br /> <br /> <br /> Alternatively, once finding the length of &lt;math&gt;AD&lt;/math&gt; one could use the Pythagorean Theorem to find &lt;math&gt;AB&lt;/math&gt; and consequently &lt;math&gt;DB&lt;/math&gt;, and then compute the ratio.<br /> <br /> ==Solution 1==<br /> Without loss of generality, let &lt;math&gt;AC = CD = 4&lt;/math&gt; and &lt;math&gt;DE = EB = 3&lt;/math&gt;. Let &lt;math&gt;\angle A = \alpha&lt;/math&gt; and &lt;math&gt;\angle B = \beta = 90^{\circ} - \alpha&lt;/math&gt;. As &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt; are isosceles, &lt;math&gt;\angle ADC = \alpha&lt;/math&gt; and &lt;math&gt;\angle BDE = \beta&lt;/math&gt;. Then &lt;math&gt;\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}&lt;/math&gt;, so &lt;math&gt;\triangle CDE&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; triangle with &lt;math&gt;CE = 5&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;CB = 5+3 = 8&lt;/math&gt;, and &lt;math&gt;\triangle ABC&lt;/math&gt; is a &lt;math&gt;1-2-\sqrt{5}&lt;/math&gt; triangle.<br /> <br /> In isosceles triangles &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt;, drop altitudes from &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; onto &lt;math&gt;AB&lt;/math&gt;; denote the feet of these altitudes by &lt;math&gt;P_C&lt;/math&gt; and &lt;math&gt;P_E&lt;/math&gt; respectively. Then &lt;math&gt;\triangle ACP_C \sim \triangle ABC&lt;/math&gt; by AAA similarity, so we get that &lt;math&gt;AP_C = P_CD = \frac{4}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD = 2 \times \frac{4}{\sqrt{5}}&lt;/math&gt;. Similarly we get &lt;math&gt;BD = 2 \times \frac{6}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD:DB = \boxed{\textbf{(A) } 2:3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;AC=CD=4x&lt;/math&gt;, and &lt;math&gt;DE=EB=3x&lt;/math&gt;. (For this solution, &lt;math&gt;A&lt;/math&gt; is above &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; is to the right of &lt;math&gt;C&lt;/math&gt;). Also let &lt;math&gt;\angle A = t^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle ACD = \left(180-2t\right)^{\circ}&lt;/math&gt;, which implies &lt;math&gt;\angle DCB = \left(2t - 90\right)^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;\angle B = \left(90-t\right)^{\circ}&lt;/math&gt;, which implies &lt;math&gt;\angle BED = 2t^{\circ}&lt;/math&gt;. This further implies that &lt;math&gt;\angle DEC = \left(180 - 2t\right)^{\circ}&lt;/math&gt;. <br /> <br /> Now we see that &lt;math&gt;\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}&lt;/math&gt;. Thus &lt;math&gt;\triangle CDE&lt;/math&gt; is a right triangle, with side lengths of &lt;math&gt;3x&lt;/math&gt;, &lt;math&gt;4x&lt;/math&gt;, and &lt;math&gt;5x&lt;/math&gt; (by the Pythagorean Theorem, or simply the Pythagorean triple &lt;math&gt;3-4-5&lt;/math&gt;). Therefore &lt;math&gt;AC=4x&lt;/math&gt; (by definition), &lt;math&gt;BC=5x+3x = 8x&lt;/math&gt;, and &lt;math&gt;AB=4\sqrt{5}x&lt;/math&gt;. Hence &lt;math&gt;\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1&lt;/math&gt; (by the double angle formula), giving &lt;math&gt;2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}&lt;/math&gt;.<br /> <br /> By the Law of Cosines in &lt;math&gt;\triangle BED&lt;/math&gt;, if &lt;math&gt;BD = d&lt;/math&gt;, we have &lt;cmath&gt;\begin{split}&amp;d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &amp;d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &amp;d = \frac{12x}{\sqrt{5}}\end{split}&lt;/cmath&gt; Now &lt;math&gt;AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}&lt;/math&gt;. Thus the answer is &lt;math&gt;\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> WLOG, let &lt;math&gt;AC=CD=4&lt;/math&gt;, and &lt;math&gt;DE=EB=3&lt;/math&gt;. &lt;math&gt;\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}&lt;/math&gt;. Because of this, &lt;math&gt;\triangle DEC&lt;/math&gt; is a 3-4-5 right triangle. Draw the altitude &lt;math&gt;DF&lt;/math&gt; of &lt;math&gt;\triangle DEC&lt;/math&gt;. &lt;math&gt;DF&lt;/math&gt; is &lt;math&gt;\frac{12}{5}&lt;/math&gt; by the base-height triangle area formula. &lt;math&gt;\triangle ABC&lt;/math&gt; is similar to &lt;math&gt;\triangle DBF&lt;/math&gt; (AA). So &lt;math&gt;\frac{DF}{AC} = \frac{BD}{AB} = \frac35&lt;/math&gt;. &lt;math&gt;DB&lt;/math&gt; is &lt;math&gt;\frac35&lt;/math&gt; of &lt;math&gt;AB&lt;/math&gt;. Therefore, &lt;math&gt;AD:DB&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) } 2:3}&lt;/math&gt;.<br /> <br /> ~Thegreatboy90<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/_0YaCyxiMBo<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2==<br /> https://youtu.be/4_x1sgcQCp4?t=4245<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=143391 2019 AMC 10B Problems/Problem 16 2021-01-27T04:53:17Z <p>Jerry122805: /* Problem */</p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with a right angle at &lt;math&gt;C&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;AC=CD,&lt;/math&gt; &lt;math&gt;DE=EB,&lt;/math&gt; and the ratio &lt;math&gt;AC:DE=4:3&lt;/math&gt;. What is the ratio &lt;math&gt;AD:DB?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2&lt;/math&gt;<br /> <br /> Alternatively, once finding the length of &lt;math&gt;AD&lt;/math&gt; one could use the Pythagorean Theorem to find &lt;math&gt;AB&lt;/math&gt; and consequently &lt;math&gt;DB&lt;/math&gt;, and then compute the ratio.<br /> <br /> ==Solution 1==<br /> Without loss of generality, let &lt;math&gt;AC = CD = 4&lt;/math&gt; and &lt;math&gt;DE = EB = 3&lt;/math&gt;. Let &lt;math&gt;\angle A = \alpha&lt;/math&gt; and &lt;math&gt;\angle B = \beta = 90^{\circ} - \alpha&lt;/math&gt;. As &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt; are isosceles, &lt;math&gt;\angle ADC = \alpha&lt;/math&gt; and &lt;math&gt;\angle BDE = \beta&lt;/math&gt;. Then &lt;math&gt;\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}&lt;/math&gt;, so &lt;math&gt;\triangle CDE&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; triangle with &lt;math&gt;CE = 5&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;CB = 5+3 = 8&lt;/math&gt;, and &lt;math&gt;\triangle ABC&lt;/math&gt; is a &lt;math&gt;1-2-\sqrt{5}&lt;/math&gt; triangle.<br /> <br /> In isosceles triangles &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt;, drop altitudes from &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; onto &lt;math&gt;AB&lt;/math&gt;; denote the feet of these altitudes by &lt;math&gt;P_C&lt;/math&gt; and &lt;math&gt;P_E&lt;/math&gt; respectively. Then &lt;math&gt;\triangle ACP_C \sim \triangle ABC&lt;/math&gt; by AAA similarity, so we get that &lt;math&gt;AP_C = P_CD = \frac{4}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD = 2 \times \frac{4}{\sqrt{5}}&lt;/math&gt;. Similarly we get &lt;math&gt;BD = 2 \times \frac{6}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD:DB = \boxed{\textbf{(A) } 2:3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;AC=CD=4x&lt;/math&gt;, and &lt;math&gt;DE=EB=3x&lt;/math&gt;. (For this solution, &lt;math&gt;A&lt;/math&gt; is above &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; is to the right of &lt;math&gt;C&lt;/math&gt;). Also let &lt;math&gt;\angle A = t^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle ACD = \left(180-2t\right)^{\circ}&lt;/math&gt;, which implies &lt;math&gt;\angle DCB = \left(2t - 90\right)^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;\angle B = \left(90-t\right)^{\circ}&lt;/math&gt;, which implies &lt;math&gt;\angle BED = 2t^{\circ}&lt;/math&gt;. This further implies that &lt;math&gt;\angle DEC = \left(180 - 2t\right)^{\circ}&lt;/math&gt;. <br /> <br /> Now we see that &lt;math&gt;\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}&lt;/math&gt;. Thus &lt;math&gt;\triangle CDE&lt;/math&gt; is a right triangle, with side lengths of &lt;math&gt;3x&lt;/math&gt;, &lt;math&gt;4x&lt;/math&gt;, and &lt;math&gt;5x&lt;/math&gt; (by the Pythagorean Theorem, or simply the Pythagorean triple &lt;math&gt;3-4-5&lt;/math&gt;). Therefore &lt;math&gt;AC=4x&lt;/math&gt; (by definition), &lt;math&gt;BC=5x+3x = 8x&lt;/math&gt;, and &lt;math&gt;AB=4\sqrt{5}x&lt;/math&gt;. Hence &lt;math&gt;\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1&lt;/math&gt; (by the double angle formula), giving &lt;math&gt;2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}&lt;/math&gt;.<br /> <br /> By the Law of Cosines in &lt;math&gt;\triangle BED&lt;/math&gt;, if &lt;math&gt;BD = d&lt;/math&gt;, we have &lt;cmath&gt;\begin{split}&amp;d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &amp;d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &amp;d = \frac{12x}{\sqrt{5}}\end{split}&lt;/cmath&gt; Now &lt;math&gt;AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}&lt;/math&gt;. Thus the answer is &lt;math&gt;\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> WLOG, let &lt;math&gt;AC=CD=4&lt;/math&gt;, and &lt;math&gt;DE=EB=3&lt;/math&gt;. &lt;math&gt;\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}&lt;/math&gt;. Because of this, &lt;math&gt;\triangle DEC&lt;/math&gt; is a 3-4-5 right triangle. Draw the altitude &lt;math&gt;DF&lt;/math&gt; of &lt;math&gt;\triangle DEC&lt;/math&gt;. &lt;math&gt;DF&lt;/math&gt; is &lt;math&gt;\frac{12}{5}&lt;/math&gt; by the base-height triangle area formula. &lt;math&gt;\triangle ABC&lt;/math&gt; is similar to &lt;math&gt;\triangle DBF&lt;/math&gt; (AA). So &lt;math&gt;\frac{DF}{AC} = \frac{BD}{AB} = \frac35&lt;/math&gt;. &lt;math&gt;DB&lt;/math&gt; is &lt;math&gt;\frac35&lt;/math&gt; of &lt;math&gt;AB&lt;/math&gt;. Therefore, &lt;math&gt;AD:DB&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) } 2:3}&lt;/math&gt;.<br /> <br /> ~Thegreatboy90<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/_0YaCyxiMBo<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2==<br /> https://youtu.be/4_x1sgcQCp4?t=4245<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_7&diff=134207 2006 AIME I Problems/Problem 7 2020-09-28T02:02:35Z <p>Jerry122805: /* Solution 3 (Bash) */</p> <hr /> <div>== Problem ==<br /> An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] &lt;math&gt; C &lt;/math&gt; to the area of shaded region &lt;math&gt; B &lt;/math&gt; is 11/5. Find the ratio of shaded region &lt;math&gt; D &lt;/math&gt; to the area of shaded region &lt;math&gt; A. &lt;/math&gt;<br /> <br /> [[Image:2006AimeA7.PNG]]<br /> <br /> == Solution 1 ==<br /> Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].<br /> <br /> Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at &lt;math&gt;0, 1, 2 \ldots&lt;/math&gt;. The base of region &lt;math&gt;\mathcal{A}&lt;/math&gt; is on the line &lt;math&gt;x = 1&lt;/math&gt;. The bigger base of region &lt;math&gt;\mathcal{D}&lt;/math&gt; is on the line &lt;math&gt;x = 7&lt;/math&gt;. <br /> Let the top side of the angle be &lt;math&gt;y = x - s&lt;/math&gt; and the bottom side be x-axis, as dividing the angle doesn't change the problem.<br /> <br /> Since the area of the triangle is equal to &lt;math&gt;\frac{1}{2}bh&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;<br /> \frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5}<br /> = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}<br /> &lt;/cmath&gt;<br /> <br /> Solve this to find that &lt;math&gt;s = \frac{5}{6}&lt;/math&gt;.<br /> <br /> Using the same reasoning as above, we get &lt;math&gt;\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}&lt;/math&gt;, which is &lt;math&gt;\boxed{408}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be &lt;math&gt;x&lt;/math&gt; and the area of it be &lt;math&gt;x^2&lt;/math&gt;. Also, let all sections of the line on the same side as the side with length &lt;math&gt;x&lt;/math&gt; on a trapezoid be equal to &lt;math&gt;1&lt;/math&gt;. <br /> <br /> Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is &lt;math&gt;(\frac{x+1}{x})^2&lt;/math&gt;. Multiplying, we get &lt;math&gt;(x+1)^2&lt;/math&gt; as the area of the triangle, so the area of the trapezoid is &lt;math&gt;2x+1&lt;/math&gt;. Repeating this process, we get that the area of B is &lt;math&gt;2x+3&lt;/math&gt;, the area of C is &lt;math&gt;2x+7&lt;/math&gt;, and the area of D is &lt;math&gt;2x+11&lt;/math&gt;.<br /> <br /> We can now use the given condition that the ratio of C and B is &lt;math&gt;\frac{11}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{11}{5} = \frac{2x+7}{2x+3}&lt;/math&gt; gives us &lt;math&gt;x = \frac{1}{6}&lt;/math&gt;<br /> <br /> So now we compute the ratio of D and A, which is &lt;math&gt;\frac{2(\frac{1}{6} + 11)}{(\frac{1}{6})^2} = \boxed{408.}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) ==<br /> <br /> Let the distances from the apex to the parallel lines be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; and the distance between the intersections be &lt;math&gt;a,b.&lt;/math&gt; We know the area ratio means &lt;math&gt;\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}&lt;/math&gt; which simplifying yields &lt;math&gt;ab = 3ay+3bx.&lt;/math&gt; The ratio we seek is &lt;math&gt;\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.&lt;/math&gt; We know that &lt;math&gt;ab = 3ay+3bx&lt;/math&gt; so the ratio we seed is &lt;math&gt;\frac{33(ay+yx)}{11xy}.&lt;/math&gt; Finally note that by similar triangles &lt;math&gt;\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.&lt;/math&gt; Therefore the ratio we seek is &lt;math&gt;\frac{66(ay)}{11xy} =\frac{66a}{11x}.&lt;/math&gt; Finally note that &lt;math&gt;ab=3ay+3bx \implies ab = 6bx \implies a = 6x&lt;/math&gt; so the final ratio is &lt;math&gt;6 \cdot 68 = \boxed{408}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_7&diff=134203 2006 AIME I Problems/Problem 7 2020-09-28T01:22:38Z <p>Jerry122805: </p> <hr /> <div>== Problem ==<br /> An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] &lt;math&gt; C &lt;/math&gt; to the area of shaded region &lt;math&gt; B &lt;/math&gt; is 11/5. Find the ratio of shaded region &lt;math&gt; D &lt;/math&gt; to the area of shaded region &lt;math&gt; A. &lt;/math&gt;<br /> <br /> [[Image:2006AimeA7.PNG]]<br /> <br /> == Solution 1 ==<br /> Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].<br /> <br /> Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at &lt;math&gt;0, 1, 2 \ldots&lt;/math&gt;. The base of region &lt;math&gt;\mathcal{A}&lt;/math&gt; is on the line &lt;math&gt;x = 1&lt;/math&gt;. The bigger base of region &lt;math&gt;\mathcal{D}&lt;/math&gt; is on the line &lt;math&gt;x = 7&lt;/math&gt;. <br /> Let the top side of the angle be &lt;math&gt;y = x - s&lt;/math&gt; and the bottom side be x-axis, as dividing the angle doesn't change the problem.<br /> <br /> Since the area of the triangle is equal to &lt;math&gt;\frac{1}{2}bh&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;<br /> \frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5}<br /> = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}<br /> &lt;/cmath&gt;<br /> <br /> Solve this to find that &lt;math&gt;s = \frac{5}{6}&lt;/math&gt;.<br /> <br /> Using the same reasoning as above, we get &lt;math&gt;\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}&lt;/math&gt;, which is &lt;math&gt;\boxed{408}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be &lt;math&gt;x&lt;/math&gt; and the area of it be &lt;math&gt;x^2&lt;/math&gt;. Also, let all sections of the line on the same side as the side with length &lt;math&gt;x&lt;/math&gt; on a trapezoid be equal to &lt;math&gt;1&lt;/math&gt;. <br /> <br /> Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is &lt;math&gt;(\frac{x+1}{x})^2&lt;/math&gt;. Multiplying, we get &lt;math&gt;(x+1)^2&lt;/math&gt; as the area of the triangle, so the area of the trapezoid is &lt;math&gt;2x+1&lt;/math&gt;. Repeating this process, we get that the area of B is &lt;math&gt;2x+3&lt;/math&gt;, the area of C is &lt;math&gt;2x+7&lt;/math&gt;, and the area of D is &lt;math&gt;2x+11&lt;/math&gt;.<br /> <br /> We can now use the given condition that the ratio of C and B is &lt;math&gt;\frac{11}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{11}{5} = \frac{2x+7}{2x+3}&lt;/math&gt; gives us &lt;math&gt;x = \frac{1}{6}&lt;/math&gt;<br /> <br /> So now we compute the ratio of D and A, which is &lt;math&gt;\frac{2(\frac{1}{6} + 11)}{(\frac{1}{6})^2} = \boxed{408.}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) ==<br /> <br /> Let the distances from the apex to the parallel lines be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; and the distance between the intersections be &lt;math&gt;a,b.&lt;/math&gt; We know the area ratio means &lt;math&gt;\displaystyle\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} = \displaystyle\frac{5}{11}&lt;/math&gt; which simplifying yields &lt;math&gt;ab = 3ay+3bx.&lt;/math&gt; The ratio we seek is &lt;math&gt;\displaystyle\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} = \displaystyle\frac{ay+yx+11ab}{xy}.&lt;/math&gt; We know that &lt;math&gt;ab = 3ay+3bx&lt;/math&gt; so the ratio we seed is &lt;math&gt;\displaystyle\frac{33(ay+yx)}{11xy}.&lt;/math&gt; Finally note that by similar triangles &lt;math&gt;\displaystyle\frac{x}{x+a} = \displaystyle\frac{y}{y+b} \implies bx = ya.&lt;/math&gt; Therefore the ratio we seek is &lt;math&gt;\displaystyle\frac{66(ay)}{11xy} = \displaystyle\frac{66a}{11x}.&lt;/math&gt; Finally note that &lt;math&gt;ab=3ay+6bx \implies ab = 6bx \implies a = 6x&lt;/math&gt; so the final ratio is &lt;math&gt;6 \cdot 68 = \boxed{408}.&lt;/math&gt; <br /> <br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=134150 2011 AIME I Problems/Problem 4 2020-09-27T17:09:24Z <p>Jerry122805: /* Solution 5 (Very fast) */</p> <hr /> <div>== Problem ==<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 /> == 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 &lt;math&gt;B&lt;/math&gt;, 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;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{056}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<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;IM \perp MC&lt;/math&gt; and &lt;math&gt;IN \perp NC&lt;/math&gt;, we have &lt;math&gt;CMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = CI&lt;/math&gt;. Since &lt;math&gt;\sin \angle MIN = \sin (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{056}&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 (LoC) 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;.<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 /> ==Video Solution==<br /> https://www.youtube.com/watch?v=vkniYGN45F4<br /> <br /> ~Shreyas S<br /> <br /> Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&amp;t=12s<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=134149 2011 AIME I Problems/Problem 4 2020-09-27T17:07:59Z <p>Jerry122805: /* Solution 4 */</p> <hr /> <div>== Problem ==<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 /> == 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 &lt;math&gt;B&lt;/math&gt;, 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;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{056}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<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;IM \perp MC&lt;/math&gt; and &lt;math&gt;IN \perp NC&lt;/math&gt;, we have &lt;math&gt;CMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = CI&lt;/math&gt;. Since &lt;math&gt;\sin \angle MIN = \sin (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{056}&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 (LoC) 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;.<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 /> == Solution 5 (Very fast) == <br /> <br /> Extend &lt;math&gt;CM&lt;/math&gt; and &lt;math&gt;CN&lt;/math&gt; to meet &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y.&lt;/math&gt; Note that &lt;math&gt;CM=MX&lt;/math&gt; and &lt;math&gt;CN = NY&lt;/math&gt; so &lt;math&gt;MN = \displaystyle\frac{XY}{2}.&lt;/math&gt; To find &lt;math&gt;XY&lt;/math&gt; we just do a system of equations and we find &lt;math&gt;XY=112&lt;/math&gt; so &lt;math&gt;XY = \boxed{056}.&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=vkniYGN45F4<br /> <br /> ~Shreyas S<br /> <br /> Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&amp;t=12s<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=133904 1990 AIME Problems/Problem 3 2020-09-21T01:53:19Z <p>Jerry122805: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 ==<br /> <br /> As in above, we have &lt;math&gt;rs = 118r - 116s.&lt;/math&gt; This means that &lt;math&gt;rs + 116s - 118r = 0.&lt;/math&gt; Using SFFT we obtain &lt;math&gt;s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.&lt;/math&gt; Since &lt;math&gt;r+116&lt;/math&gt; is always positive, we know thta &lt;math&gt;s-118&lt;/math&gt; must be negative. Therefore the maximum value of &lt;math&gt;s&lt;/math&gt; must be &lt;math&gt;\boxed{117}&lt;/math&gt; which indeed yields an integral value of &lt;math&gt;r.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=133903 1990 AIME Problems/Problem 3 2020-09-21T01:53:07Z <p>Jerry122805: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 ==<br /> <br /> As in above, we have <br /> <br /> == Solution 3 ==<br /> <br /> As in above, we have &lt;math&gt;rs = 118r - 116s.&lt;/math&gt; This means that &lt;math&gt;rs + 116s - 118r = 0.&lt;/math&gt; Using SFFT we obtain &lt;math&gt;s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.&lt;/math&gt; Since &lt;math&gt;r+116&lt;/math&gt; is always positive, we know thta &lt;math&gt;s-118&lt;/math&gt; must be negative. Therefore the maximum value of &lt;math&gt;s&lt;/math&gt; must be &lt;math&gt;\boxed{117}&lt;/math&gt; which indeed yields an integral value of &lt;math&gt;r.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=133902 1990 AIME Problems/Problem 3 2020-09-21T01:52:47Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 ==<br /> <br /> As in above, we have <br /> <br /> == Solution 3 ==<br /> <br /> As in above, we have &lt;math&gt;rs = 118r - 116s.&lt;/math&gt; This means that &lt;math&gt;rs + 116s - 118r = 0.&lt;/math&gt; Using SFFT we obtain &lt;math&gt;s(r+116) - 118(r+116) = -118 \cdot 116 \implies &lt;/math&gt;(s-118)(r+116) = -118 \cdot 116.&lt;math&gt; Since &lt;/math&gt;r+116&lt;math&gt; is always positive, we know thta &lt;/math&gt;s-118&lt;math&gt; must be negative. Therefore the maximum value of &lt;/math&gt;s&lt;math&gt; must be &lt;/math&gt;\boxed{117}&lt;math&gt; which indeed yields an integral value of &lt;/math&gt;r.$<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_4&diff=133651 1989 AIME Problems/Problem 4 2020-09-14T20:13:40Z <p>Jerry122805: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;a&lt;b&lt;c&lt;d&lt;e&lt;/math&gt; are [[consecutive]] [[positive]] [[integer]]s such that &lt;math&gt;b+c+d&lt;/math&gt; is a [[perfect square]] and &lt;math&gt;a+b+c+d+e&lt;/math&gt; is a [[perfect cube]], what is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know &lt;math&gt;b + c + d = 3c&lt;/math&gt; and &lt;math&gt;a + b + c + d + e = 5c&lt;/math&gt;. Thus, &lt;math&gt;c&lt;/math&gt; must be in the form of &lt;math&gt;3 \cdot x^2&lt;/math&gt; based upon the first part and in the form of &lt;math&gt;5^2 \cdot y^3&lt;/math&gt; based upon the second part, with &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; denoting an [[integer]]s. &lt;math&gt;c&lt;/math&gt; is minimized if it’s [[prime factorization]] contains only &lt;math&gt;3,5&lt;/math&gt;, and since there is a cubed term in &lt;math&gt;5^2 \cdot y^3&lt;/math&gt;, &lt;math&gt;3^3&lt;/math&gt; must be a factor of &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;3^35^2 = \boxed{675}&lt;/math&gt;, which works as the solution.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; equal &lt;math&gt;a+1&lt;/math&gt;, &lt;math&gt;a+2&lt;/math&gt;, &lt;math&gt;a+3&lt;/math&gt;, and &lt;math&gt;a+4&lt;/math&gt;, respectively. Call the square and cube &lt;math&gt;k^2&lt;/math&gt; and &lt;math&gt;m^3&lt;/math&gt;, where both k and m are integers. Then:<br /> <br /> &lt;math&gt;5a + 10 = m^3&lt;/math&gt;<br /> <br /> Now we know &lt;math&gt;m^3&lt;/math&gt; is a multiple of 125 and &lt;math&gt;m&lt;/math&gt; is a multiple of 5. The lower &lt;math&gt;m&lt;/math&gt; is, the lower the value of &lt;math&gt;c&lt;/math&gt; will be. Start from 5 and add 5 each time.<br /> <br /> &lt;math&gt;m = 5&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 10&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 15&lt;/math&gt; gives a solution for k.<br /> <br /> <br /> &lt;math&gt;10 + 5a = 15^3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;2 + a = 675&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;c = \boxed{675}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 == <br /> <br /> <br /> Let the numbers be &lt;math&gt;a,a+1,a+2,a+3,a+4.&lt;/math&gt; When then know &lt;math&gt;3a+6&lt;/math&gt; is a perfect cube and &lt;math&gt;5a+10&lt;/math&gt; is perfect cube. Since &lt;math&gt;5a+10&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt; we know that &lt;math&gt;5a+10 = (5k)^3&lt;/math&gt; since otherwise we get a contradiction. This means &lt;math&gt;a = 25k^3 - 2&lt;/math&gt; in which plugging into the other expression we know &lt;math&gt;3(25k^3 - 2) + 6 = 75k^3&lt;/math&gt; is a perfect square. We know &lt;math&gt;75 = 5^2 \cdot 3&lt;/math&gt; so we let &lt;math&gt;k = 3&lt;/math&gt; to obtain the perfect square. This means that &lt;math&gt;c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_4&diff=133650 1989 AIME Problems/Problem 4 2020-09-14T20:06:43Z <p>Jerry122805: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;a&lt;b&lt;c&lt;d&lt;e&lt;/math&gt; are [[consecutive]] [[positive]] [[integer]]s such that &lt;math&gt;b+c+d&lt;/math&gt; is a [[perfect square]] and &lt;math&gt;a+b+c+d+e&lt;/math&gt; is a [[perfect cube]], what is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know &lt;math&gt;b + c + d = 3c&lt;/math&gt; and &lt;math&gt;a + b + c + d + e = 5c&lt;/math&gt;. Thus, &lt;math&gt;c&lt;/math&gt; must be in the form of &lt;math&gt;3 \cdot x^2&lt;/math&gt; based upon the first part and in the form of &lt;math&gt;5^2 \cdot y^3&lt;/math&gt; based upon the second part, with &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; denoting an [[integer]]s. &lt;math&gt;c&lt;/math&gt; is minimized if it’s [[prime factorization]] contains only &lt;math&gt;3,5&lt;/math&gt;, and since there is a cubed term in &lt;math&gt;5^2 \cdot y^3&lt;/math&gt;, &lt;math&gt;3^3&lt;/math&gt; must be a factor of &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;3^35^2 = \boxed{675}&lt;/math&gt;, which works as the solution.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; equal &lt;math&gt;a+1&lt;/math&gt;, &lt;math&gt;a+2&lt;/math&gt;, &lt;math&gt;a+3&lt;/math&gt;, and &lt;math&gt;a+4&lt;/math&gt;, respectively. Call the square and cube &lt;math&gt;k^2&lt;/math&gt; and &lt;math&gt;m^3&lt;/math&gt;, where both k and m are integers. Then:<br /> <br /> &lt;math&gt;5a + 10 = m^3&lt;/math&gt;<br /> <br /> Now we know &lt;math&gt;m^3&lt;/math&gt; is a multiple of 125 and &lt;math&gt;m&lt;/math&gt; is a multiple of 5. The lower &lt;math&gt;m&lt;/math&gt; is, the lower the value of &lt;math&gt;c&lt;/math&gt; will be. Start from 5 and add 5 each time.<br /> <br /> &lt;math&gt;m = 5&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 10&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 15&lt;/math&gt; gives a solution for k.<br /> <br /> <br /> &lt;math&gt;10 + 5a = 15^3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;2 + a = 675&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;c = \boxed{675}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 == <br /> <br /> <br /> Let the numbers be &lt;math&gt;a,a+1,a+2,a+3,a+4.&lt;/math&gt; When then know &lt;math&gt;3a+6&lt;/math&gt; is a perfect cube and &lt;math&gt;5a+10&lt;/math&gt; is perfect cube. Since &lt;math&gt;5a+10&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt; we know that &lt;math&gt;5a+10 = (5k)^3&lt;/math&gt; since otherwise we get a contradiction. This means &lt;math&gt;a = 25k^3 - 2&lt;/math&gt; in which plugging into the other expression we know &lt;math&gt;3(25k^3 - 2) + 6 = 75k^3&lt;/math&gt; is a perfect square. We know &lt;math&gt;75 = 5^2 \cdot 3&lt;/math&gt; so we let &lt;math&gt;k = 3&lt;/math&gt; to obtain the perfect square. This means that &lt;math&gt;c = a+2 = 925 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_4&diff=133649 1989 AIME Problems/Problem 4 2020-09-14T20:06:30Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;a&lt;b&lt;c&lt;d&lt;e&lt;/math&gt; are [[consecutive]] [[positive]] [[integer]]s such that &lt;math&gt;b+c+d&lt;/math&gt; is a [[perfect square]] and &lt;math&gt;a+b+c+d+e&lt;/math&gt; is a [[perfect cube]], what is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know &lt;math&gt;b + c + d = 3c&lt;/math&gt; and &lt;math&gt;a + b + c + d + e = 5c&lt;/math&gt;. Thus, &lt;math&gt;c&lt;/math&gt; must be in the form of &lt;math&gt;3 \cdot x^2&lt;/math&gt; based upon the first part and in the form of &lt;math&gt;5^2 \cdot y^3&lt;/math&gt; based upon the second part, with &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; denoting an [[integer]]s. &lt;math&gt;c&lt;/math&gt; is minimized if it’s [[prime factorization]] contains only &lt;math&gt;3,5&lt;/math&gt;, and since there is a cubed term in &lt;math&gt;5^2 \cdot y^3&lt;/math&gt;, &lt;math&gt;3^3&lt;/math&gt; must be a factor of &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;3^35^2 = \boxed{675}&lt;/math&gt;, which works as the solution.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; equal &lt;math&gt;a+1&lt;/math&gt;, &lt;math&gt;a+2&lt;/math&gt;, &lt;math&gt;a+3&lt;/math&gt;, and &lt;math&gt;a+4&lt;/math&gt;, respectively. Call the square and cube &lt;math&gt;k^2&lt;/math&gt; and &lt;math&gt;m^3&lt;/math&gt;, where both k and m are integers. Then:<br /> <br /> &lt;math&gt;5a + 10 = m^3&lt;/math&gt;<br /> <br /> Now we know &lt;math&gt;m^3&lt;/math&gt; is a multiple of 125 and &lt;math&gt;m&lt;/math&gt; is a multiple of 5. The lower &lt;math&gt;m&lt;/math&gt; is, the lower the value of &lt;math&gt;c&lt;/math&gt; will be. Start from 5 and add 5 each time.<br /> <br /> &lt;math&gt;m = 5&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 10&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 15&lt;/math&gt; gives a solution for k.<br /> <br /> <br /> &lt;math&gt;10 + 5a = 15^3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;2 + a = 675&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;c = \boxed{675}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> == Solution 3 == <br /> <br /> <br /> Let the numbers be &lt;math&gt;a,a+1,a+2,a+3,a+4.&lt;/math&gt; When then know &lt;math&gt;3a+6&lt;/math&gt; is a perfect cube and &lt;math&gt;5a+10&lt;/math&gt; is perfect cube. Since &lt;math&gt;5a+10&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt; we know that &lt;math&gt;5a+10 = (5k)^3&lt;/math&gt; since otherwise we get a contradiction. This means &lt;math&gt;a = 25k^3 - 2&lt;/math&gt; in which plugging into the other expression we know &lt;math&gt;3(25k^3 - 2) + 6 = 75k^3&lt;/math&gt; is a perfect square. We know &lt;math&gt;75 = 5^2 \cdot 3&lt;/math&gt; so we let &lt;math&gt;k = 3&lt;/math&gt; to obtain the perfect square. This means that &lt;math&gt;c = a+2 = 925 \cdot 27 - 2)+2 = 25\ cdot 27 = \boxed{675}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_5&diff=133561 2000 AIME II Problems/Problem 5 2020-09-13T01:13:17Z <p>Jerry122805: Undo revision 133184 by Xxhalo711 (talk)</p> <hr /> <div>== Problem ==<br /> Given eight distinguishable rings, let &lt;math&gt;n&lt;/math&gt; be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> There are &lt;math&gt;\binom{8}{5}&lt;/math&gt; ways to choose the rings, and there are &lt;math&gt;5!&lt;/math&gt; distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just &lt;math&gt;\binom {8}{3}&lt;/math&gt;.<br /> <br /> Multiplying gives the answer: &lt;math&gt;\binom{8}{5}\binom{8}{3}5! = 376320&lt;/math&gt;, and the three leftmost digits are &lt;math&gt;\boxed{376}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_4&diff=133560 2004 AIME II Problems/Problem 4 2020-09-13T01:10:07Z <p>Jerry122805: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many [[positive integer]]s less than 10,000 have at most two different [[digit]]s?<br /> <br /> == Solution ==<br /> First, let's count numbers with only a single digit. We have nine of these for each length, and four lengths, so 36 total numbers.<br /> <br /> Now, let's count those with two distinct digits. We handle the cases &quot;0 included&quot; and &quot;0 not included&quot; separately.<br /> <br /> There are &lt;math&gt;{9 \choose 2}&lt;/math&gt; ways to choose two digits, &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Given two digits, there are &lt;math&gt;2^n - 2&lt;/math&gt; ways to arrange them in an &lt;math&gt;n&lt;/math&gt;-digit number, for a total of &lt;math&gt;(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22&lt;/math&gt; such numbers (or we can list them: &lt;math&gt;AB, BA, AAB, ABA, BAA, ABB, BAB, BBA, AAAB, AABA, ABAA,&lt;/math&gt; &lt;math&gt;BAAA, AABB, ABAB, BAAB, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA&lt;/math&gt;). Thus, we have &lt;math&gt;{9 \choose 2} \cdot 22 = 36\cdot22 = 792&lt;/math&gt; numbers of this form.<br /> <br /> Now, suppose 0 is one of our digits. We have nine choices for the other digit. For each choice, we have &lt;math&gt;2^{n - 1} - 1&lt;/math&gt; &lt;math&gt;n&lt;/math&gt;-digit numbers we can form, for a total of &lt;math&gt;(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11&lt;/math&gt; such numbers (or we can list them: &lt;math&gt;A0, A00, A0A, AA0, A000, AA00, A0A0, A00A, AAA0, AA0A, A0AA&lt;/math&gt;). This gives us &lt;math&gt;9\cdot 11 = 99&lt;/math&gt; numbers of this form.<br /> <br /> Thus, in total, we have &lt;math&gt;36 + 792 + 99 = \boxed{927}&lt;/math&gt; such numbers.<br /> <br /> <br /> == Solution 2 ==<br /> <br /> We use casework on the number of digits for this problem.<br /> <br /> If the number has a single digit, namely the number &lt;math&gt;n \in [1,9],&lt;/math&gt; we can clearly all such &lt;math&gt;n&lt;/math&gt; work.<br /> <br /> If the number has two digits, or the number &lt;math&gt;n \in [10,99]&lt;/math&gt; we can clearly see all such &lt;math&gt;n&lt;/math&gt; work.<br /> <br /> If the number &lt;math&gt;n&lt;/math&gt; has three digits, there are a total of &lt;math&gt;900&lt;/math&gt; three digit numbers, and there are &lt;math&gt;9 \cdot 9 \cdot 8&lt;/math&gt; numbers that have all distinct digits so there are &lt;math&gt;900 - 9 \cdot 9 \cdot 8&lt;/math&gt; total three digit numbers that work.<br /> <br /> If the number &lt;math&gt;n&lt;/math&gt; has four digits, then we have a total of &lt;math&gt;9 + \displaystyle\binom{9}{2}\left(\displaystyle\binom{4}{1}+ \displaystyle\binom{4}{2} + \displaystyle\binom{4}{3}\right) + 9 \cdot \left(\displaystyle\binom{3}{1}+\displaystyle\binom{3}{2} + \displaystyle\binom{3}{3}\right)&lt;/math&gt; so in total we get &lt;math&gt;\boxed{927}&lt;/math&gt; numbers that work.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|num-b=3|num-a=5|n=II}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_12&diff=133512 2008 AIME II Problems/Problem 12 2020-09-11T20:40:05Z <p>Jerry122805: /* Solution 3 (mad bash) */</p> <hr /> <div>== Problem ==<br /> There are two distinguishable flagpoles, and there are &lt;math&gt;19&lt;/math&gt; flags, of which &lt;math&gt;10&lt;/math&gt; are identical blue flags, and &lt;math&gt;9&lt;/math&gt; are identical green flags. Let &lt;math&gt;N&lt;/math&gt; be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. 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 /> === Solution 1 ===<br /> The well known problem of ordering &lt;math&gt;x&lt;/math&gt; elements of a string of &lt;math&gt;y&lt;/math&gt; elements such that none of the &lt;math&gt;x&lt;/math&gt; elements are next to each other has &lt;math&gt;{y-x+1\choose x}&lt;/math&gt; solutions. (1)<br /> <br /> We generalize for &lt;math&gt;a&lt;/math&gt; blues and &lt;math&gt;b&lt;/math&gt; greens. Consider a string of &lt;math&gt;a+b&lt;/math&gt; elements such that we want to choose the greens such that none of them are next to each other. We would also like to choose a place where we can divide this string into two strings such that the left one represents the first pole, and the right one represents the second pole, in order up the pole according to position on the string. <br /> <br /> However, this method does not account for the fact that the first pole could end with a green, and the second pole could start with a green, since the original string assumed that no greens could be consecutive. We solve this problem by introducing an extra blue, such that we choose our divider by choosing one of these &lt;math&gt;a+1&lt;/math&gt; blues, and not including that one as a flag on either pole.<br /> <br /> From (1), we now have &lt;math&gt;{a+2\choose b}&lt;/math&gt; ways to order the string such that no greens are next to each other, and &lt;math&gt;a+1&lt;/math&gt; ways to choose the extra blue that will divide the string into the two poles: or &lt;math&gt;(a+1){a+2\choose b}&lt;/math&gt; orderings in total.<br /> <br /> However, we have overcounted the solutions where either pole has no flags, we have to count these separately. This is the same as choosing our extra blue as one of the two ends, and ordering the other &lt;math&gt;a&lt;/math&gt; blues and &lt;math&gt;b&lt;/math&gt; greens such that no greens are next to each other: for a total of &lt;math&gt;2{a+1\choose b}&lt;/math&gt; such orderings.<br /> <br /> Thus, we have &lt;math&gt;(a+1){a+2\choose b}-2{a+1\choose b}&lt;/math&gt; orderings that satisfy the conditions in the problem: plugging in &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, we get &lt;math&gt;2310 \equiv \boxed{310} \pmod{1000}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Split the problem into two cases:<br /> <br /> Case 1 - Both poles have blue flags:<br /> There are 9 ways to place the 10 blue flags on the poles. In each of these configurations, there are 12 spots that a green flag could go. (either in between two blues or at the tops or bottoms of the poles) Then, since there are 9 green flags, all of which must be used, there are &lt;math&gt;{12\choose9}&lt;/math&gt; possiblities for each of the 9 ways to place the blue flags. Total: &lt;math&gt;{12\choose9}*9&lt;/math&gt; possibilities.<br /> <br /> Case 2 - One pole has no blue flags:<br /> Since each pole is non empty, the pole without a blue flag must have one green flag. The other pole has 10 blue flags and, by the argument used in case 1, there are &lt;math&gt;{11\choose8}&lt;/math&gt; possibilities, and since the poles are distinguishable, there are a total of &lt;math&gt;2*{11\choose8}&lt;/math&gt; possiblities for this case.<br /> <br /> Finally, we have &lt;math&gt;9*{12\choose9}+2*{11\choose8}=2310 \equiv \boxed{310} \pmod{1000}&lt;/math&gt; as our answer.<br /> <br /> === Solution 3 (mad bash) ===<br /> <br /> Call the two flagpoles Flagpole A and Flagpole B.<br /> Case 1:<br /> Flag distribution: 1|18<br /> B|variable: &lt;math&gt;\dbinom{10}{9}=10&lt;/math&gt; ways<br /> G|variable: &lt;math&gt;\dbinom{11}{8}=165&lt;/math&gt; ways<br /> Case 2:<br /> Flag distribution: 2|17<br /> BB|variable: &lt;math&gt;\dbinom{9}{9}=1&lt;/math&gt; way<br /> BG|variable (two ways to arrange the first sequence): &lt;math&gt;\dbinom{10}{8} \times 2 = 90&lt;/math&gt; ways<br /> GG|variable: can't happen<br /> Case 3: <br /> Flag distribution: 3|16<br /> BBB|variable: can't happen<br /> BBG|variable (3 legal ways to arrange): &lt;math&gt;\dbinom{9}{8} \times 3 = 27&lt;/math&gt; ways<br /> GBG|variable (1 legal way to arrange): &lt;math&gt;\dbinom{10}{7} = 120&lt;/math&gt; ways<br /> GGG|variable: clearly can't happen<br /> Case 4:<br /> Flag distribution: 4|15<br /> BBBB|variable: can't happen<br /> BBBG|variable (4 legal ways to arrange): &lt;math&gt;\dbinom{8}{8} \times 4 = 4&lt;/math&gt; ways<br /> GBBG|variable (3 legal ways to arrange): &lt;math&gt;\dbinom{9}{7} \times 3 = 108&lt;/math&gt; ways<br /> GGBG|variable: can't happen<br /> GGGG|variable: can't happen<br /> Case 5:<br /> Flag distribution: 5|14<br /> BBBBB|variable: can't happen<br /> BBBBG|variable: can't happen<br /> BBGBG|variable (6 legal ways to arrange): &lt;math&gt;\dbinom{8}{7} \times 6 = 48&lt;/math&gt; ways<br /> GBGBG|variable (1 legal way to arrange): &lt;math&gt;\dbinom{9}{6} = 84&lt;/math&gt; ways<br /> GGGGB|variable: can't happen<br /> GGGGG|variable: can't happen <br /> Case 6:<br /> Flag distribution: 6|13<br /> BBBBBB|variable: can't happen<br /> BBBBBG|variable: can't happen<br /> BBBBGG|variable (10 legal ways to arrange): &lt;math&gt;\dbinom{7}{7} \times 10 = 10&lt;/math&gt; ways<br /> BBBGGG|variable (4 legal ways to arrange): &lt;math&gt;\dbinom{8}{6} \times 4 = 112&lt;/math&gt; ways<br /> BBGGGG|variable: can't happen<br /> BGGGGG|variable: can't happen<br /> GGGGGG|variable: can't happen<br /> Case 7: <br /> Flag distribution: 7|12<br /> BBBBBBB|variable: can't happen<br /> BBBBBBG|variable: can't happen<br /> BBBBBGG|variable: can't happen<br /> BBBBGGG|variable (10 legal ways to arrange): &lt;math&gt;\dbinom{7}{6} \times 10 = 70&lt;/math&gt; ways<br /> BBBGGGG|variable (1 legal way to arrange): &lt;math&gt;\dbinom{8}{5} = 56&lt;/math&gt; ways<br /> rest can't happen<br /> Case 8:<br /> Flag distribution: 8|11<br /> BBBBBBBB|variable: can't happen<br /> BBBBBBBG|variable: can't happen<br /> BBBBBBGG|variable: can't happen<br /> BBBBBGGG|variable (20 legal ways to arrange): &lt;math&gt;\dbinom{6}{6} \times 20 = 20&lt;/math&gt; ways<br /> BBBBGGGG|variable: (5 legal ways to arrange): &lt;math&gt;\dbinom{7}{5} \times 5 = 105&lt;/math&gt; ways<br /> others can't happen<br /> Case 9:<br /> Flag distribution: 9|10<br /> BBBBBGGGG|variable (15 legal ways to arrange): &lt;math&gt;\dbinom{6}{5} \times 15 = 90&lt;/math&gt; ways<br /> BBBBGGGGG|variable (1 legal way to arrange): &lt;math&gt;\dbinom{7}{4} = 35&lt;/math&gt; ways<br /> others can't happen<br /> So our total number of ways is &lt;math&gt;2&lt;/math&gt; times &lt;math&gt;10+165+1+90+27+120+4+108+48+84+10+112+70+56+20+105+90+35&lt;/math&gt; (since the two flagpoles are distinguishable) which is &lt;math&gt;2310&lt;/math&gt; ways. We have to find the last 3 digits, so our final answer is &lt;math&gt;310&lt;/math&gt;.<br /> <br /> Solution by fidgetboss_4000<br /> <br /> Note: Do not attempt unless you are good at bashing.<br /> <br /> <br /> <br /> <br /> === Solution 4 (Vandermonde's) ===<br /> <br /> Let the number of blue flags on the first flag be &lt;math&gt;b&lt;/math&gt; and define &lt;math&gt;g&lt;/math&gt; similarly. Now note the bound &lt;math&gt;g \le b+1.&lt;/math&gt; Now we cannot have any two consecutive greens. This condition is the same as &quot;there are &lt;math&gt;b+1&lt;/math&gt; spaces between each blue flag. How many ways can we put the &lt;math&gt;g&lt;/math&gt; green flags in the &lt;math&gt;b+1&lt;/math&gt; spaces?&quot; Therefore given &lt;math&gt;b&lt;/math&gt; blue flags on the first flag and &lt;math&gt;g&lt;/math&gt; green flags on the first flag we have a total of &lt;math&gt;\displaystyle\binom{b+1}{g}.&lt;/math&gt; Similarly there are &lt;math&gt;\displaystyle\binom{11-b}{9-g}&lt;/math&gt; ways for the second flag. Summing over all possible possibilites we see the sum is &lt;math&gt;\sum_{g = 0}^{b+1} \sum_{n=0}^{10} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g}.&lt;/math&gt; Swapping the sum we have &lt;math&gt;\sum_{g = 0}^{b+1} \sum_{n=0}^{10} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g} = \sum_{n=0}^{10} \sum_{g = 0}^{b+1} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g}.&lt;/math&gt; Then applying vandermondes yields &lt;math&gt; \sum_{n=0}^{10} \displaystyle\binom{12}{9}&lt;/math&gt; so the sum is &lt;math&gt;11 \cdot \displaystyle\binom{12}{9}.&lt;/math&gt; However, this answer overcounts. We cannot have no flags on a pole, so we subtract &lt;math&gt;2 \cdot \displaystyle\binom{11}{9}&lt;/math&gt; since we can have empty flags for the first or second flag. Therefore the answer is &lt;math&gt;11 \cdot \displaystyle\binom{12}{9} - 2 \cdot \displaystyle\binom{11}{9} \pmod{1000} \equiv \boxed{310} \pmod{1000}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=133394 2011 AIME II Problems/Problem 6 2020-09-09T19:56:21Z <p>Jerry122805: /* Solution 6 */</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution 1==<br /> Rearranging the [[inequality]] we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{080}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding &lt;math&gt;\binom83 = 56&lt;/math&gt; ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding &lt;math&gt;\binom 63 = 20&lt;/math&gt; ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding &lt;math&gt;\binom43 = 4&lt;/math&gt; ways and there cannot be 3 holders between a and b so our total is 56+20+4=&lt;math&gt;\fbox{080}&lt;/math&gt;.<br /> <br /> ==Solution 3 (&lt;i&gt;Slightly&lt;/i&gt; bashy)==<br /> We first start out when the value of &lt;math&gt;a=1&lt;/math&gt;. <br /> <br /> Doing casework, we discover that &lt;math&gt;d=5,6,7,8,9,10&lt;/math&gt;. We quickly find a pattern.<br /> <br /> Now, doing this for the rest of the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, we see that the answer is simply:<br /> <br /> <br /> &lt;math&gt;(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)&lt;/math&gt;<br /> &lt;math&gt;+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}&lt;/math&gt;<br /> <br /> ==Solution 4 (quick)==<br /> <br /> Notice that if &lt;math&gt;a+d&gt;b+c&lt;/math&gt;, then &lt;math&gt;(11-a)+(11-d)&lt;(11-b)+(11-c)&lt;/math&gt;, so there is a 1-to-1 correspondence between the number of ordered quadruples with &lt;math&gt;a+d&gt;b+c&lt;/math&gt; and the number of ordered quadruples with &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. <br /> <br /> Quick counting gives that the number of ordered quadruples with &lt;math&gt;a+d=b+c&lt;/math&gt; is 50. To count this, consider our numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10&lt;/math&gt;. Notice that if, for example, &lt;math&gt;a+d=b+c=8&lt;/math&gt;, that the average of &lt;math&gt;a,d&lt;/math&gt; and &lt;math&gt;b,c&lt;/math&gt; must both be &lt;math&gt;4&lt;/math&gt;. In this way, there is a symmetry for this case, centered at &lt;math&gt;4&lt;/math&gt;. If instead, say, &lt;math&gt;a+d=b+c=7&lt;/math&gt;, an odd number, then there is symmetry with &lt;math&gt;(a,d);(b,c)&lt;/math&gt; about &lt;math&gt;3.5&lt;/math&gt;. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at &lt;math&gt;2.5,3,8,8.5&lt;/math&gt;, there is &lt;math&gt;1&lt;/math&gt; case for each and so on, until centered at &lt;math&gt;5.5&lt;/math&gt;, there are &lt;math&gt;10&lt;/math&gt; possible cases. Adding these all, we have &lt;math&gt;2(1+3+6)+10=50&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;\frac{\binom{10}{4}-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is &lt;math&gt;(10*9*8*7/24)=210&lt;/math&gt; <br /> <br /> Next, intuitively, the number of quadruples where &lt;math&gt;a+d&gt;b+c&lt;/math&gt; is equal to the number of quadruples where &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when &lt;math&gt;a-d&lt;/math&gt; ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where &lt;math&gt;a+d=b+c&lt;/math&gt;. <br /> <br /> So the answer will be &lt;math&gt;\frac{210-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> <br /> <br /> == Solution 6 ==<br /> <br /> <br /> Let &lt;math&gt;b = a+x&lt;/math&gt; and &lt;math&gt;c = a+x+y&lt;/math&gt; and &lt;math&gt;d = a+x+y+z&lt;/math&gt; for positive integers &lt;math&gt;x,y,z.&lt;/math&gt; In order to satisfy the other condition we need &lt;math&gt;z &gt; x&lt;/math&gt; so we let &lt;math&gt;z = x+k.&lt;/math&gt; Now the only other condition we need to satisfy so &lt;math&gt;a+2x+y+k \le 10.&lt;/math&gt; This condition can be transformed into &lt;math&gt;a+2x+y+k+b = 11&lt;/math&gt; for positive &lt;math&gt;a,x,y,k,b.&lt;/math&gt; Now we use generating functions to finish. We find the generating function of the whole expression is &lt;math&gt;(x + x^2 + \cdots)^4 \cdot (x^2+x^4 + \cdots)&lt;/math&gt; and we are looking for the &lt;math&gt;x^{11}&lt;/math&gt; coefficient. This simplifies to finding the &lt;math&gt;x^5&lt;/math&gt; coefficient of &lt;math&gt;(1+x+\cdots)^4 \cdot (1+x^2+\cdots) =\frac{1}{(1-x)^4} \cdot\frac{1}{1+x}.&lt;/math&gt; Now this expression simplifies to &lt;cmath&gt;\left(\binom{4}{4}+\binom{5}{4} + \cdots +\binom{4+k}{4}x^k\right)(1-x+x^2-x^3 \cdots).&lt;/cmath&gt; The &lt;math&gt;x^5&lt;/math&gt; coefficient ends up to be &lt;math&gt;\binom{9}{4} -\binom{8}{4} +\binom{7}{4} -\binom{6}{4} +\binom{5}{4} -\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = \boxed{080}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=133393 2011 AIME II Problems/Problem 6 2020-09-09T19:55:59Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution 1==<br /> Rearranging the [[inequality]] we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{080}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding &lt;math&gt;\binom83 = 56&lt;/math&gt; ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding &lt;math&gt;\binom 63 = 20&lt;/math&gt; ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding &lt;math&gt;\binom43 = 4&lt;/math&gt; ways and there cannot be 3 holders between a and b so our total is 56+20+4=&lt;math&gt;\fbox{080}&lt;/math&gt;.<br /> <br /> ==Solution 3 (&lt;i&gt;Slightly&lt;/i&gt; bashy)==<br /> We first start out when the value of &lt;math&gt;a=1&lt;/math&gt;. <br /> <br /> Doing casework, we discover that &lt;math&gt;d=5,6,7,8,9,10&lt;/math&gt;. We quickly find a pattern.<br /> <br /> Now, doing this for the rest of the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, we see that the answer is simply:<br /> <br /> <br /> &lt;math&gt;(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)&lt;/math&gt;<br /> &lt;math&gt;+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}&lt;/math&gt;<br /> <br /> ==Solution 4 (quick)==<br /> <br /> Notice that if &lt;math&gt;a+d&gt;b+c&lt;/math&gt;, then &lt;math&gt;(11-a)+(11-d)&lt;(11-b)+(11-c)&lt;/math&gt;, so there is a 1-to-1 correspondence between the number of ordered quadruples with &lt;math&gt;a+d&gt;b+c&lt;/math&gt; and the number of ordered quadruples with &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. <br /> <br /> Quick counting gives that the number of ordered quadruples with &lt;math&gt;a+d=b+c&lt;/math&gt; is 50. To count this, consider our numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10&lt;/math&gt;. Notice that if, for example, &lt;math&gt;a+d=b+c=8&lt;/math&gt;, that the average of &lt;math&gt;a,d&lt;/math&gt; and &lt;math&gt;b,c&lt;/math&gt; must both be &lt;math&gt;4&lt;/math&gt;. In this way, there is a symmetry for this case, centered at &lt;math&gt;4&lt;/math&gt;. If instead, say, &lt;math&gt;a+d=b+c=7&lt;/math&gt;, an odd number, then there is symmetry with &lt;math&gt;(a,d);(b,c)&lt;/math&gt; about &lt;math&gt;3.5&lt;/math&gt;. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at &lt;math&gt;2.5,3,8,8.5&lt;/math&gt;, there is &lt;math&gt;1&lt;/math&gt; case for each and so on, until centered at &lt;math&gt;5.5&lt;/math&gt;, there are &lt;math&gt;10&lt;/math&gt; possible cases. Adding these all, we have &lt;math&gt;2(1+3+6)+10=50&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;\frac{\binom{10}{4}-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is &lt;math&gt;(10*9*8*7/24)=210&lt;/math&gt; <br /> <br /> Next, intuitively, the number of quadruples where &lt;math&gt;a+d&gt;b+c&lt;/math&gt; is equal to the number of quadruples where &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when &lt;math&gt;a-d&lt;/math&gt; ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where &lt;math&gt;a+d=b+c&lt;/math&gt;. <br /> <br /> So the answer will be &lt;math&gt;\frac{210-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> <br /> <br /> == Solution 6 ==<br /> <br /> <br /> Let &lt;math&gt;b = a+x&lt;/math&gt; and &lt;math&gt;c = a+x+y&lt;/math&gt; and &lt;math&gt;d = a+x+y+z&lt;/math&gt; for positive integers &lt;math&gt;x,y,z.&lt;/math&gt; In order to satisfy the other condition we need &lt;math&gt;z &gt; x&lt;/math&gt; so we let &lt;math&gt;z = x+k.&lt;/math&gt; Now the only other condition we need to satisfy so &lt;math&gt;a+2x+y+k \le 10.&lt;/math&gt; This condition can be transformed into &lt;math&gt;a+2x+y+k+b = 11&lt;/math&gt; for positive &lt;math&gt;a,x,y,k,b.&lt;/math&gt; Now we use generating functions to finish. We find the generating function of the whole expression is &lt;math&gt;(x + x^2 + \cdots)^4 \cdot (x^2+x^4 + \cdots)&lt;/math&gt; and we are looking for the &lt;math&gt;x^{11}&lt;/math&gt; coefficient. This simplifies to finding the &lt;math&gt;x^5&lt;/math&gt; coefficient of &lt;math&gt;(1+x+\cdots)^4 \cdot (1+x^2+\cdots) = \displaystyle\frac{1}{(1-x)^4} \cdot \displaystyle\frac{1}{1+x}.&lt;/math&gt; Now this expression simplifies to &lt;math&gt;&lt;/math&gt;\left(\displaystyle\binom{4}{4}+\displaystyle\binom{5}{4} + \cdots + \displaystyle\binom{4+k}{4}x^k\right)(1-x+x^2-x^3 \cdots).&lt;math&gt; The &lt;/math&gt;x^5&lt;math&gt; coefficient ends up to be &lt;/math&gt;\displaystyle\binom{9}{4} - \displaystyle\binom{8}{4} + \displaystyle\binom{7}{4} - \displaystyle\binom{6}{4} + \displaystyle\binom{5}{4} -\displaystyle\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = \boxed{080}.$<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_6&diff=133390 2007 AIME I Problems/Problem 6 2020-09-09T15:40:24Z <p>Jerry122805: /* Solution 5 (Recursionish) */</p> <hr /> <div>== Problem ==<br /> A frog is placed at the [[origin]] on the [[number line]], and moves according to the following rule: in a given move, the frog advances to either the closest [[point]] with a greater [[integer]] [[coordinate]] that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A ''move sequence'' is a [[sequence]] of coordinates which correspond to valid moves, beginning with 0, and ending with 39. For example, &lt;math&gt;0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39&lt;/math&gt; is a move sequence. How many move sequences are possible for the frog?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let us keep a careful tree of the possible number of paths around every multiple of &lt;math&gt;13&lt;/math&gt;.<br /> <br /> From &lt;math&gt;0 \Rightarrow 13&lt;/math&gt;, we can end at either &lt;math&gt;12&lt;/math&gt; (mult. of 3) or &lt;math&gt;13&lt;/math&gt; (mult. of 13). <br /> <br /> *Only &lt;math&gt;1&lt;/math&gt; path leads to &lt;math&gt;12&lt;/math&gt;<br /> **Continuing from &lt;math&gt;12&lt;/math&gt;, there is &lt;math&gt;1 \cdot 1 = 1&lt;/math&gt; way to continue to &lt;math&gt;24&lt;/math&gt;<br /> **There are &lt;math&gt;1 \cdot \left(\frac{24-15}{3} + 1\right) = 4&lt;/math&gt; ways to reach &lt;math&gt;26&lt;/math&gt;.<br /> *There are &lt;math&gt;\frac{12 - 0}{3} + 1 = 5&lt;/math&gt; ways to reach &lt;math&gt;13&lt;/math&gt;. <br /> ** Continuing from &lt;math&gt;13&lt;/math&gt;, there are &lt;math&gt;5 \cdot 1 = 5&lt;/math&gt; ways to get to &lt;math&gt;24&lt;/math&gt;<br /> **There are &lt;math&gt;5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25&lt;/math&gt; ways (the first 1 to make it inclusive, the second to also jump from &lt;math&gt;13 \Rightarrow 26&lt;/math&gt;) to get to &lt;math&gt;26&lt;/math&gt;. <br /> <br /> Regrouping, work from &lt;math&gt;24 | 26\Rightarrow 39&lt;/math&gt;<br /> *There are &lt;math&gt;1 + 5 = 6&lt;/math&gt; ways to get to &lt;math&gt;24&lt;/math&gt; <br /> ** Continuing from &lt;math&gt;24&lt;/math&gt;, there are &lt;math&gt;6 \cdot \left(\frac{39 - 27}{3}\right) = 24&lt;/math&gt; ways to continue to &lt;math&gt;39&lt;/math&gt;.<br /> *There are &lt;math&gt;4 + 25 = 29&lt;/math&gt; ways to reach &lt;math&gt;26&lt;/math&gt;. <br /> ** Continuing from &lt;math&gt;26&lt;/math&gt;, there are &lt;math&gt;29 \cdot \left(\frac{39-27}{3} + 1\right) = 145&lt;/math&gt; (note that the 1 is not to inclusive, but to count &lt;math&gt;26 \Rightarrow 39&lt;/math&gt;). <br /> <br /> In total, we get &lt;math&gt;145 + 24 = 169&lt;/math&gt;.<br /> <br /> &lt;br&gt;<br /> <br /> In summary, we can draw the following tree, where in &lt;math&gt;(x,y)&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; represents the current position on the number line, and &lt;math&gt;y&lt;/math&gt; represents the number of paths to get there:<br /> <br /> {| class=&quot;wikitable&quot;<br /> |-<br /> | width=&quot;50%&quot; |<br /> *&lt;math&gt;(12,1)&lt;/math&gt;<br /> **&lt;math&gt;(24,1)&lt;/math&gt;<br /> ***&lt;math&gt;(39,4)&lt;/math&gt;<br /> **&lt;math&gt;(26,4)&lt;/math&gt;<br /> ***&lt;math&gt;(39,20)&lt;/math&gt;<br /> |<br /> *&lt;math&gt;(13,5)&lt;/math&gt;<br /> **&lt;math&gt;(24,5)&lt;/math&gt;<br /> ***&lt;math&gt;(39,20)&lt;/math&gt;<br /> **&lt;math&gt;(26,25)&lt;/math&gt;<br /> ***&lt;math&gt;(39,125)&lt;/math&gt;<br /> |}<br /> <br /> Again, this totals &lt;math&gt;4 + 20 + 20 + 125 = 169&lt;/math&gt;.<br /> <br /> === Solution 2 === <br /> We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else.<br /> <br /> For the first stage the possible paths are &lt;math&gt;(0,13)&lt;/math&gt;, &lt;math&gt;(0,3,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,9,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,9,12,13)&lt;/math&gt;, and &lt;math&gt;(0,3,6,9,12)&lt;/math&gt;. That is a total of 6.<br /> <br /> For the second stage the possible paths are &lt;math&gt;(26)&lt;/math&gt;, &lt;math&gt;(15,26)&lt;/math&gt;, &lt;math&gt;(15,18,26)&lt;/math&gt;, &lt;math&gt;(15,18,21,26)&lt;/math&gt;, &lt;math&gt;(15,18,21,24,26)&lt;/math&gt;, and &lt;math&gt;(15,18,21,24)&lt;/math&gt;. That is a total of 6.<br /> <br /> For the third stage the possible paths are &lt;math&gt;(39)&lt;/math&gt;, &lt;math&gt;(27,39)&lt;/math&gt;, &lt;math&gt;(27,30,39)&lt;/math&gt;, &lt;math&gt;(27,30,33,39)&lt;/math&gt;, and &lt;math&gt;(27,30,33,36,39)&lt;/math&gt;. That is a total of 5.<br /> <br /> However, we cannot jump from &lt;math&gt;12 \Rightarrow 26&lt;/math&gt; (this eliminates 5 paths) or &lt;math&gt;24 \Rightarrow 39&lt;/math&gt; (this eliminates 6 paths), so we must subtract &lt;math&gt;6 + 5 = 11&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;6*6*5 - 11=169&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Another way would be to use a table representing the number of ways to reach a certain number<br /> <br /> &lt;math&gt;\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}<br /> 0 &amp; 3 &amp; 6 &amp; 9 &amp; 12 &amp; 13 &amp; 15 &amp; 18 &amp; 21 &amp; 24 &amp; 26 &amp; 27 &amp; 30 &amp; 33 &amp; 36 \\ <br /> \hline<br /> 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 5 &amp; 6 &amp; 6 &amp; 6 &amp; 6 &amp; 29 &amp; 35 &amp; 35 &amp; 35 &amp; 35 \\<br /> \end{tabular}&lt;/math&gt;<br /> <br /> How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for &lt;math&gt;26&lt;/math&gt;, we can reach it from &lt;math&gt;13, 15, 18, 21, 24&lt;/math&gt;, so we add all those values to get the value for &lt;math&gt;26&lt;/math&gt;. For &lt;math&gt;27&lt;/math&gt;, it is only reachable from &lt;math&gt;24&lt;/math&gt; or &lt;math&gt;26&lt;/math&gt;, so we have &lt;math&gt;29 + 6 = 35&lt;/math&gt;.<br /> <br /> The answer for &lt;math&gt;39&lt;/math&gt; can be computed in a similar way to get &lt;math&gt;35 * 4 + 29 = \boxed{169}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> I believe this is an easier way of organizing the solution to reduce the possibility of mistakes.<br /> This is a highly visual solution, so it's much easier to record than a tree or table. <br /> <br /> Here is my diagram to help you see what I did: https://drive.google.com/file/d/1Gk_cziYvoeg--uVTap5FGOds3SEfiDAW/view?usp=sharing<br /> <br /> <br /> Using graph paper, draw a number line from 0-39.<br /> On one line, dot every multiple of 3.<br /> Then on a line below it, dot every multiple of 13.<br /> This way you can clearly see which goes before or after which.<br /> <br /> <br /> To make it easier to understand, I'll compare these jumps to a train system. Imagine that every multiple of 3 is a short transit, while the multiples of 13 are long transits; because of the possibility to skip a large section in one move. As we continue, picture each &quot;transit&quot; of 13 to be an option, like a switch. If you are the frog that are riding these trains, you would probably think like this: &quot;I could use the first long transit, skip the second, and use the third. Or, I could skip both first and second and use only the third. etc.) From now on I'll be calling the multiples of 13: 13, 26, and 39, as stations 1, 2, and 3 for clarity.<br /> <br /> Thinking like this organizes the problem efficiently into &lt;math&gt;2^3&lt;/math&gt; cases, where you could choose which &quot;long transits&quot; to ride. We will start with the harder cases and move downwards. Write out each of the 8 cases to record each one. Once we do the hardest case, which is the first one, every preceding one will be easier with the knowledge we gather.<br /> <br /> <br /> Case #1: 111<br /> <br /> You have 5 locations to choose to jump to station 1. From the start (0), 3, 6, 9, or 12. The same goes for station #2, with 13, 15, 18, 21, or 24. However, station #3 is tricky. You can jump from 26, 27, 30, or 33, but if you jumped from 36, then that could qualify as skipping jumping station #3, because station #3 is both a multiple of 3 and a multiple of 13. For example, if you jumped from 36 as your last move, then those moves are essentially the same as case 110. <br /> <br /> So, &lt;math&gt;5*5*4=100&lt;/math&gt;. <br /> <br /> <br /> Case #2: 110 <br /> <br /> Start with what you did on case #1, but when you reach station 2 continue only jumping multiples of 3 to 39.<br /> <br /> &lt;math&gt;5*5=25&lt;/math&gt;<br /> <br /> <br /> Case #3: 101<br /> <br /> We have the 5 original choices, and when you reach station #1, move to 27 on the number line, slightly ahead of station #2. Here is another tricky part. Since we did not start on station 2 like we did in case #1, there are only 3 cases instead of 4.<br /> <br /> &lt;math&gt;5*3=15&lt;/math&gt;<br /> <br /> <br /> Case #4: 100<br /> <br /> This one is simple.<br /> <br /> &lt;math&gt;5&lt;/math&gt;<br /> <br /> <br /> Case #5: 011<br /> <br /> Jump multiples of 3 to get to 15, where you will have 4 locations to jump to station 2, and another 4 choices to reach station 3.<br /> <br /> &lt;math&gt;4*4=16&lt;/math&gt;<br /> <br /> <br /> Case #6: 010<br /> <br /> This one is also simple. Once you reach 15 there are only 4 choices.<br /> <br /> &lt;math&gt;4&lt;/math&gt;<br /> <br /> <br /> Case #7: 001<br /> <br /> It only gets simpler. You can only jump at 27, 30, or 33.<br /> <br /> &lt;math&gt;3&lt;/math&gt;<br /> <br /> <br /> Case #8: 000<br /> <br /> Simple jumps.<br /> <br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> <br /> You have reached the end! &lt;math&gt;100+25+15+5+16+4+3+1=\boxed{169}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> <br /> == Solution 5 (Recursionish) ==<br /> <br /> Let &lt;math&gt;f(n)&lt;/math&gt; be the number of ways one can get to &lt;math&gt;39&lt;/math&gt; starting at position &lt;math&gt;n.&lt;/math&gt; We wish to compute &lt;math&gt;f(0).&lt;/math&gt; Now it's just a long simplifications until you get to &lt;math&gt;f(36) = 1.&lt;/math&gt; We have &lt;cmath&gt;f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).&lt;/cmath&gt;<br /> <br /> Most of these steps are valid since at any &lt;math&gt;n&lt;/math&gt; that is a multiple of &lt;math&gt;3&lt;/math&gt; we can either go to the next multiple of &lt;math&gt;3&lt;/math&gt; or we can skip to the next multiple of &lt;math&gt;13&lt;/math&gt; which is simply &lt;math&gt;13.&lt;/math&gt; <br /> <br /> From these equations we have deduced &lt;math&gt;f(0) = f(15) + 5f(12).&lt;/math&gt; Continuing we have &lt;cmath&gt;f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).&lt;/cmath&gt;<br /> <br /> Finally, note that &lt;math&gt;f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5&lt;/math&gt; since at any point we can either go to the next multiple of &lt;math&gt;3&lt;/math&gt; or go to the next multiple of &lt;math&gt;13&lt;/math&gt; which happens to be &lt;math&gt;39.&lt;/math&gt; Therefore &lt;math&gt;f(26) = 5.&lt;/math&gt; Similarly we find &lt;math&gt;f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4&lt;/math&gt; so the end answer is &lt;math&gt;5 \cdot 29 + 6 \cdot 4 = \boxed{169}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_6&diff=133389 2007 AIME I Problems/Problem 6 2020-09-09T15:40:07Z <p>Jerry122805: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A frog is placed at the [[origin]] on the [[number line]], and moves according to the following rule: in a given move, the frog advances to either the closest [[point]] with a greater [[integer]] [[coordinate]] that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A ''move sequence'' is a [[sequence]] of coordinates which correspond to valid moves, beginning with 0, and ending with 39. For example, &lt;math&gt;0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39&lt;/math&gt; is a move sequence. How many move sequences are possible for the frog?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let us keep a careful tree of the possible number of paths around every multiple of &lt;math&gt;13&lt;/math&gt;.<br /> <br /> From &lt;math&gt;0 \Rightarrow 13&lt;/math&gt;, we can end at either &lt;math&gt;12&lt;/math&gt; (mult. of 3) or &lt;math&gt;13&lt;/math&gt; (mult. of 13). <br /> <br /> *Only &lt;math&gt;1&lt;/math&gt; path leads to &lt;math&gt;12&lt;/math&gt;<br /> **Continuing from &lt;math&gt;12&lt;/math&gt;, there is &lt;math&gt;1 \cdot 1 = 1&lt;/math&gt; way to continue to &lt;math&gt;24&lt;/math&gt;<br /> **There are &lt;math&gt;1 \cdot \left(\frac{24-15}{3} + 1\right) = 4&lt;/math&gt; ways to reach &lt;math&gt;26&lt;/math&gt;.<br /> *There are &lt;math&gt;\frac{12 - 0}{3} + 1 = 5&lt;/math&gt; ways to reach &lt;math&gt;13&lt;/math&gt;. <br /> ** Continuing from &lt;math&gt;13&lt;/math&gt;, there are &lt;math&gt;5 \cdot 1 = 5&lt;/math&gt; ways to get to &lt;math&gt;24&lt;/math&gt;<br /> **There are &lt;math&gt;5 \cdot \left(\frac{24-15}{3} + 1 + 1\right) = 25&lt;/math&gt; ways (the first 1 to make it inclusive, the second to also jump from &lt;math&gt;13 \Rightarrow 26&lt;/math&gt;) to get to &lt;math&gt;26&lt;/math&gt;. <br /> <br /> Regrouping, work from &lt;math&gt;24 | 26\Rightarrow 39&lt;/math&gt;<br /> *There are &lt;math&gt;1 + 5 = 6&lt;/math&gt; ways to get to &lt;math&gt;24&lt;/math&gt; <br /> ** Continuing from &lt;math&gt;24&lt;/math&gt;, there are &lt;math&gt;6 \cdot \left(\frac{39 - 27}{3}\right) = 24&lt;/math&gt; ways to continue to &lt;math&gt;39&lt;/math&gt;.<br /> *There are &lt;math&gt;4 + 25 = 29&lt;/math&gt; ways to reach &lt;math&gt;26&lt;/math&gt;. <br /> ** Continuing from &lt;math&gt;26&lt;/math&gt;, there are &lt;math&gt;29 \cdot \left(\frac{39-27}{3} + 1\right) = 145&lt;/math&gt; (note that the 1 is not to inclusive, but to count &lt;math&gt;26 \Rightarrow 39&lt;/math&gt;). <br /> <br /> In total, we get &lt;math&gt;145 + 24 = 169&lt;/math&gt;.<br /> <br /> &lt;br&gt;<br /> <br /> In summary, we can draw the following tree, where in &lt;math&gt;(x,y)&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; represents the current position on the number line, and &lt;math&gt;y&lt;/math&gt; represents the number of paths to get there:<br /> <br /> {| class=&quot;wikitable&quot;<br /> |-<br /> | width=&quot;50%&quot; |<br /> *&lt;math&gt;(12,1)&lt;/math&gt;<br /> **&lt;math&gt;(24,1)&lt;/math&gt;<br /> ***&lt;math&gt;(39,4)&lt;/math&gt;<br /> **&lt;math&gt;(26,4)&lt;/math&gt;<br /> ***&lt;math&gt;(39,20)&lt;/math&gt;<br /> |<br /> *&lt;math&gt;(13,5)&lt;/math&gt;<br /> **&lt;math&gt;(24,5)&lt;/math&gt;<br /> ***&lt;math&gt;(39,20)&lt;/math&gt;<br /> **&lt;math&gt;(26,25)&lt;/math&gt;<br /> ***&lt;math&gt;(39,125)&lt;/math&gt;<br /> |}<br /> <br /> Again, this totals &lt;math&gt;4 + 20 + 20 + 125 = 169&lt;/math&gt;.<br /> <br /> === Solution 2 === <br /> We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else.<br /> <br /> For the first stage the possible paths are &lt;math&gt;(0,13)&lt;/math&gt;, &lt;math&gt;(0,3,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,9,13)&lt;/math&gt;, &lt;math&gt;(0,3,6,9,12,13)&lt;/math&gt;, and &lt;math&gt;(0,3,6,9,12)&lt;/math&gt;. That is a total of 6.<br /> <br /> For the second stage the possible paths are &lt;math&gt;(26)&lt;/math&gt;, &lt;math&gt;(15,26)&lt;/math&gt;, &lt;math&gt;(15,18,26)&lt;/math&gt;, &lt;math&gt;(15,18,21,26)&lt;/math&gt;, &lt;math&gt;(15,18,21,24,26)&lt;/math&gt;, and &lt;math&gt;(15,18,21,24)&lt;/math&gt;. That is a total of 6.<br /> <br /> For the third stage the possible paths are &lt;math&gt;(39)&lt;/math&gt;, &lt;math&gt;(27,39)&lt;/math&gt;, &lt;math&gt;(27,30,39)&lt;/math&gt;, &lt;math&gt;(27,30,33,39)&lt;/math&gt;, and &lt;math&gt;(27,30,33,36,39)&lt;/math&gt;. That is a total of 5.<br /> <br /> However, we cannot jump from &lt;math&gt;12 \Rightarrow 26&lt;/math&gt; (this eliminates 5 paths) or &lt;math&gt;24 \Rightarrow 39&lt;/math&gt; (this eliminates 6 paths), so we must subtract &lt;math&gt;6 + 5 = 11&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;6*6*5 - 11=169&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Another way would be to use a table representing the number of ways to reach a certain number<br /> <br /> &lt;math&gt;\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}<br /> 0 &amp; 3 &amp; 6 &amp; 9 &amp; 12 &amp; 13 &amp; 15 &amp; 18 &amp; 21 &amp; 24 &amp; 26 &amp; 27 &amp; 30 &amp; 33 &amp; 36 \\ <br /> \hline<br /> 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 5 &amp; 6 &amp; 6 &amp; 6 &amp; 6 &amp; 29 &amp; 35 &amp; 35 &amp; 35 &amp; 35 \\<br /> \end{tabular}&lt;/math&gt;<br /> <br /> How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for &lt;math&gt;26&lt;/math&gt;, we can reach it from &lt;math&gt;13, 15, 18, 21, 24&lt;/math&gt;, so we add all those values to get the value for &lt;math&gt;26&lt;/math&gt;. For &lt;math&gt;27&lt;/math&gt;, it is only reachable from &lt;math&gt;24&lt;/math&gt; or &lt;math&gt;26&lt;/math&gt;, so we have &lt;math&gt;29 + 6 = 35&lt;/math&gt;.<br /> <br /> The answer for &lt;math&gt;39&lt;/math&gt; can be computed in a similar way to get &lt;math&gt;35 * 4 + 29 = \boxed{169}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> I believe this is an easier way of organizing the solution to reduce the possibility of mistakes.<br /> This is a highly visual solution, so it's much easier to record than a tree or table. <br /> <br /> Here is my diagram to help you see what I did: https://drive.google.com/file/d/1Gk_cziYvoeg--uVTap5FGOds3SEfiDAW/view?usp=sharing<br /> <br /> <br /> Using graph paper, draw a number line from 0-39.<br /> On one line, dot every multiple of 3.<br /> Then on a line below it, dot every multiple of 13.<br /> This way you can clearly see which goes before or after which.<br /> <br /> <br /> To make it easier to understand, I'll compare these jumps to a train system. Imagine that every multiple of 3 is a short transit, while the multiples of 13 are long transits; because of the possibility to skip a large section in one move. As we continue, picture each &quot;transit&quot; of 13 to be an option, like a switch. If you are the frog that are riding these trains, you would probably think like this: &quot;I could use the first long transit, skip the second, and use the third. Or, I could skip both first and second and use only the third. etc.) From now on I'll be calling the multiples of 13: 13, 26, and 39, as stations 1, 2, and 3 for clarity.<br /> <br /> Thinking like this organizes the problem efficiently into &lt;math&gt;2^3&lt;/math&gt; cases, where you could choose which &quot;long transits&quot; to ride. We will start with the harder cases and move downwards. Write out each of the 8 cases to record each one. Once we do the hardest case, which is the first one, every preceding one will be easier with the knowledge we gather.<br /> <br /> <br /> Case #1: 111<br /> <br /> You have 5 locations to choose to jump to station 1. From the start (0), 3, 6, 9, or 12. The same goes for station #2, with 13, 15, 18, 21, or 24. However, station #3 is tricky. You can jump from 26, 27, 30, or 33, but if you jumped from 36, then that could qualify as skipping jumping station #3, because station #3 is both a multiple of 3 and a multiple of 13. For example, if you jumped from 36 as your last move, then those moves are essentially the same as case 110. <br /> <br /> So, &lt;math&gt;5*5*4=100&lt;/math&gt;. <br /> <br /> <br /> Case #2: 110 <br /> <br /> Start with what you did on case #1, but when you reach station 2 continue only jumping multiples of 3 to 39.<br /> <br /> &lt;math&gt;5*5=25&lt;/math&gt;<br /> <br /> <br /> Case #3: 101<br /> <br /> We have the 5 original choices, and when you reach station #1, move to 27 on the number line, slightly ahead of station #2. Here is another tricky part. Since we did not start on station 2 like we did in case #1, there are only 3 cases instead of 4.<br /> <br /> &lt;math&gt;5*3=15&lt;/math&gt;<br /> <br /> <br /> Case #4: 100<br /> <br /> This one is simple.<br /> <br /> &lt;math&gt;5&lt;/math&gt;<br /> <br /> <br /> Case #5: 011<br /> <br /> Jump multiples of 3 to get to 15, where you will have 4 locations to jump to station 2, and another 4 choices to reach station 3.<br /> <br /> &lt;math&gt;4*4=16&lt;/math&gt;<br /> <br /> <br /> Case #6: 010<br /> <br /> This one is also simple. Once you reach 15 there are only 4 choices.<br /> <br /> &lt;math&gt;4&lt;/math&gt;<br /> <br /> <br /> Case #7: 001<br /> <br /> It only gets simpler. You can only jump at 27, 30, or 33.<br /> <br /> &lt;math&gt;3&lt;/math&gt;<br /> <br /> <br /> Case #8: 000<br /> <br /> Simple jumps.<br /> <br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> <br /> You have reached the end! &lt;math&gt;100+25+15+5+16+4+3+1=\boxed{169}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> <br /> == Solution 5 (Recursionish) ==<br /> <br /> Let &lt;math&gt;f(n)&lt;/math&gt; be the number of ways one can get to &lt;math&gt;39&lt;/math&gt; starting at position &lt;math&gt;n.&lt;/math&gt; We wish to compute &lt;math&gt;f(0).&lt;/math&gt; Now it's just a long simplifications until you get to &lt;math&gt;f(36) = 1.&lt;/math&gt; We have &lt;math&gt;&lt;/math&gt;f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).&lt;math&gt; &lt;/math&gt;<br /> <br /> Most of these steps are valid since at any &lt;math&gt;n&lt;/math&gt; that is a multiple of &lt;math&gt;3&lt;/math&gt; we can either go to the next multiple of &lt;math&gt;3&lt;/math&gt; or we can skip to the next multiple of &lt;math&gt;13&lt;/math&gt; which is simply &lt;math&gt;13.&lt;/math&gt; <br /> <br /> From these equations we have deduced &lt;math&gt;f(0) = f(15) + 5f(12).&lt;/math&gt; Continuing we have &lt;cmath&gt;f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).&lt;/cmath&gt;<br /> <br /> Finally, note that &lt;math&gt;f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5&lt;/math&gt; since at any point we can either go to the next multiple of &lt;math&gt;3&lt;/math&gt; or go to the next multiple of &lt;math&gt;13&lt;/math&gt; which happens to be &lt;math&gt;39.&lt;/math&gt; Therefore &lt;math&gt;f(26) = 5.&lt;/math&gt; Similarly we find &lt;math&gt;f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4&lt;/math&gt; so the end answer is &lt;math&gt;5 \cdot 29 + 6 \cdot 4 = \boxed{169}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=133359 2005 AIME II Problems/Problem 11 2020-09-08T14:52:22Z <p>Jerry122805: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> <br /> Note: In order for &lt;math&gt;a_{m} = 0&lt;/math&gt; we need &lt;math&gt;a_{m-1}a_{m}=3&lt;/math&gt; simply by the recursion definition.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=JfxNr7lv7iQ<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=133357 1992 AIME Problems/Problem 8 2020-09-08T14:30:06Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> For any sequence of real numbers &lt;math&gt;A=(a_1,a_2,a_3,\ldots)&lt;/math&gt;, define &lt;math&gt;\Delta A^{}_{}&lt;/math&gt; to be the sequence &lt;math&gt;(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)&lt;/math&gt;, whose &lt;math&gt;n^{\mbox{th}}_{}&lt;/math&gt; term is &lt;math&gt;a_{n+1}-a_n^{}&lt;/math&gt;. Suppose that all of the terms of the sequence &lt;math&gt;\Delta(\Delta A^{}_{})&lt;/math&gt; are &lt;math&gt;1^{}_{}&lt;/math&gt;, and that &lt;math&gt;a_{19}=a_{92}^{}=0&lt;/math&gt;. Find &lt;math&gt;a_1^{}&lt;/math&gt;.<br /> <br /> == Solution 1 (uses calculus)==<br /> Note that the &lt;math&gt;\Delta&lt;/math&gt;s are reminiscent of differentiation; from the condition &lt;math&gt;\Delta(\Delta{A}) = 1&lt;/math&gt;, we are led to consider the differential equation<br /> &lt;cmath&gt; \frac{d^2 A}{dn^2} = 1 &lt;/cmath&gt;<br /> This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br /> &lt;cmath&gt; a_{n} = \frac{1}{2}(n-19)(n-92) &lt;/cmath&gt;<br /> as we must have roots at &lt;math&gt;n = 19&lt;/math&gt; and &lt;math&gt;n = 92&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;\Delta^1 A=\Delta A&lt;/math&gt;, and &lt;math&gt;\Delta^n A=\Delta(\Delta^{(n-1)}A)&lt;/math&gt;.<br /> <br /> Note that in every sequence of &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...&lt;/math&gt;<br /> <br /> Then &lt;math&gt;a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\Delta a_1 =a_2 -a_1&lt;/math&gt;, &lt;math&gt;a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The sequence &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; is the second finite difference sequence, and the first &lt;math&gt;k-1&lt;/math&gt; terms of this sequence can be computed in terms of the original sequence as shown below.<br /> <br /> &lt;math&gt;\begin{array}{rcl}<br /> a_3+a_1-2a_2&amp;=&amp;1\\<br /> a_4+a_2-2a_3&amp;=&amp;1\\<br /> &amp;\vdots\\<br /> a_k + a_{k-2} - 2a_{k-1} &amp;= &amp;1\\<br /> a_{k+1} + a_{k-1} - 2a_k &amp;=&amp; 1.\\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> Adding the above &lt;math&gt;k-1&lt;/math&gt; equations we find that<br /> <br /> &lt;cmath&gt;(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}&lt;/cmath&gt;<br /> <br /> We can sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;18&lt;/math&gt;, finding<br /> &lt;cmath&gt;18(a_1-a_2) - a_1 = 153.\tag{2} &lt;/cmath&gt;<br /> <br /> We can also sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;91&lt;/math&gt;, finding<br /> &lt;cmath&gt;91(a_1-a_2) - a_1 = 4095.\tag{3}&lt;/cmath&gt;<br /> Finally, &lt;math&gt;18\cdot (3) - 91\cdot(2)&lt;/math&gt; gives &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> Kris17<br /> <br /> == Solution 4 ==<br /> Since all terms of &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; are 1, we know that &lt;math&gt;\Delta A&lt;/math&gt; looks like &lt;math&gt;(k,k+1,k+2,...)&lt;/math&gt; for some &lt;math&gt;k&lt;/math&gt;. This means &lt;math&gt;A&lt;/math&gt; looks like &lt;math&gt;(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)&lt;/math&gt;. More specifically, &lt;math&gt;A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a_{19}=a_{92}=0&lt;/math&gt;, we have the following linear system: &lt;cmath&gt;a_1+91k=-4095&lt;/cmath&gt; &lt;cmath&gt;a_1+18k=-153&lt;/cmath&gt; From this, we can easily find that &lt;math&gt;k=-54&lt;/math&gt; and &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> Solution by Zeroman<br /> <br /> <br /> <br /> == Solution 5 ==<br /> <br /> Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that &lt;math&gt;f(19) = f(92) = 0&lt;/math&gt; so the quadratic is &lt;math&gt;f(x) = a(x-19)(x-92)&lt;/math&gt; for some constant &lt;math&gt;a.&lt;/math&gt; Now we use the conditions that the finite difference is &lt;math&gt;1&lt;/math&gt; to find &lt;math&gt;a.&lt;/math&gt; We know &lt;math&gt;f(19) = 0&lt;/math&gt; and &lt;math&gt;f(20) = -72a&lt;/math&gt; and &lt;math&gt;f(18) = 74a.&lt;/math&gt; Therefore applying finite differences once yields the sequence &lt;math&gt;-74a,-72a&lt;/math&gt; and then applying finite differences one more time yields &lt;math&gt;2a&lt;/math&gt; so &lt;math&gt;a =\frac{1}{2}.&lt;/math&gt; Therefore &lt;math&gt;f(1) = 9 \cdot 91 = \boxed{819}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1992|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_8&diff=133356 1992 AIME Problems/Problem 8 2020-09-08T14:29:53Z <p>Jerry122805: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> For any sequence of real numbers &lt;math&gt;A=(a_1,a_2,a_3,\ldots)&lt;/math&gt;, define &lt;math&gt;\Delta A^{}_{}&lt;/math&gt; to be the sequence &lt;math&gt;(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)&lt;/math&gt;, whose &lt;math&gt;n^{\mbox{th}}_{}&lt;/math&gt; term is &lt;math&gt;a_{n+1}-a_n^{}&lt;/math&gt;. Suppose that all of the terms of the sequence &lt;math&gt;\Delta(\Delta A^{}_{})&lt;/math&gt; are &lt;math&gt;1^{}_{}&lt;/math&gt;, and that &lt;math&gt;a_{19}=a_{92}^{}=0&lt;/math&gt;. Find &lt;math&gt;a_1^{}&lt;/math&gt;.<br /> <br /> == Solution 1 (uses calculus)==<br /> Note that the &lt;math&gt;\Delta&lt;/math&gt;s are reminiscent of differentiation; from the condition &lt;math&gt;\Delta(\Delta{A}) = 1&lt;/math&gt;, we are led to consider the differential equation<br /> &lt;cmath&gt; \frac{d^2 A}{dn^2} = 1 &lt;/cmath&gt;<br /> This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; <br /> &lt;cmath&gt; a_{n} = \frac{1}{2}(n-19)(n-92) &lt;/cmath&gt;<br /> as we must have roots at &lt;math&gt;n = 19&lt;/math&gt; and &lt;math&gt;n = 92&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;\Delta^1 A=\Delta A&lt;/math&gt;, and &lt;math&gt;\Delta^n A=\Delta(\Delta^{(n-1)}A)&lt;/math&gt;.<br /> <br /> Note that in every sequence of &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...&lt;/math&gt;<br /> <br /> Then &lt;math&gt;a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\Delta a_1 =a_2 -a_1&lt;/math&gt;, &lt;math&gt;a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153&lt;/math&gt;<br /> <br /> &lt;math&gt;a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The sequence &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; is the second finite difference sequence, and the first &lt;math&gt;k-1&lt;/math&gt; terms of this sequence can be computed in terms of the original sequence as shown below.<br /> <br /> &lt;math&gt;\begin{array}{rcl}<br /> a_3+a_1-2a_2&amp;=&amp;1\\<br /> a_4+a_2-2a_3&amp;=&amp;1\\<br /> &amp;\vdots\\<br /> a_k + a_{k-2} - 2a_{k-1} &amp;= &amp;1\\<br /> a_{k+1} + a_{k-1} - 2a_k &amp;=&amp; 1.\\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> Adding the above &lt;math&gt;k-1&lt;/math&gt; equations we find that<br /> <br /> &lt;cmath&gt;(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}&lt;/cmath&gt;<br /> <br /> We can sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;18&lt;/math&gt;, finding<br /> &lt;cmath&gt;18(a_1-a_2) - a_1 = 153.\tag{2} &lt;/cmath&gt;<br /> <br /> We can also sum equation &lt;math&gt;(1)&lt;/math&gt; from &lt;math&gt;k=1&lt;/math&gt; to &lt;math&gt;91&lt;/math&gt;, finding<br /> &lt;cmath&gt;91(a_1-a_2) - a_1 = 4095.\tag{3}&lt;/cmath&gt;<br /> Finally, &lt;math&gt;18\cdot (3) - 91\cdot(2)&lt;/math&gt; gives &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> <br /> Kris17<br /> <br /> == Solution 4 ==<br /> Since all terms of &lt;math&gt;\Delta(\Delta A)&lt;/math&gt; are 1, we know that &lt;math&gt;\Delta A&lt;/math&gt; looks like &lt;math&gt;(k,k+1,k+2,...)&lt;/math&gt; for some &lt;math&gt;k&lt;/math&gt;. This means &lt;math&gt;A&lt;/math&gt; looks like &lt;math&gt;(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)&lt;/math&gt;. More specifically, &lt;math&gt;A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a_{19}=a_{92}=0&lt;/math&gt;, we have the following linear system: &lt;cmath&gt;a_1+91k=-4095&lt;/cmath&gt; &lt;cmath&gt;a_1+18k=-153&lt;/cmath&gt; From this, we can easily find that &lt;math&gt;k=-54&lt;/math&gt; and &lt;math&gt;a_1=\boxed{819}&lt;/math&gt;.<br /> Solution by Zeroman<br /> <br /> <br /> <br /> == Solution 5 ==<br /> <br /> Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that &lt;math&gt;f(19) = f(92) = 0&lt;/math&gt; so the quadratic is &lt;math&gt;f(x) = a(x-19)(x-92)&lt;/math&gt; for some constant &lt;math&gt;a.&lt;/math&gt; Now we use the conditions that the finite difference is &lt;math&gt;1&lt;/math&gt; to find &lt;math&gt;a.&lt;/math&gt; We know &lt;math&gt;f(19) = 0&lt;/math&gt; and &lt;math&gt;f(20) = -72a&lt;/math&gt; and &lt;math&gt;f(18) = 74a.&lt;/math&gt; Therefore applying finite differences once yields the sequence &lt;math&gt;-74a,-72a&lt;/math&gt; and then applying finite differences one more time yields &lt;math&gt;2a&lt;/math&gt; so &lt;math&gt;a = \displaystyle\frac{1}{2}.&lt;/math&gt; Therefore &lt;math&gt;f() = 9 \cdot 91 = \boxed{819}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1992|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_2&diff=133355 2012 AIME I Problems/Problem 2 2020-09-08T14:19:50Z <p>Jerry122805: /* Solution 1 */</p> <hr /> <div>== Problem 2 ==<br /> The terms of an arithmetic sequence add to &lt;math&gt;715&lt;/math&gt;. The first term of the sequence is increased by &lt;math&gt;1&lt;/math&gt;, the second term is increased by &lt;math&gt;3&lt;/math&gt;, the third term is increased by &lt;math&gt;5&lt;/math&gt;, and in general, the &lt;math&gt;k&lt;/math&gt;th term is increased by the &lt;math&gt;k&lt;/math&gt;th odd positive integer. The terms of the new sequence add to &lt;math&gt;836&lt;/math&gt;. Find the sum of the first, last, and middle terms of the original sequence.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> If the sum of the original sequence is &lt;math&gt;\sum_{i=1}^{n} a_i&lt;/math&gt; then the sum of the new sequence can be expressed as &lt;math&gt;\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.&lt;/math&gt; Therefore, &lt;math&gt;836 = n^2 + 715 \rightarrow n=11.&lt;/math&gt; Now the middle term of the original sequence is simply the average of all the terms, or &lt;math&gt;\frac{715}{11} = 65,&lt;/math&gt; and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or &lt;math&gt;\boxed{195}.&lt;/math&gt;<br /> <br /> \\<br /> <br /> Alternatively, notice that in the original sequence, &lt;math&gt;11a_1 + 55d = 715&lt;/math&gt;, from which &lt;math&gt;a_1 + 5d = 65&lt;/math&gt;. Since we are tasked to find &lt;math&gt;a_1 + a_6 + a_{11} = 3(a_1 + 5d)&lt;/math&gt;, the desired answer is &lt;math&gt;3 \cdot 65 = \boxed{195}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> After the adding of the odd numbers, the total of the sequence increases by &lt;math&gt;836 - 715 = 121 = 11^2&lt;/math&gt;. Since the sum of the first &lt;math&gt;n&lt;/math&gt; positive odd numbers is &lt;math&gt;n^2&lt;/math&gt;, there must be &lt;math&gt;11&lt;/math&gt; terms in the sequence, so the mean of the sequence is &lt;math&gt;\dfrac{715}{11} = 65&lt;/math&gt;. Since the first, last, and middle terms are centered around the mean, our final answer is &lt;math&gt;65 \cdot 3 = \boxed{195}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2012aimei/298<br /> <br /> ~ dolphin7<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=T8Ox412AkZc<br /> ~Shreyas S<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_2&diff=133354 2012 AIME I Problems/Problem 2 2020-09-08T14:19:33Z <p>Jerry122805: /* Solution 1 */</p> <hr /> <div>== Problem 2 ==<br /> The terms of an arithmetic sequence add to &lt;math&gt;715&lt;/math&gt;. The first term of the sequence is increased by &lt;math&gt;1&lt;/math&gt;, the second term is increased by &lt;math&gt;3&lt;/math&gt;, the third term is increased by &lt;math&gt;5&lt;/math&gt;, and in general, the &lt;math&gt;k&lt;/math&gt;th term is increased by the &lt;math&gt;k&lt;/math&gt;th odd positive integer. The terms of the new sequence add to &lt;math&gt;836&lt;/math&gt;. Find the sum of the first, last, and middle terms of the original sequence.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> If the sum of the original sequence is &lt;math&gt;\sum_{i=1}^{n} a_i&lt;/math&gt; then the sum of the new sequence can be expressed as &lt;math&gt;\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.&lt;/math&gt; Therefore, &lt;math&gt;836 = n^2 + 715 \rightarrow n=11.&lt;/math&gt; Now the middle term of the original sequence is simply the average of all the terms, or &lt;math&gt;\frac{715}{11} = 65,&lt;/math&gt; and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or &lt;math&gt;\boxed{195.}&lt;/math&gt;<br /> <br /> \\<br /> <br /> Alternatively, notice that in the original sequence, &lt;math&gt;11a_1 + 55d = 715&lt;/math&gt;, from which &lt;math&gt;a_1 + 5d = 65&lt;/math&gt;. Since we are tasked to find &lt;math&gt;a_1 + a_6 + a_{11} = 3(a_1 + 5d)&lt;/math&gt;, the desired answer is &lt;math&gt;3 \cdot 65 = \boxed{195}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> After the adding of the odd numbers, the total of the sequence increases by &lt;math&gt;836 - 715 = 121 = 11^2&lt;/math&gt;. Since the sum of the first &lt;math&gt;n&lt;/math&gt; positive odd numbers is &lt;math&gt;n^2&lt;/math&gt;, there must be &lt;math&gt;11&lt;/math&gt; terms in the sequence, so the mean of the sequence is &lt;math&gt;\dfrac{715}{11} = 65&lt;/math&gt;. Since the first, last, and middle terms are centered around the mean, our final answer is &lt;math&gt;65 \cdot 3 = \boxed{195}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2012aimei/298<br /> <br /> ~ dolphin7<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=T8Ox412AkZc<br /> ~Shreyas S<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_13&diff=133335 2009 AIME I Problems/Problem 13 2020-09-07T23:50:26Z <p>Jerry122805: /* Solution 3 (BS Solution) */</p> <hr /> <div>== Problem ==<br /> The terms of the sequence &lt;math&gt;(a_i)&lt;/math&gt; defined by &lt;math&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt; are positive integers. Find the minimum possible value of &lt;math&gt;a_1 + a_2&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> This question is guessable but let's prove our answer<br /> <br /> &lt;cmath&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}(1 + a_{n + 1})= a_n + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009&lt;/cmath&gt;<br /> <br /> <br /> lets put &lt;math&gt;n+1&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009&lt;/cmath&gt;<br /> <br /> <br /> and set them equal now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> let's rewrite it<br /> <br /> <br /> &lt;cmath&gt;(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> Let's make it look nice and let &lt;math&gt;b_n=a_{n + 2}-a_n&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;(b_{n+1})(a_{n + 2}+1)= b_n&lt;/cmath&gt;<br /> <br /> <br /> Since &lt;math&gt;b_n&lt;/math&gt; and &lt;math&gt;b_{n+1}&lt;/math&gt; are integers, we can see &lt;math&gt;b_n&lt;/math&gt; is divisible by &lt;math&gt;b_{n+1}&lt;/math&gt;<br /> <br /> <br /> But we can't have an infinite sequence of proper factors, unless &lt;math&gt;b_n=0&lt;/math&gt;<br /> <br /> <br /> Thus, &lt;math&gt;a_{n + 2}-a_n=0&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}=a_n&lt;/cmath&gt;<br /> <br /> <br /> So now, we know &lt;math&gt;a_3=a_1&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{3} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}+a_{1}a_{2} = a_1 + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}a_{2} = 2009&lt;/cmath&gt;<br /> <br /> <br /> To minimize &lt;math&gt;a_{1}+a_{2}&lt;/math&gt;, we need &lt;math&gt;41&lt;/math&gt; and &lt;math&gt;49&lt;/math&gt;<br /> <br /> <br /> Thus, our answer &lt;math&gt;= 41+49=\boxed {090}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> If &lt;math&gt;a_{n} \ne \frac {2009}{a_{n+1}}&lt;/math&gt;, then either<br /> &lt;cmath&gt;a_{n} = \frac {a_{n}}{1} &lt; \frac {a_{n} + 2009}{1 + a_{n+1}} &lt; \frac {2009}{a_{n+1}}&lt;/cmath&gt;<br /> <br /> or<br /> <br /> &lt;cmath&gt;\frac {2009}{a_{n+1}} &lt; \frac {2009 + a_{n}}{a_{n+1} + 1} &lt; \frac {a_{n}}{1} = a_{n}&lt;/cmath&gt;<br /> <br /> All the integers between &lt;math&gt;a_{n}&lt;/math&gt; and &lt;math&gt;\frac {2009}{a_{n+1}}&lt;/math&gt; would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.<br /> <br /> So &lt;math&gt;a_{n} = \frac {2009}{a_{n+1}}&lt;/math&gt;, which &lt;math&gt;a_{n} \cdot a_{n+1} = 2009&lt;/math&gt;. When &lt;math&gt;n = 1&lt;/math&gt;, &lt;math&gt;a_{1} \cdot a_{2} = 2009&lt;/math&gt;. The smallest sum of two factors which have a product of &lt;math&gt;2009&lt;/math&gt; is &lt;math&gt;41 + 49=\boxed {090}&lt;/math&gt;<br /> <br /> <br /> <br /> === Solution 3 (BS Solution) ===<br /> <br /> Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_{1} &amp;= a \\<br /> a_{2} &amp;= b \\<br /> a_{3} &amp;=\frac{a+2009}{1+b} \\<br /> a_{4} &amp;=\frac{(b+1)(b+2009)}{a+b+2010} \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> The terms get more and more wacky, so we just solve for &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;a_{1}=a_{3}&lt;/math&gt; and &lt;math&gt;a_{2}=a_{4}.&lt;/math&gt;<br /> <br /> Solving we find both equations end up to the equation &lt;math&gt;ab=2009&lt;/math&gt; in which we see to minimize we see that &lt;math&gt;a = 49&lt;/math&gt; and &lt;math&gt;b=41&lt;/math&gt; or vice versa for an answer of &lt;math&gt;\boxed{90}.&lt;/math&gt; This solution is VERY non rigorous and not recommended.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_13&diff=133334 2009 AIME I Problems/Problem 13 2020-09-07T23:49:26Z <p>Jerry122805: /* Solution 3 (BS Solution) */</p> <hr /> <div>== Problem ==<br /> The terms of the sequence &lt;math&gt;(a_i)&lt;/math&gt; defined by &lt;math&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt; are positive integers. Find the minimum possible value of &lt;math&gt;a_1 + a_2&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> This question is guessable but let's prove our answer<br /> <br /> &lt;cmath&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}(1 + a_{n + 1})= a_n + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009&lt;/cmath&gt;<br /> <br /> <br /> lets put &lt;math&gt;n+1&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009&lt;/cmath&gt;<br /> <br /> <br /> and set them equal now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> let's rewrite it<br /> <br /> <br /> &lt;cmath&gt;(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> Let's make it look nice and let &lt;math&gt;b_n=a_{n + 2}-a_n&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;(b_{n+1})(a_{n + 2}+1)= b_n&lt;/cmath&gt;<br /> <br /> <br /> Since &lt;math&gt;b_n&lt;/math&gt; and &lt;math&gt;b_{n+1}&lt;/math&gt; are integers, we can see &lt;math&gt;b_n&lt;/math&gt; is divisible by &lt;math&gt;b_{n+1}&lt;/math&gt;<br /> <br /> <br /> But we can't have an infinite sequence of proper factors, unless &lt;math&gt;b_n=0&lt;/math&gt;<br /> <br /> <br /> Thus, &lt;math&gt;a_{n + 2}-a_n=0&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}=a_n&lt;/cmath&gt;<br /> <br /> <br /> So now, we know &lt;math&gt;a_3=a_1&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{3} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}+a_{1}a_{2} = a_1 + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}a_{2} = 2009&lt;/cmath&gt;<br /> <br /> <br /> To minimize &lt;math&gt;a_{1}+a_{2}&lt;/math&gt;, we need &lt;math&gt;41&lt;/math&gt; and &lt;math&gt;49&lt;/math&gt;<br /> <br /> <br /> Thus, our answer &lt;math&gt;= 41+49=\boxed {090}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> If &lt;math&gt;a_{n} \ne \frac {2009}{a_{n+1}}&lt;/math&gt;, then either<br /> &lt;cmath&gt;a_{n} = \frac {a_{n}}{1} &lt; \frac {a_{n} + 2009}{1 + a_{n+1}} &lt; \frac {2009}{a_{n+1}}&lt;/cmath&gt;<br /> <br /> or<br /> <br /> &lt;cmath&gt;\frac {2009}{a_{n+1}} &lt; \frac {2009 + a_{n}}{a_{n+1} + 1} &lt; \frac {a_{n}}{1} = a_{n}&lt;/cmath&gt;<br /> <br /> All the integers between &lt;math&gt;a_{n}&lt;/math&gt; and &lt;math&gt;\frac {2009}{a_{n+1}}&lt;/math&gt; would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.<br /> <br /> So &lt;math&gt;a_{n} = \frac {2009}{a_{n+1}}&lt;/math&gt;, which &lt;math&gt;a_{n} \cdot a_{n+1} = 2009&lt;/math&gt;. When &lt;math&gt;n = 1&lt;/math&gt;, &lt;math&gt;a_{1} \cdot a_{2} = 2009&lt;/math&gt;. The smallest sum of two factors which have a product of &lt;math&gt;2009&lt;/math&gt; is &lt;math&gt;41 + 49=\boxed {090}&lt;/math&gt;<br /> <br /> <br /> <br /> === Solution 3 (BS Solution) ===<br /> <br /> Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_{1} &amp;= a \\<br /> a_{2} &amp;= b \\<br /> a_{3} &amp;=\frac{a+2009}{1+b} \\<br /> a_{4} &amp;=\frac{(b+1)(b+2009)}{a+b+2010} \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> The terms get more and more wacky, so we just solve for &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;a_{1}=a_{3}&lt;/math&gt; and &lt;math&gt;a_{2}=a_{4}.&lt;/math&gt;<br /> <br /> Solving we find both equations end up to the equation &lt;math&gt;ab=2009&lt;/math&gt; in which we see to minimize we see that &lt;math&gt;a = 49&lt;/math&gt; and &lt;math&gt;b=41.&lt;/math&gt; This solution is VERY non rigorous and not recommended.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_13&diff=133333 2009 AIME I Problems/Problem 13 2020-09-07T23:49:14Z <p>Jerry122805: /* Solution 3 (BS Solution) */</p> <hr /> <div>== Problem ==<br /> The terms of the sequence &lt;math&gt;(a_i)&lt;/math&gt; defined by &lt;math&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt; are positive integers. Find the minimum possible value of &lt;math&gt;a_1 + a_2&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> This question is guessable but let's prove our answer<br /> <br /> &lt;cmath&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}(1 + a_{n + 1})= a_n + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009&lt;/cmath&gt;<br /> <br /> <br /> lets put &lt;math&gt;n+1&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009&lt;/cmath&gt;<br /> <br /> <br /> and set them equal now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> let's rewrite it<br /> <br /> <br /> &lt;cmath&gt;(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> Let's make it look nice and let &lt;math&gt;b_n=a_{n + 2}-a_n&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;(b_{n+1})(a_{n + 2}+1)= b_n&lt;/cmath&gt;<br /> <br /> <br /> Since &lt;math&gt;b_n&lt;/math&gt; and &lt;math&gt;b_{n+1}&lt;/math&gt; are integers, we can see &lt;math&gt;b_n&lt;/math&gt; is divisible by &lt;math&gt;b_{n+1}&lt;/math&gt;<br /> <br /> <br /> But we can't have an infinite sequence of proper factors, unless &lt;math&gt;b_n=0&lt;/math&gt;<br /> <br /> <br /> Thus, &lt;math&gt;a_{n + 2}-a_n=0&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}=a_n&lt;/cmath&gt;<br /> <br /> <br /> So now, we know &lt;math&gt;a_3=a_1&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{3} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}+a_{1}a_{2} = a_1 + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}a_{2} = 2009&lt;/cmath&gt;<br /> <br /> <br /> To minimize &lt;math&gt;a_{1}+a_{2}&lt;/math&gt;, we need &lt;math&gt;41&lt;/math&gt; and &lt;math&gt;49&lt;/math&gt;<br /> <br /> <br /> Thus, our answer &lt;math&gt;= 41+49=\boxed {090}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> If &lt;math&gt;a_{n} \ne \frac {2009}{a_{n+1}}&lt;/math&gt;, then either<br /> &lt;cmath&gt;a_{n} = \frac {a_{n}}{1} &lt; \frac {a_{n} + 2009}{1 + a_{n+1}} &lt; \frac {2009}{a_{n+1}}&lt;/cmath&gt;<br /> <br /> or<br /> <br /> &lt;cmath&gt;\frac {2009}{a_{n+1}} &lt; \frac {2009 + a_{n}}{a_{n+1} + 1} &lt; \frac {a_{n}}{1} = a_{n}&lt;/cmath&gt;<br /> <br /> All the integers between &lt;math&gt;a_{n}&lt;/math&gt; and &lt;math&gt;\frac {2009}{a_{n+1}}&lt;/math&gt; would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.<br /> <br /> So &lt;math&gt;a_{n} = \frac {2009}{a_{n+1}}&lt;/math&gt;, which &lt;math&gt;a_{n} \cdot a_{n+1} = 2009&lt;/math&gt;. When &lt;math&gt;n = 1&lt;/math&gt;, &lt;math&gt;a_{1} \cdot a_{2} = 2009&lt;/math&gt;. The smallest sum of two factors which have a product of &lt;math&gt;2009&lt;/math&gt; is &lt;math&gt;41 + 49=\boxed {090}&lt;/math&gt;<br /> <br /> <br /> <br /> === Solution 3 (BS Solution) ===<br /> <br /> Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms.<br /> &lt;math&gt;<br /> \begin{align*}<br /> a_{1} &amp;= a \\<br /> a_{2} &amp;= b \\<br /> a_{3} &amp;=\frac{a+2009}{1+b} \\<br /> a_{4} &amp;=\frac{(b+1)(b+2009)}{a+b+2010} \\<br /> \end{align*}<br /> &lt;/math&gt;<br /> The terms get more and more wacky, so we just solve for &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;a_{1}=a_{3}&lt;/math&gt; and &lt;math&gt;a_{2}=a_{4}.&lt;/math&gt;<br /> <br /> Solving we find both equations end up to the equation &lt;math&gt;ab=2009&lt;/math&gt; in which we see to minimize we see that &lt;math&gt;a = 49&lt;/math&gt; and &lt;math&gt;b=41.&lt;/math&gt; This solution is VERY non rigorous and not recommended.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_13&diff=133332 2009 AIME I Problems/Problem 13 2020-09-07T23:49:05Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The terms of the sequence &lt;math&gt;(a_i)&lt;/math&gt; defined by &lt;math&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt; are positive integers. Find the minimum possible value of &lt;math&gt;a_1 + a_2&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> This question is guessable but let's prove our answer<br /> <br /> &lt;cmath&gt;a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}(1 + a_{n + 1})= a_n + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009&lt;/cmath&gt;<br /> <br /> <br /> lets put &lt;math&gt;n+1&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009&lt;/cmath&gt;<br /> <br /> <br /> and set them equal now<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> let's rewrite it<br /> <br /> <br /> &lt;cmath&gt;(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n&lt;/cmath&gt;<br /> <br /> <br /> Let's make it look nice and let &lt;math&gt;b_n=a_{n + 2}-a_n&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;(b_{n+1})(a_{n + 2}+1)= b_n&lt;/cmath&gt;<br /> <br /> <br /> Since &lt;math&gt;b_n&lt;/math&gt; and &lt;math&gt;b_{n+1}&lt;/math&gt; are integers, we can see &lt;math&gt;b_n&lt;/math&gt; is divisible by &lt;math&gt;b_{n+1}&lt;/math&gt;<br /> <br /> <br /> But we can't have an infinite sequence of proper factors, unless &lt;math&gt;b_n=0&lt;/math&gt;<br /> <br /> <br /> Thus, &lt;math&gt;a_{n + 2}-a_n=0&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{n + 2}=a_n&lt;/cmath&gt;<br /> <br /> <br /> So now, we know &lt;math&gt;a_3=a_1&lt;/math&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{3} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1} = \frac {a_1 + 2009} {1 + a_{2}}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}+a_{1}a_{2} = a_1 + 2009&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a_{1}a_{2} = 2009&lt;/cmath&gt;<br /> <br /> <br /> To minimize &lt;math&gt;a_{1}+a_{2}&lt;/math&gt;, we need &lt;math&gt;41&lt;/math&gt; and &lt;math&gt;49&lt;/math&gt;<br /> <br /> <br /> Thus, our answer &lt;math&gt;= 41+49=\boxed {090}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> If &lt;math&gt;a_{n} \ne \frac {2009}{a_{n+1}}&lt;/math&gt;, then either<br /> &lt;cmath&gt;a_{n} = \frac {a_{n}}{1} &lt; \frac {a_{n} + 2009}{1 + a_{n+1}} &lt; \frac {2009}{a_{n+1}}&lt;/cmath&gt;<br /> <br /> or<br /> <br /> &lt;cmath&gt;\frac {2009}{a_{n+1}} &lt; \frac {2009 + a_{n}}{a_{n+1} + 1} &lt; \frac {a_{n}}{1} = a_{n}&lt;/cmath&gt;<br /> <br /> All the integers between &lt;math&gt;a_{n}&lt;/math&gt; and &lt;math&gt;\frac {2009}{a_{n+1}}&lt;/math&gt; would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.<br /> <br /> So &lt;math&gt;a_{n} = \frac {2009}{a_{n+1}}&lt;/math&gt;, which &lt;math&gt;a_{n} \cdot a_{n+1} = 2009&lt;/math&gt;. When &lt;math&gt;n = 1&lt;/math&gt;, &lt;math&gt;a_{1} \cdot a_{2} = 2009&lt;/math&gt;. The smallest sum of two factors which have a product of &lt;math&gt;2009&lt;/math&gt; is &lt;math&gt;41 + 49=\boxed {090}&lt;/math&gt;<br /> <br /> <br /> <br /> === Solution 3 (BS Solution) ===<br /> <br /> Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms.<br /> <br /> \begin{align*}<br /> a_{1} &amp;= a \\<br /> a_{2} &amp;= b \\<br /> a_{3} &amp;= \displaystyle\frac{a+2009}{1+b} \\<br /> a_{4} &amp;= \displaystyle\frac{(b+1)(b+2009)}{a+b+2010} \\<br /> \end{align*}<br /> <br /> The terms get more and more wacky, so we just solve for &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;a_{1}=a_{3}&lt;/math&gt; and &lt;math&gt;a_{2}=a_{4}.&lt;/math&gt;<br /> <br /> Solving we find both equations end up to the equation &lt;math&gt;ab=2009&lt;/math&gt; in which we see to minimize we see that &lt;math&gt;a = 49&lt;/math&gt; and &lt;math&gt;b=41.&lt;/math&gt; This solution is VERY non rigorous and not recommended.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_9&diff=133271 2011 AIME II Problems/Problem 9 2020-09-06T19:24:27Z <p>Jerry122805: /* Solution */</p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;x_1, x_2, ... , x_6&lt;/math&gt; be non-negative real numbers such that &lt;math&gt;x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1&lt;/math&gt;, and &lt;math&gt;x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive relatively prime integers such that &lt;math&gt;\frac{p}{q}&lt;/math&gt; is the maximum possible value of<br /> &lt;math&gt;x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2&lt;/math&gt;. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Note that neither the constraint nor the expression we need to maximize involves products &lt;math&gt;x_i x_j&lt;/math&gt; with &lt;math&gt;i \equiv j \pmod 3&lt;/math&gt;. Factoring out say &lt;math&gt;x_1&lt;/math&gt; and &lt;math&gt;x_4&lt;/math&gt; we see that the constraint is &lt;math&gt;x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}&lt;/math&gt;, while the expression we want to maximize is &lt;math&gt;x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)&lt;/math&gt;. Adding the left side of the constraint to the expression, we get: &lt;math&gt;(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)&lt;/math&gt;. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most &lt;math&gt;\frac1{27}&lt;/math&gt;. Since we have added at least &lt;math&gt;\frac{1}{540}&lt;/math&gt; the desired maximum is at most &lt;math&gt;\frac1{27} - \frac1{540} =\frac{19}{540}&lt;/math&gt;. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to &lt;math&gt;\displaystyle\frac1{540}&lt;/math&gt; with &lt;math&gt;x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \displaystyle\frac13&lt;/math&gt;&amp;mdash;for example, by choosing &lt;math&gt;x_1&lt;/math&gt; and &lt;math&gt;x_2&lt;/math&gt; small enough&amp;mdash;so our answer is &lt;math&gt;540 + 19 = \fbox{559}.&lt;/math&gt;<br /> <br /> An example is:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x_3 &amp;= x_6 = \frac16 \\<br /> x_1 &amp;= x_2 = \frac{5 - \sqrt{20}}{30} \\<br /> x_5 &amp;= x_4 = \frac{5 + \sqrt{20}}{30}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Another example is <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x_1 = x_3 = \frac{1}{3} \\<br /> x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \\<br /> x_4 &amp;= x_6 = 0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> ==Solution 2 (Not legit)==<br /> There's a symmetry between &lt;math&gt;x_1, x_3, x_5&lt;/math&gt; and &lt;math&gt;x_2,x_4,x_6&lt;/math&gt;. Therefore, a good guess is that &lt;math&gt;a = x_1 = x_3 = x_5&lt;/math&gt; and &lt;math&gt;b = x_2 = x_4 = x_6&lt;/math&gt;, at which point we know that &lt;math&gt;a+b = 1/3&lt;/math&gt;, &lt;math&gt;a^3+b^3 \geq 1/540&lt;/math&gt;, and we are trying to maximize &lt;math&gt;3a^2b+3ab^2&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540&lt;/cmath&gt;, which is the answer.<br /> <br /> This solution is extremely lucky; if you attempt to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; you receive complex answers (which contradict the problem statement), but the final answer is correct.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_9&diff=133270 2011 AIME II Problems/Problem 9 2020-09-06T19:23:54Z <p>Jerry122805: /* Solution */</p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;x_1, x_2, ... , x_6&lt;/math&gt; be non-negative real numbers such that &lt;math&gt;x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1&lt;/math&gt;, and &lt;math&gt;x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive relatively prime integers such that &lt;math&gt;\frac{p}{q}&lt;/math&gt; is the maximum possible value of<br /> &lt;math&gt;x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2&lt;/math&gt;. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Note that neither the constraint nor the expression we need to maximize involves products &lt;math&gt;x_i x_j&lt;/math&gt; with &lt;math&gt;i \equiv j \pmod 3&lt;/math&gt;. Factoring out say &lt;math&gt;x_1&lt;/math&gt; and &lt;math&gt;x_4&lt;/math&gt; we see that the constraint is &lt;math&gt;x_1(x_3x_5) + x_4(x_2x_6) \ge {\scriptstyle\frac1{540}}&lt;/math&gt;, while the expression we want to maximize is &lt;math&gt;x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)&lt;/math&gt;. Adding the left side of the constraint to the expression, we get: &lt;math&gt;(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)&lt;/math&gt;. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most &lt;math&gt;\displaystyle\frac1{27}&lt;/math&gt;. Since we have added at least &lt;math&gt;\displaystyle\frac{1}{540}&lt;/math&gt; the desired maximum is at most &lt;math&gt;\displaystyle\frac1{27} - \\displaystylefrac1{540} = \displaystyle\frac{19}{540}&lt;/math&gt;. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to &lt;math&gt;\scriptstyle\frac1{540}&lt;/math&gt; with &lt;math&gt;x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \scriptstyle\frac13&lt;/math&gt;&amp;mdash;for example, by choosing &lt;math&gt;x_1&lt;/math&gt; and &lt;math&gt;x_2&lt;/math&gt; small enough&amp;mdash;so our answer is &lt;math&gt;540 + 19 = \fbox{559}.&lt;/math&gt;<br /> <br /> An example is:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x_3 &amp;= x_6 = \frac16 \\<br /> x_1 &amp;= x_2 = \frac{5 - \sqrt{20}}{30} \\<br /> x_5 &amp;= x_4 = \frac{5 + \sqrt{20}}{30}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Another example is <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x_1 = x_3 = \frac{1}{3} \\<br /> x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \\<br /> x_4 &amp;= x_6 = 0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> ==Solution 2 (Not legit)==<br /> There's a symmetry between &lt;math&gt;x_1, x_3, x_5&lt;/math&gt; and &lt;math&gt;x_2,x_4,x_6&lt;/math&gt;. Therefore, a good guess is that &lt;math&gt;a = x_1 = x_3 = x_5&lt;/math&gt; and &lt;math&gt;b = x_2 = x_4 = x_6&lt;/math&gt;, at which point we know that &lt;math&gt;a+b = 1/3&lt;/math&gt;, &lt;math&gt;a^3+b^3 \geq 1/540&lt;/math&gt;, and we are trying to maximize &lt;math&gt;3a^2b+3ab^2&lt;/math&gt;. Then,<br /> <br /> &lt;cmath&gt;3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540&lt;/cmath&gt;, which is the answer.<br /> <br /> This solution is extremely lucky; if you attempt to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; you receive complex answers (which contradict the problem statement), but the final answer is correct.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_4&diff=133204 1988 AIME Problems/Problem 4 2020-09-05T21:42:38Z <p>Jerry122805: /* Solution */</p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; for &lt;math&gt;i = 1, 2, \dots, n&lt;/math&gt;. Suppose further that<br /> &lt;math&gt;<br /> |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.<br /> &lt;/math&gt;<br /> What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> == Solution ==<br /> <br /> Since &lt;math&gt;|x_i| &lt; 1&lt;/math&gt; then<br /> <br /> &lt;cmath&gt;|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| &lt; n.&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;n \ge 20&lt;/math&gt;. We now just need to find an example where &lt;math&gt;n = 20&lt;/math&gt;: suppose &lt;math&gt;x_{2k-1} = \frac{19}{20}&lt;/math&gt; and &lt;math&gt;x_{2k} = -\frac{19}{20}&lt;/math&gt;; then on the left hand side we have &lt;math&gt;\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19&lt;/math&gt;. On the right hand side, we have &lt;math&gt;19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19&lt;/math&gt;, and so the equation can hold for &lt;math&gt;n = \boxed{020}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 (Motivating solution)==<br /> <br /> First off, one can test &lt;math&gt;1,-1,1&lt;/math&gt; and find that the &lt;math&gt;LHS&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the RHS is &lt;math&gt;1.&lt;/math&gt; Similarly testing &lt;math&gt;1,-1,-1,1&lt;/math&gt; yields &lt;math&gt;4&lt;/math&gt; on the LHS and &lt;math&gt;0&lt;/math&gt; on the RHS. It seems for every negative we gain twice of that negative on the LHS. However, when we test something like &lt;math&gt;1,-1,-1,-1,1&lt;/math&gt; we find the LHS to be &lt;math&gt;5&lt;/math&gt; and the RHS to be &lt;math&gt;1.&lt;/math&gt; What happened? There were more negatives than positives. Why does this mean that the LHS doesn't grow? There aren't enough positives to &quot;cancel out!&quot; Therefore if for every negative we need a positive for it to cancel out to grow. We can make the &lt;math&gt;LHS&lt;/math&gt; grow by approximately &lt;math&gt;2&lt;/math&gt; every time we choose a negative and a positive, so if we have &lt;math&gt;20&lt;/math&gt; numbers, namely &lt;math&gt;10&lt;/math&gt; positive and &lt;math&gt;10&lt;/math&gt; negative we can obtain the desired answer.<br /> <br /> == See also ==<br /> *[[Triangle Inequality]]<br /> {{AIME box|year=1988|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_12&diff=133180 2004 AIME I Problems/Problem 12 2020-09-05T18:41:43Z <p>Jerry122805: /* Alternate way of computing the end result */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; S &lt;/math&gt; be the set of [[ordered pair]]s &lt;math&gt; (x, y) &lt;/math&gt; such that &lt;math&gt; 0 &lt; x \le 1, 0&lt;y\le 1, &lt;/math&gt; and &lt;math&gt; \left[\log_2{\left(\frac 1x\right)}\right] &lt;/math&gt; and &lt;math&gt; \left[\log_5{\left(\frac 1y\right)}\right] &lt;/math&gt; are both even. Given that the area of the graph of &lt;math&gt; S &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers, find &lt;math&gt; m+n. &lt;/math&gt; The notation &lt;math&gt; [z] &lt;/math&gt; denotes the [[floor function|greatest integer]] that is less than or equal to &lt;math&gt; z. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Likewise:<br /> &lt;math&gt;\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Graphing this yields a series of [[rectangle]]s which become smaller as you move toward the [[origin]]. The &lt;math&gt;x&lt;/math&gt; interval of each box is given by the [[geometric sequence]] &lt;math&gt;\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots&lt;/math&gt;, and the &lt;math&gt;y&lt;/math&gt; interval is given by &lt;math&gt;\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots&lt;/math&gt;<br /> <br /> Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:<br /> <br /> &lt;cmath&gt;\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},&lt;/cmath&gt; and the answer is &lt;math&gt;m+n = 5 + 9 = \boxed{014}&lt;/math&gt;.<br /> <br /> <br /> === Alternate way of computing the end result ===<br /> <br /> Note that actually you can just multiply &lt;math&gt;(1 -\frac{1}{2} +\frac{1}{4} -\frac{1}{8} \cdots)(1 -\frac{1}{5} +\frac{1}{25} -\frac{1}{125} \cdots) =\frac{3}{2} \cdot\frac{5}{6} =\frac{5}{9} \implies m+n = \boxed{014}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_12&diff=133179 2004 AIME I Problems/Problem 12 2020-09-05T18:40:55Z <p>Jerry122805: /* Alternate way of computing the end result */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; S &lt;/math&gt; be the set of [[ordered pair]]s &lt;math&gt; (x, y) &lt;/math&gt; such that &lt;math&gt; 0 &lt; x \le 1, 0&lt;y\le 1, &lt;/math&gt; and &lt;math&gt; \left[\log_2{\left(\frac 1x\right)}\right] &lt;/math&gt; and &lt;math&gt; \left[\log_5{\left(\frac 1y\right)}\right] &lt;/math&gt; are both even. Given that the area of the graph of &lt;math&gt; S &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers, find &lt;math&gt; m+n. &lt;/math&gt; The notation &lt;math&gt; [z] &lt;/math&gt; denotes the [[floor function|greatest integer]] that is less than or equal to &lt;math&gt; z. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Likewise:<br /> &lt;math&gt;\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Graphing this yields a series of [[rectangle]]s which become smaller as you move toward the [[origin]]. The &lt;math&gt;x&lt;/math&gt; interval of each box is given by the [[geometric sequence]] &lt;math&gt;\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots&lt;/math&gt;, and the &lt;math&gt;y&lt;/math&gt; interval is given by &lt;math&gt;\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots&lt;/math&gt;<br /> <br /> Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:<br /> <br /> &lt;cmath&gt;\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},&lt;/cmath&gt; and the answer is &lt;math&gt;m+n = 5 + 9 = \boxed{014}&lt;/math&gt;.<br /> <br /> <br /> === Alternate way of computing the end result ===<br /> <br /> Note that actually you can just multiply &lt;math&gt;(1 -\frac{1}{2} +\frac{1}{4} -\frac{1}{8} \cdots)(1 -\frac{1}{5} +\frac{1}{25} -\frac{1}{125} \cdots) =\frac{3}{2} \cdot\frac{5}{6} =\frac{5}{9} \implies m+n \boxed{014}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_12&diff=133178 2004 AIME I Problems/Problem 12 2020-09-05T18:40:35Z <p>Jerry122805: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; S &lt;/math&gt; be the set of [[ordered pair]]s &lt;math&gt; (x, y) &lt;/math&gt; such that &lt;math&gt; 0 &lt; x \le 1, 0&lt;y\le 1, &lt;/math&gt; and &lt;math&gt; \left[\log_2{\left(\frac 1x\right)}\right] &lt;/math&gt; and &lt;math&gt; \left[\log_5{\left(\frac 1y\right)}\right] &lt;/math&gt; are both even. Given that the area of the graph of &lt;math&gt; S &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers, find &lt;math&gt; m+n. &lt;/math&gt; The notation &lt;math&gt; [z] &lt;/math&gt; denotes the [[floor function|greatest integer]] that is less than or equal to &lt;math&gt; z. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Likewise:<br /> &lt;math&gt;\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor&lt;/math&gt; is even when<br /> <br /> &lt;cmath&gt;y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots&lt;/cmath&gt;<br /> <br /> Graphing this yields a series of [[rectangle]]s which become smaller as you move toward the [[origin]]. The &lt;math&gt;x&lt;/math&gt; interval of each box is given by the [[geometric sequence]] &lt;math&gt;\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots&lt;/math&gt;, and the &lt;math&gt;y&lt;/math&gt; interval is given by &lt;math&gt;\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots&lt;/math&gt;<br /> <br /> Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:<br /> <br /> &lt;cmath&gt;\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},&lt;/cmath&gt; and the answer is &lt;math&gt;m+n = 5 + 9 = \boxed{014}&lt;/math&gt;.<br /> <br /> <br /> === Alternate way of computing the end result ===<br /> <br /> Note that actually you can just multiply &lt;math&gt;(1 - \displaystyle\frac{1}{2} + \displaystyle\frac{1}{4} - \displaystyle\frac{1}{8} \cdots)(1 - \displaystyle\frac{1}{5} + \displaystyle\frac{1}{25} - \displaystyle\frac{1}{125} \cdots) = \displaystyle\frac{3}{2} \cdot \displaystyle\frac{5}{6} = \displaystyle\frac{5}{9} = \boxed{014}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=133117 2012 AIME II Problems/Problem 9 2020-09-05T02:30:07Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2==<br /> <br /> As mentioned above, the first term is clearly &lt;math&gt;\frac{3}{2}.&lt;/math&gt; For the second term, we first wish to find &lt;math&gt;\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.&lt;/math&gt; Now we first square the first equation getting &lt;math&gt;\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.&lt;/math&gt; Squaring the second equation yields &lt;math&gt;\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.&lt;/math&gt; Let &lt;math&gt;\cos^2x = a&lt;/math&gt; and &lt;math&gt;\cos^2y = b.&lt;/math&gt; We have the system of equations<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 1-a &amp;= 9-9b \\<br /> 4a &amp;= b \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Multiplying the first equation by &lt;math&gt;4&lt;/math&gt; yields &lt;math&gt;4-4a = 36 - 36b&lt;/math&gt; and so &lt;math&gt;4-b =36 - 36b \implies b =\frac{32}{35}.&lt;/math&gt; We then find &lt;math&gt;a =\frac{8}{35}.&lt;/math&gt; Therefore the second fraction ends up being &lt;math&gt;\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}&lt;/math&gt; so that means our desired sum is &lt;math&gt;\frac{49}{58}&lt;/math&gt; so the desired sum is &lt;math&gt;\boxed{107}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=133115 2012 AIME II Problems/Problem 9 2020-09-05T00:02:44Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2==<br /> <br /> As mentioned above, the first term is clearly &lt;math&gt;\frac{3}{2}.&lt;/math&gt; For the second term, we first wish to find &lt;math&gt;\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.&lt;/math&gt; Now we first square the first equation getting &lt;math&gt;\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.&lt;/math&gt; Squaring the second equation yields &lt;math&gt;\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.&lt;/math&gt; Let &lt;math&gt;\cos^2x = a&lt;/math&gt; and &lt;math&gt;\cos^2y = b.&lt;/math&gt; We have the system of equations<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 1-a &amp;= 9-9b \\<br /> 4a &amp;= b \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Multiplying the first equation by &lt;math&gt;4&lt;/math&gt; yields &lt;math&gt;4-4a = 36 - 36b&lt;/math&gt; and so &lt;math&gt;4-b =36 - 36b \implies b =\frac{32}{35}.&lt;/math&gt; We then find &lt;math&gt;a =\frac{8}{35}.&lt;/math&gt; Therefore the second fraction ends up being &lt;math&gt;\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}&lt;/math&gt; so that means our desired sum is &lt;math&gt;\boxed{\frac{49}{58}}&lt;/math&gt; so the desired sum is &lt;math&gt;\boxed{107}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=133114 2012 AIME II Problems/Problem 9 2020-09-05T00:02:11Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> <br /> === Solution 2===<br /> <br /> As mentioned above, the first term is clearly &lt;math&gt;\frac{3}{2}.&lt;/math&gt; For the second term, we first wish to find &lt;math&gt;\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.&lt;/math&gt; Now we first square the first equation getting &lt;math&gt;\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.&lt;/math&gt; Squaring the second equation yields &lt;math&gt;\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.&lt;/math&gt; Let &lt;math&gt;\cos^2x = a&lt;/math&gt; and &lt;math&gt;\cos^2y = b.&lt;/math&gt; We have the system of equations<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 1-a &amp;= 9-9b \\<br /> 4a &amp;= b \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Multiplying the first equation by &lt;math&gt;4&lt;/math&gt; yields &lt;math&gt;4-4a = 36 - 36b&lt;/math&gt; and so &lt;math&gt;4-b =36 - 36b \implies b =\frac{32}{35}.&lt;/math&gt; We then find &lt;math&gt;a =\frac{8}{35}.&lt;/math&gt; Therefore the second fraction ends up being &lt;math&gt;\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}&lt;/math&gt; so that means our desired sum is &lt;math&gt;\boxed{\frac{49}{58}}&lt;/math&gt; so the desired sum is &lt;math&gt;\boxed{107}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=133113 2012 AIME II Problems/Problem 9 2020-09-05T00:02:01Z <p>Jerry122805: /* Solution 2 */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> <br /> === Solution 2===<br /> <br /> As mentioned above, the first term is clearly &lt;math&gt;\frac{3}{2}.&lt;/math&gt; For the second term, we first wish to find &lt;math&gt;\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.&lt;/math&gt; Now we first square the first equation getting &lt;math&gt;\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.&lt;/math&gt; Squaring the second equation yields &lt;math&gt;\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.&lt;/math&gt; Let &lt;math&gt;\cos^2x = a&lt;/math&gt; and &lt;math&gt;\cos^2y = b.&lt;/math&gt; We have the system of equations<br /> &lt;math&gt;<br /> \begin{align*}<br /> 1-a &amp;= 9-9b \\<br /> 4a &amp;= b \\<br /> \end{align*}<br /> &lt;/math&gt;<br /> Multiplying the first equation by &lt;math&gt;4&lt;/math&gt; yields &lt;math&gt;4-4a = 36 - 36b&lt;/math&gt; and so &lt;math&gt;4-b =36 - 36b \implies b =\frac{32}{35}.&lt;/math&gt; We then find &lt;math&gt;a =\frac{8}{35}.&lt;/math&gt; Therefore the second fraction ends up being &lt;math&gt;\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}&lt;/math&gt; so that means our desired sum is &lt;math&gt;\boxed{\frac{49}{58}}&lt;/math&gt; so the desired sum is &lt;math&gt;\boxed{107}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_9&diff=133112 2012 AIME II Problems/Problem 9 2020-09-05T00:01:47Z <p>Jerry122805: /* Now Back to the Solution! */</p> <hr /> <div>== Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> Examine the first term in the expression we want to evaluate, &lt;math&gt;\frac{\sin 2x}{\sin 2y}&lt;/math&gt;, separately from the second term, &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> === The First Term ===<br /> <br /> Using the identity &lt;math&gt;\sin 2\theta = 2\sin\theta\cos\theta&lt;/math&gt;, we have:<br /> <br /> &lt;math&gt;\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}&lt;/math&gt;<br /> <br /> === The Second Term ===<br /> <br /> Let the equation &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; be equation 1, and let the equation &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt; be equation 2.<br /> Hungry for the widely-used identity &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt;, we cross multiply equation 1 by &lt;math&gt;\sin y&lt;/math&gt; and multiply equation 2 by &lt;math&gt;\cos y&lt;/math&gt;.<br /> <br /> Equation 1 then becomes:<br /> <br /> &lt;math&gt;\sin x = 3\sin y&lt;/math&gt;.<br /> <br /> Equation 2 then becomes:<br /> <br /> &lt;math&gt;\cos x = \frac{1}{2} \cos y&lt;/math&gt;<br /> <br /> Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt;<br /> <br /> Applying the identity &lt;math&gt;\cos^2 y = 1 - \sin^2 y&lt;/math&gt; (which is similar to &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; but a bit different), we can change &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} \cos^2 y&lt;/math&gt; into:<br /> <br /> &lt;math&gt;1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{3}{4} = \frac{35}{4} \sin^2 y &lt;/math&gt;.<br /> <br /> So, &lt;math&gt;\sin^2 y = \frac{3}{35}&lt;/math&gt;.<br /> <br /> Squaring Equation 1 (leading to &lt;math&gt;\sin^2 x = 9\sin^2 y&lt;/math&gt;), we can solve for &lt;math&gt;\sin^2 x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}&lt;/math&gt;<br /> <br /> Using the identity &lt;math&gt;\cos 2\theta = 1 - 2\sin^2\theta&lt;/math&gt;, we can solve for &lt;math&gt;\frac{\cos 2x}{\cos 2y}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}&lt;/math&gt;.<br /> <br /> === Now Back to the Solution! ===<br /> <br /> Finally, &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}&lt;/math&gt;.<br /> <br /> So, the answer is &lt;math&gt;49+58=\boxed{107}&lt;/math&gt;.<br /> <br /> <br /> === Solution 2===<br /> <br /> As mentioned above, the first term is clearly &lt;math&gt;\displaystyle\frac{3}{2}.&lt;/math&gt; For the second term, we first wish to find &lt;math&gt;\displaystyle\frac{\cos 2x}{\cos 2y} = \displaystyle\frac{2\cos^2 x - 1}{2 \cos^2y -1}.&lt;/math&gt; Now we first square the first equation getting &lt;math&gt;\displaystyle\frac{\sin^2x}{\sin^2y} = \displaystyle\frac{1-\cos^2x}{1 - \cos^2y} =9.&lt;/math&gt; Squaring the second equation yields &lt;math&gt;\displaystyle\frac{\cos^2x}{\cos^2y} = \displaystyle\frac{1}{4}.&lt;/math&gt; Let &lt;math&gt;\cos^2x = a&lt;/math&gt; and &lt;math&gt;\cos^2y = b.&lt;/math&gt; We have the system of equations<br /> <br /> \begin{align*}<br /> 1-a &amp;= 9-9b \\<br /> 4a &amp;= b \\<br /> \end{align*}<br /> <br /> Multiplying the first equation by &lt;math&gt;4&lt;/math&gt; yields &lt;math&gt;4-4a = 36 - 36b&lt;/math&gt; and so &lt;math&gt;4-b =36 - 36b \implies b = \displaystyle\frac{32}{35}.&lt;/math&gt; We then find &lt;math&gt;a = \displaystyle\frac{8}{35}.&lt;/math&gt; Therefore the second fraction ends up being &lt;math&gt;\displaystyle\frac{\displaystyle\frac{64}{35}-1}{\displaystyle\frac{16}{35}-1} = -\displaystyle\frac{19}{29}&lt;/math&gt; so that means our desired sum is &lt;math&gt;\boxed{\frac{49}{58}}&lt;/math&gt; so the desired sum is &lt;math&gt;\boxed{107}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133107 1997 AIME Problems/Problem 11 2020-09-04T19:46:21Z <p>Jerry122805: /* Solution 6 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> === Solution 6 ===<br /> <br /> We write &lt;math&gt;x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = \sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt; Therefore we have the equation &lt;math&gt;x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.&lt;/math&gt; Finishing, we have &lt;math&gt;\lfloor 100x \rfloor = \boxed{241}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133106 1997 AIME Problems/Problem 11 2020-09-04T19:46:07Z <p>Jerry122805: /* Solution6 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> == Solution 6 ==<br /> <br /> We write &lt;math&gt;x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = \sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt; Therefore we have the equation &lt;math&gt;x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.&lt;/math&gt; Finishing, we have &lt;math&gt;\lfloor 100x \rfloor = \boxed{241}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133105 1997 AIME Problems/Problem 11 2020-09-04T19:45:55Z <p>Jerry122805: /* Solution6 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> == Solution6 ==<br /> <br /> We write &lt;math&gt;x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = \sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt; Therefore we have the equation &lt;math&gt;x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.&lt;/math&gt; Finishing, we have &lt;math&gt;\lfloor 100x \rfloor = \boxed{241}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133104 1997 AIME Problems/Problem 11 2020-09-04T19:44:26Z <p>Jerry122805: /* Solution6 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> == Solution6 ==<br /> <br /> We write &lt;math&gt;x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = \sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} \sin n^{\circ}+\cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133103 1997 AIME Problems/Problem 11 2020-09-04T19:43:55Z <p>Jerry122805: /* Solution6 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> == Solution6 ==<br /> <br /> We write &lt;math&gt;x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = \sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\frac{\sum_{n=1}^{4} \sin n^{\circ}+\cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \sin n^{\circ}}{\sum_{n=1}^{44} sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&diff=133102 1997 AIME Problems/Problem 11 2020-09-04T19:43:06Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>== Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Note that <br /> &lt;math&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}&lt;/math&gt;<br /> <br /> Now use the sum-product formula &lt;math&gt;\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})&lt;/math&gt; <br /> We want to pair up &lt;math&gt;[1, 44]&lt;/math&gt;, &lt;math&gt;[2, 43]&lt;/math&gt;, &lt;math&gt;[3, 42]&lt;/math&gt;, etc. from the numerator and &lt;math&gt;[46, 89]&lt;/math&gt;, &lt;math&gt;[47, 88]&lt;/math&gt;, &lt;math&gt;[48, 87]&lt;/math&gt; etc. from the denominator. Then we get:<br /> &lt;cmath&gt;\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}&lt;/cmath&gt;<br /> <br /> To calculate this number, use the half angle formula. Since &lt;math&gt;\cos(\frac{x}{2}) = \pm \sqrt{\frac{\cos x + 1}{2}}&lt;/math&gt;, then our number becomes: &lt;cmath&gt;\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}&lt;/cmath&gt; in which we drop the negative roots (as it is clear cosine of &lt;math&gt;22.5&lt;/math&gt; and &lt;math&gt;67.5&lt;/math&gt; are positive). We can easily simplify this:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &amp;=&amp; \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &amp;=&amp; \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &amp;=&amp; \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &amp;=&amp; \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &amp;=&amp; \sqrt{2}+1<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> And hence our answer is &lt;math&gt;\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} x &amp;=&amp; \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\<br /> &amp;=&amp; \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Using the identity &lt;math&gt;\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}&lt;/math&gt; &lt;math&gt;\Longrightarrow \sin x + \cos x&lt;/math&gt; &lt;math&gt;= \sin x + \sin (90-x)&lt;/math&gt; &lt;math&gt;= 2 \sin 45 \cos (45-x)&lt;/math&gt; &lt;math&gt;= \sqrt{2} \cos (45-x)&lt;/math&gt;, that [[summation]] reduces to<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\<br /> &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)<br /> \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> <br /> This fraction is equivalent to &lt;math&gt;x&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> x &amp;=&amp; \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\<br /> \frac {1}{\sqrt {2}} &amp;=&amp; x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\<br /> x &amp;=&amp; \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\<br /> \lfloor 100x \rfloor &amp;=&amp; \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> A slight variant of the above solution, note that <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &amp;=&amp; \sum_{n=1}^{44} \sin n + \sin(90-n)\\<br /> &amp;=&amp; \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\<br /> \sum_{n=1}^{44} \sin n &amp;=&amp; (\sqrt{2}-1)\sum_{n=1}^{44} \cos n<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> This is the [[ratio]] we are looking for. &lt;math&gt;x&lt;/math&gt; reduces to &lt;math&gt;\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1&lt;/math&gt;, and &lt;math&gt;\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written &lt;math&gt;- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}&lt;/math&gt; (by [[De Moivre's Theorem]] with geometric series)<br /> <br /> &lt;math&gt;= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}&lt;/math&gt; (after multiplying by [[complex conjugate]])<br /> <br /> &lt;math&gt;= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> &lt;math&gt;= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}&lt;/math&gt;<br /> <br /> Using the [[trigonometric identities|tangent half-angle formula]], this becomes &lt;math&gt;\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)&lt;/math&gt;.<br /> <br /> Dividing the two parts and multiplying each part by 4, the fraction is &lt;math&gt;\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}&lt;/math&gt;.<br /> <br /> Although an exact value for &lt;math&gt;\cot (1/2^\circ)&lt;/math&gt; in terms of radicals will be difficult, this is easily known: it is really large!<br /> <br /> So treat it as though it were &lt;math&gt;\infty&lt;/math&gt;. The fraction is approximated by &lt;math&gt;\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Consider the sum &lt;math&gt;\sum_{n = 1}^{44} \text{cis } n^\circ&lt;/math&gt;. The fraction is given by the real part divided by the imaginary part.<br /> <br /> The sum can be written as &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt;. <br /> Consider the rhombus &lt;math&gt;OABC&lt;/math&gt; on the complex plane such that &lt;math&gt;O&lt;/math&gt; is the origin, &lt;math&gt;A&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; represents &lt;math&gt;\text{cis } n^\circ&lt;/math&gt;. Simple geometry shows that &lt;math&gt;\angle BOA = 22.5-k^\circ&lt;/math&gt;, so the angle that &lt;math&gt;\text{cis } n^\circ + \text{cis } 45-n^\circ&lt;/math&gt; makes with the real axis is simply &lt;math&gt;22.5^\circ&lt;/math&gt;. So &lt;math&gt;\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)&lt;/math&gt; is the sum of collinear complex numbers, so the angle the sum makes with the real axis is &lt;math&gt;22.5^\circ&lt;/math&gt;. So our answer is &lt;math&gt;\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}&lt;/math&gt;. <br /> <br /> Note that the &lt;math&gt;\cot(22.5^\circ) = \sqrt2 + 1&lt;/math&gt; can be shown easily through half-angle formula.<br /> <br /> <br /> <br /> == Solution6 ==<br /> <br /> We write &lt;math&gt;x = \displaystyle\frac{\sum_{n=46}^{89} sin n^{\circ}}{\sum_{n=1}^{44} sin n^{\circ}}&lt;/math&gt; since &lt;math&gt;\cos x = sin (90^{\circ}-x).&lt;/math&gt; Now we by the sine angle sum we know that &lt;math&gt;\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).&lt;/math&gt; So the expression simplifies to &lt;math&gt;\sin 45^{\circ}\left(\displaystyle\frac{\sum_{n=1}^{4} sin n^{\circ}+\cos n^{\circ}}{\sum_{n=1}^{44} sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\displaystyle\frac{\sum_{n=46}^{89} sin n^{\circ}}{\sum_{n=1}^{44} sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=133010 2014 AIME I Problems/Problem 9 2020-09-02T21:11:55Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>== Problem 9 ==<br /> <br /> Let &lt;math&gt;x_1&lt;x_2&lt;x_3&lt;/math&gt; be the three real roots of the equation &lt;math&gt;\sqrt{2014}x^3-4029x^2+2=0&lt;/math&gt;. Find &lt;math&gt;x_2(x_1+x_3)&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> <br /> Substituting &lt;math&gt;n&lt;/math&gt; for &lt;math&gt;2014&lt;/math&gt;, we get &lt;math&gt;\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0&lt;/math&gt;. Noting that &lt;math&gt;nx^2 - 1&lt;/math&gt; factors as a difference of squares to &lt;math&gt;(\sqrt{n}x - 1)(\sqrt{n}x+1)&lt;/math&gt;, we can factor the left side as &lt;math&gt;(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))&lt;/math&gt;. This means that &lt;math&gt;\frac{1}{\sqrt{n}}&lt;/math&gt; is a root, and the other two roots are the roots of &lt;math&gt;x^2 - 2\sqrt{n}x - 2&lt;/math&gt;. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to &lt;math&gt;2\sqrt{n}&lt;/math&gt;, so the positive root must be greater than &lt;math&gt;2\sqrt{n}&lt;/math&gt; in order to produce this sum when added to a negative value. Since &lt;math&gt;0 &lt; \frac{1}{\sqrt{2014}} &lt; 2\sqrt{2014}&lt;/math&gt; is clearly true, &lt;math&gt;x_2 = \frac{1}{\sqrt{2014}}&lt;/math&gt; and &lt;math&gt;x_1 + x_3 = 2\sqrt{2014}&lt;/math&gt;. Multiplying these values together, we find that &lt;math&gt;x_2(x_1+x_3) = \boxed{002}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> From Vieta's formulae, we know that &lt;math&gt;x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}, x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}},&lt;/math&gt; and &lt;math&gt;x_1x_2 + x_2x_3 + x_1x_3 = 0.&lt;/math&gt; Thus, we know that &lt;math&gt;x_2(x_1 + x_3) = -x_1x_3&lt;/math&gt;.<br /> <br /> Now consider the polynomial with roots &lt;math&gt;x_1x_2, x_2x_3,&lt;/math&gt; and &lt;math&gt;x_1x_3&lt;/math&gt;. Expanding the polynomial &lt;math&gt;(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)&lt;/math&gt;, we get the polynomial &lt;math&gt;x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2.&lt;/math&gt; Substituting the values obtained from Vieta's formulae, we find that this polynomial is &lt;math&gt;x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}&lt;/math&gt;. We know &lt;math&gt;x_1x_3&lt;/math&gt; is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to &lt;math&gt;1007x^3 - 4029x - 2 = 0&lt;/math&gt;.<br /> <br /> Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the &lt;math&gt;x^3&lt;/math&gt; term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that &lt;math&gt;x = -2&lt;/math&gt; is a solution. Factoring it out, we get that &lt;math&gt;1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)&lt;/math&gt;. Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, &lt;math&gt;x_1x_3 = -2&lt;/math&gt;, so &lt;math&gt;-x_1x_3 = \boxed{002}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Observing the equation, we notice that the coefficient for the middle term &lt;math&gt;-4029&lt;/math&gt; is equal to &lt;math&gt;-2{\sqrt{2014}}^2-1&lt;/math&gt;. Also notice that the coefficient for the &lt;math&gt;{x^3}&lt;/math&gt; term is &lt;math&gt;\sqrt{2014}&lt;/math&gt;. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the &lt;math&gt;x&lt;/math&gt; term of the binomial would have a coefficient of &lt;math&gt;\sqrt{2014}&lt;/math&gt;. Similarly, the &lt;math&gt;x&lt;/math&gt; term of the trinomial would also have a coefficient of &lt;math&gt;\sqrt{2014}&lt;/math&gt;. The factored form of the expression would look something like the following:<br /> &lt;math&gt;({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)&lt;/math&gt; where &lt;math&gt;{a, b,c}&lt;/math&gt; are all positive integers (because the &lt;math&gt;{x^2}&lt;/math&gt; term of the original expression is negative, and the constant term is positive), and &lt;math&gt;{ab=2}&lt;/math&gt;.<br /> <br /> Multiplying this expression out gives &lt;math&gt;{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}&lt;/math&gt;. Equating this with the original expression gives &lt;math&gt;{2014n+a}=-4029&lt;/math&gt;. The only positive integer solutions of this expression is &lt;math&gt;(n, a)=(1, 2015)&lt;/math&gt; or &lt;math&gt;(2, 1)&lt;/math&gt;. If &lt;math&gt;(n, a)=(1, 2015)&lt;/math&gt; then setting &lt;math&gt;{an{\sqrt{2014}}-b{\sqrt{2014}}}=0&lt;/math&gt; yields &lt;math&gt;{b=2015}&lt;/math&gt; and therefore &lt;math&gt;{ab=2015^2}&lt;/math&gt; which clearly isn't equal to &lt;math&gt;2&lt;/math&gt; as the constant term. Therefore, &lt;math&gt;(n, a)=(2, 1)&lt;/math&gt; and the factored form of the expression is:<br /> &lt;math&gt;({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)&lt;/math&gt;. Therefore, one of the three roots of the original expression is &lt;math&gt;{x=\dfrac{1}{\sqrt{2014}}}&lt;/math&gt;.<br /> Using the quadratic formula yields the other two roots as &lt;math&gt;{x={\sqrt{2014}}+{\sqrt{2016}}}&lt;/math&gt; and &lt;math&gt;{x={\sqrt{2014}}-{\sqrt{2016}}}&lt;/math&gt;. Arranging the roots in ascending order (in the order &lt;math&gt;x_1&lt;x_2&lt;x_3&lt;/math&gt;), &lt;math&gt;{\sqrt{2014}}-{\sqrt{2016}}&lt;\dfrac{1}{\sqrt{2014}}&lt;{\sqrt{2014}}+{\sqrt{2016}}&lt;/math&gt;.<br /> Therefore, &lt;math&gt;x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> By Vieta's, we are seeking to find &lt;math&gt;x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}&lt;/math&gt;. Substitute &lt;math&gt;n=-x_1x_3&lt;/math&gt; and &lt;math&gt;x_2=\frac{2}{\sqrt{2014}n}&lt;/math&gt;. Substituting this back into the original equation, we have &lt;math&gt;\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0&lt;/math&gt;, so &lt;math&gt;2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0&lt;/math&gt;. Hence, &lt;math&gt;8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}&lt;/math&gt;, and &lt;math&gt;n\equiv 2\pmod{1007}&lt;/math&gt;. But since &lt;math&gt;n\le 999&lt;/math&gt; because it is our desired answer, the only possible value for &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{002}&lt;/math&gt;<br /> BEST PROOOFFFF<br /> Stormersyle &amp; mathleticguyyy<br /> <br /> <br /> == Solution 5 ==<br /> <br /> Let &lt;math&gt;y =\frac{x}{\sqrt{2014}}.&lt;/math&gt; The original equation simplifies to &lt;math&gt;\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029x + 4028=0.&lt;/math&gt; Here we clearly see that &lt;math&gt;y=1&lt;/math&gt; is a root. Divideing &lt;math&gt;y-1&lt;/math&gt; from the sum we find that &lt;math&gt;(y-1)(y^2-4028y-4028)=0.&lt;/math&gt; From simple bounding we see that &lt;math&gt;y=1&lt;/math&gt; is the middle root. Therefore &lt;math&gt;x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=133009 2014 AIME I Problems/Problem 9 2020-09-02T21:11:40Z <p>Jerry122805: /* Solution 4 */</p> <hr /> <div>== Problem 9 ==<br /> <br /> Let &lt;math&gt;x_1&lt;x_2&lt;x_3&lt;/math&gt; be the three real roots of the equation &lt;math&gt;\sqrt{2014}x^3-4029x^2+2=0&lt;/math&gt;. Find &lt;math&gt;x_2(x_1+x_3)&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> <br /> Substituting &lt;math&gt;n&lt;/math&gt; for &lt;math&gt;2014&lt;/math&gt;, we get &lt;math&gt;\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0&lt;/math&gt;. Noting that &lt;math&gt;nx^2 - 1&lt;/math&gt; factors as a difference of squares to &lt;math&gt;(\sqrt{n}x - 1)(\sqrt{n}x+1)&lt;/math&gt;, we can factor the left side as &lt;math&gt;(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))&lt;/math&gt;. This means that &lt;math&gt;\frac{1}{\sqrt{n}}&lt;/math&gt; is a root, and the other two roots are the roots of &lt;math&gt;x^2 - 2\sqrt{n}x - 2&lt;/math&gt;. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to &lt;math&gt;2\sqrt{n}&lt;/math&gt;, so the positive root must be greater than &lt;math&gt;2\sqrt{n}&lt;/math&gt; in order to produce this sum when added to a negative value. Since &lt;math&gt;0 &lt; \frac{1}{\sqrt{2014}} &lt; 2\sqrt{2014}&lt;/math&gt; is clearly true, &lt;math&gt;x_2 = \frac{1}{\sqrt{2014}}&lt;/math&gt; and &lt;math&gt;x_1 + x_3 = 2\sqrt{2014}&lt;/math&gt;. Multiplying these values together, we find that &lt;math&gt;x_2(x_1+x_3) = \boxed{002}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> From Vieta's formulae, we know that &lt;math&gt;x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}, x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}},&lt;/math&gt; and &lt;math&gt;x_1x_2 + x_2x_3 + x_1x_3 = 0.&lt;/math&gt; Thus, we know that &lt;math&gt;x_2(x_1 + x_3) = -x_1x_3&lt;/math&gt;.<br /> <br /> Now consider the polynomial with roots &lt;math&gt;x_1x_2, x_2x_3,&lt;/math&gt; and &lt;math&gt;x_1x_3&lt;/math&gt;. Expanding the polynomial &lt;math&gt;(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)&lt;/math&gt;, we get the polynomial &lt;math&gt;x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2.&lt;/math&gt; Substituting the values obtained from Vieta's formulae, we find that this polynomial is &lt;math&gt;x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}&lt;/math&gt;. We know &lt;math&gt;x_1x_3&lt;/math&gt; is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to &lt;math&gt;1007x^3 - 4029x - 2 = 0&lt;/math&gt;.<br /> <br /> Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the &lt;math&gt;x^3&lt;/math&gt; term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that &lt;math&gt;x = -2&lt;/math&gt; is a solution. Factoring it out, we get that &lt;math&gt;1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)&lt;/math&gt;. Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, &lt;math&gt;x_1x_3 = -2&lt;/math&gt;, so &lt;math&gt;-x_1x_3 = \boxed{002}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Observing the equation, we notice that the coefficient for the middle term &lt;math&gt;-4029&lt;/math&gt; is equal to &lt;math&gt;-2{\sqrt{2014}}^2-1&lt;/math&gt;. Also notice that the coefficient for the &lt;math&gt;{x^3}&lt;/math&gt; term is &lt;math&gt;\sqrt{2014}&lt;/math&gt;. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the &lt;math&gt;x&lt;/math&gt; term of the binomial would have a coefficient of &lt;math&gt;\sqrt{2014}&lt;/math&gt;. Similarly, the &lt;math&gt;x&lt;/math&gt; term of the trinomial would also have a coefficient of &lt;math&gt;\sqrt{2014}&lt;/math&gt;. The factored form of the expression would look something like the following:<br /> &lt;math&gt;({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)&lt;/math&gt; where &lt;math&gt;{a, b,c}&lt;/math&gt; are all positive integers (because the &lt;math&gt;{x^2}&lt;/math&gt; term of the original expression is negative, and the constant term is positive), and &lt;math&gt;{ab=2}&lt;/math&gt;.<br /> <br /> Multiplying this expression out gives &lt;math&gt;{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}&lt;/math&gt;. Equating this with the original expression gives &lt;math&gt;{2014n+a}=-4029&lt;/math&gt;. The only positive integer solutions of this expression is &lt;math&gt;(n, a)=(1, 2015)&lt;/math&gt; or &lt;math&gt;(2, 1)&lt;/math&gt;. If &lt;math&gt;(n, a)=(1, 2015)&lt;/math&gt; then setting &lt;math&gt;{an{\sqrt{2014}}-b{\sqrt{2014}}}=0&lt;/math&gt; yields &lt;math&gt;{b=2015}&lt;/math&gt; and therefore &lt;math&gt;{ab=2015^2}&lt;/math&gt; which clearly isn't equal to &lt;math&gt;2&lt;/math&gt; as the constant term. Therefore, &lt;math&gt;(n, a)=(2, 1)&lt;/math&gt; and the factored form of the expression is:<br /> &lt;math&gt;({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)&lt;/math&gt;. Therefore, one of the three roots of the original expression is &lt;math&gt;{x=\dfrac{1}{\sqrt{2014}}}&lt;/math&gt;.<br /> Using the quadratic formula yields the other two roots as &lt;math&gt;{x={\sqrt{2014}}+{\sqrt{2016}}}&lt;/math&gt; and &lt;math&gt;{x={\sqrt{2014}}-{\sqrt{2016}}}&lt;/math&gt;. Arranging the roots in ascending order (in the order &lt;math&gt;x_1&lt;x_2&lt;x_3&lt;/math&gt;), &lt;math&gt;{\sqrt{2014}}-{\sqrt{2016}}&lt;\dfrac{1}{\sqrt{2014}}&lt;{\sqrt{2014}}+{\sqrt{2016}}&lt;/math&gt;.<br /> Therefore, &lt;math&gt;x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> By Vieta's, we are seeking to find &lt;math&gt;x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}&lt;/math&gt;. Substitute &lt;math&gt;n=-x_1x_3&lt;/math&gt; and &lt;math&gt;x_2=\frac{2}{\sqrt{2014}n}&lt;/math&gt;. Substituting this back into the original equation, we have &lt;math&gt;\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0&lt;/math&gt;, so &lt;math&gt;2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0&lt;/math&gt;. Hence, &lt;math&gt;8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}&lt;/math&gt;, and &lt;math&gt;n\equiv 2\pmod{1007}&lt;/math&gt;. But since &lt;math&gt;n\le 999&lt;/math&gt; because it is our desired answer, the only possible value for &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{002}&lt;/math&gt;<br /> BEST PROOOFFFF<br /> Stormersyle &amp; mathleticguyyy<br /> <br /> <br /> == Solution 5 ==<br /> <br /> Let &lt;math&gt;y = \displaystyle\frac{x}{\sqrt{2014}}.&lt;/math&gt; The original equation simplifies to &lt;math&gt;\displaystyle\frac{y^3}{2014} - \displaystyle\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029x + 4028=0.&lt;/math&gt; Here we clearly see that &lt;math&gt;y=1&lt;/math&gt; is a root. Divideing &lt;math&gt;y-1&lt;/math&gt; from the sum we find that &lt;math&gt;(y-1)(y^2-4028y-4028.&lt;/math&gt; From simple bounding we see that &lt;math&gt;y=1&lt;/math&gt; is the middle root. Therefore &lt;math&gt;x_{2}(x_{1}+x_{3}) = \displaystyle\frac{1}{\sqrt{2014}} \cdot \displaystyle\frac{4028}{\sqrt{2014}} = \boxed{002}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_6&diff=133008 2005 AIME I Problems/Problem 6 2020-09-02T20:44:01Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; P &lt;/math&gt; be the [[product]] of the [[nonreal]] [[root]]s of &lt;math&gt; x^4-4x^3+6x^2-4x=2005. &lt;/math&gt; Find &lt;math&gt; \lfloor P\rfloor. &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The left-hand side of that [[equation]] is nearly equal to &lt;math&gt;(x - 1)^4&lt;/math&gt;. Thus, we add 1 to each side in order to complete the fourth power and get<br /> &lt;math&gt;(x - 1)^4 = 2006&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;r = \sqrt{2006}&lt;/math&gt; be the positive [[real]] fourth root of 2006. Then the roots of the above equation are &lt;math&gt;x = 1 + i^n r&lt;/math&gt; for &lt;math&gt;n = 0, 1, 2, 3&lt;/math&gt;. The two non-real members of this set are &lt;math&gt;1 + ir&lt;/math&gt; and &lt;math&gt;1 - ir&lt;/math&gt;. Their product is &lt;math&gt;P = 1 + r^2 = 1 + \sqrt{2006}&lt;/math&gt;. &lt;math&gt;44^2 = 1936 &lt; 2006 &lt; 2025 = 45^2&lt;/math&gt; so &lt;math&gt;\lfloor P \rfloor = 1 + 44 = \boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Starting like before,<br /> &lt;math&gt;(x-1)^4= 2006&lt;/math&gt;<br /> This time we apply differences of squares.<br /> &lt;math&gt;(x-1)^4-2006=0&lt;/math&gt; so<br /> &lt;math&gt;((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0&lt;/math&gt;<br /> If you think of each part of the product as a quadratic, then &lt;math&gt;((x-1)^2+\sqrt{2006})&lt;/math&gt; is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just &lt;math&gt;x^2&lt;/math&gt; translated down and right.<br /> Therefore the products of the roots of &lt;math&gt;((x-1)^2+\sqrt{2006})&lt;/math&gt; or &lt;math&gt; P=1+\sqrt{2006}&lt;/math&gt; so<br /> <br /> &lt;math&gt;\lfloor P \rfloor = 1 + 44 = \boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; are both roots. Synthetic division gives<br /> &lt;math&gt;(x^2-2x)(x^2-2x+2)=2005&lt;/math&gt;. We now have our quadratic substitution of<br /> &lt;math&gt;y=x^2-2x+1=(x-1)^2&lt;/math&gt;, giving us<br /> &lt;math&gt;(y-1)(y+1)=2005&lt;/math&gt;. From here we proceed as in Solution 1 to get &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 4 == <br /> <br /> Realizing that if we add 1 to both sides we get &lt;math&gt;x^4-4x^3+6x^2-4x+1=2006&lt;/math&gt; which can be factored as &lt;math&gt;(x-1)^4=2006&lt;/math&gt;. Then we can substitute &lt;math&gt;(x-1)&lt;/math&gt; with &lt;math&gt;y&lt;/math&gt; which leaves us with &lt;math&gt;y^4=2006&lt;/math&gt;. Now subtracting 2006 from both sides we get some difference of squares &lt;math&gt;y^4-2006=0 \rightarrow (y-\sqrt{2006})(y+\sqrt{2006})(y^2+\sqrt{2006})=0&lt;/math&gt;. The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve &lt;math&gt;y^2+\sqrt{2006}=0&lt;/math&gt;, we can substitute &lt;math&gt;(x-1)&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt; giving us &lt;math&gt;(x-1)^2+\sqrt{2006}=0&lt;/math&gt;, expanding this we get &lt;math&gt;x^2-2x+1+\sqrt{2006}=0&lt;/math&gt;. We know that the product of a quadratics roots is &lt;math&gt;\frac{c}{a}&lt;/math&gt; which leaves us with &lt;math&gt;\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> <br /> As in solution 1, we find that &lt;math&gt;(x-1)^4 = 2006&lt;/math&gt;. Now &lt;math&gt;x-1=\pm \sqrt{2006}&lt;/math&gt; so &lt;math&gt;x_1 = 1+\sqrt{2006}&lt;/math&gt; and &lt;math&gt;x_2 = 1-\sqrt{2006}&lt;/math&gt; are the real roots of the equation. Multiplying, we get &lt;math&gt;x_1 x_2 = 1 - \sqrt{2006}&lt;/math&gt;. Now transforming the original function and using Vieta's formula, &lt;math&gt;x^4-4x^3+6x^2-4x-2005=0&lt;/math&gt; so &lt;math&gt;x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005&lt;/math&gt;. We find that the product of the nonreal roots is &lt;math&gt;x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8&lt;/math&gt; and we get &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> <br /> <br /> Note: &lt;math&gt;\frac{2005}{\sqrt{2006}-1}=\frac{2005(1+\sqrt{2006})}{2005} = 1+\sqrt{2006}.&lt;/math&gt;<br /> <br /> ==Solution 6 (De Moivre's Theorem)==<br /> <br /> As all the other solutions, we find that &lt;math&gt;(x-1)^4 = 2006&lt;/math&gt;. Thus &lt;math&gt;x=\sqrt{2006}+1&lt;/math&gt;. Thus &lt;math&gt;x= \sqrt{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1&lt;/math&gt; when &lt;math&gt;k=0,1,2,3&lt;/math&gt;. The complex values of &lt;math&gt;x&lt;/math&gt; are the ones where &lt;math&gt;i\sin(\frac{2\pi(k)}{4})&lt;/math&gt; does not equal 0. These complex roots are &lt;math&gt;1+\sqrt{2006}(i)&lt;/math&gt; and &lt;math&gt;1-\sqrt{2006}(i)&lt;/math&gt;. The product of these two nonreal roots is (&lt;math&gt;1+\sqrt{2006}(i)&lt;/math&gt;)(&lt;math&gt;1-\sqrt{2006}(i)&lt;/math&gt;) which is equal to &lt;math&gt;1+\sqrt {2006}&lt;/math&gt;. The floor of that value is &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_6&diff=133007 2005 AIME I Problems/Problem 6 2020-09-02T20:43:39Z <p>Jerry122805: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; P &lt;/math&gt; be the [[product]] of the [[nonreal]] [[root]]s of &lt;math&gt; x^4-4x^3+6x^2-4x=2005. &lt;/math&gt; Find &lt;math&gt; \lfloor P\rfloor. &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The left-hand side of that [[equation]] is nearly equal to &lt;math&gt;(x - 1)^4&lt;/math&gt;. Thus, we add 1 to each side in order to complete the fourth power and get<br /> &lt;math&gt;(x - 1)^4 = 2006&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;r = \sqrt{2006}&lt;/math&gt; be the positive [[real]] fourth root of 2006. Then the roots of the above equation are &lt;math&gt;x = 1 + i^n r&lt;/math&gt; for &lt;math&gt;n = 0, 1, 2, 3&lt;/math&gt;. The two non-real members of this set are &lt;math&gt;1 + ir&lt;/math&gt; and &lt;math&gt;1 - ir&lt;/math&gt;. Their product is &lt;math&gt;P = 1 + r^2 = 1 + \sqrt{2006}&lt;/math&gt;. &lt;math&gt;44^2 = 1936 &lt; 2006 &lt; 2025 = 45^2&lt;/math&gt; so &lt;math&gt;\lfloor P \rfloor = 1 + 44 = \boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Starting like before,<br /> &lt;math&gt;(x-1)^4= 2006&lt;/math&gt;<br /> This time we apply differences of squares.<br /> &lt;math&gt;(x-1)^4-2006=0&lt;/math&gt; so<br /> &lt;math&gt;((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0&lt;/math&gt;<br /> If you think of each part of the product as a quadratic, then &lt;math&gt;((x-1)^2+\sqrt{2006})&lt;/math&gt; is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just &lt;math&gt;x^2&lt;/math&gt; translated down and right.<br /> Therefore the products of the roots of &lt;math&gt;((x-1)^2+\sqrt{2006})&lt;/math&gt; or &lt;math&gt; P=1+\sqrt{2006}&lt;/math&gt; so<br /> <br /> &lt;math&gt;\lfloor P \rfloor = 1 + 44 = \boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; are both roots. Synthetic division gives<br /> &lt;math&gt;(x^2-2x)(x^2-2x+2)=2005&lt;/math&gt;. We now have our quadratic substitution of<br /> &lt;math&gt;y=x^2-2x+1=(x-1)^2&lt;/math&gt;, giving us<br /> &lt;math&gt;(y-1)(y+1)=2005&lt;/math&gt;. From here we proceed as in Solution 1 to get &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 4 == <br /> <br /> Realizing that if we add 1 to both sides we get &lt;math&gt;x^4-4x^3+6x^2-4x+1=2006&lt;/math&gt; which can be factored as &lt;math&gt;(x-1)^4=2006&lt;/math&gt;. Then we can substitute &lt;math&gt;(x-1)&lt;/math&gt; with &lt;math&gt;y&lt;/math&gt; which leaves us with &lt;math&gt;y^4=2006&lt;/math&gt;. Now subtracting 2006 from both sides we get some difference of squares &lt;math&gt;y^4-2006=0 \rightarrow (y-\sqrt{2006})(y+\sqrt{2006})(y^2+\sqrt{2006})=0&lt;/math&gt;. The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve &lt;math&gt;y^2+\sqrt{2006}=0&lt;/math&gt;, we can substitute &lt;math&gt;(x-1)&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt; giving us &lt;math&gt;(x-1)^2+\sqrt{2006}=0&lt;/math&gt;, expanding this we get &lt;math&gt;x^2-2x+1+\sqrt{2006}=0&lt;/math&gt;. We know that the product of a quadratics roots is &lt;math&gt;\frac{c}{a}&lt;/math&gt; which leaves us with &lt;math&gt;\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> <br /> As in solution 1, we find that &lt;math&gt;(x-1)^4 = 2006&lt;/math&gt;. Now &lt;math&gt;x-1=\pm \sqrt{2006}&lt;/math&gt; so &lt;math&gt;x_1 = 1+\sqrt{2006}&lt;/math&gt; and &lt;math&gt;x_2 = 1-\sqrt{2006}&lt;/math&gt; are the real roots of the equation. Multiplying, we get &lt;math&gt;x_1 x_2 = 1 - \sqrt{2006}&lt;/math&gt;. Now transforming the original function and using Vieta's formula, &lt;math&gt;x^4-4x^3+6x^2-4x-2005=0&lt;/math&gt; so &lt;math&gt;x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005&lt;/math&gt;. We find that the product of the nonreal roots is &lt;math&gt;x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8&lt;/math&gt; and we get &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> <br /> <br /> Note: &lt;math&gt;\displaystyle\frac{2005}{\sqrt{2006}-1}=\displaystyle\frac{2005(1+\sqrt{2006}}{2005} = 1+\sqrt{2006}.&lt;/math&gt;<br /> <br /> ==Solution 6 (De Moivre's Theorem)==<br /> <br /> As all the other solutions, we find that &lt;math&gt;(x-1)^4 = 2006&lt;/math&gt;. Thus &lt;math&gt;x=\sqrt{2006}+1&lt;/math&gt;. Thus &lt;math&gt;x= \sqrt{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1&lt;/math&gt; when &lt;math&gt;k=0,1,2,3&lt;/math&gt;. The complex values of &lt;math&gt;x&lt;/math&gt; are the ones where &lt;math&gt;i\sin(\frac{2\pi(k)}{4})&lt;/math&gt; does not equal 0. These complex roots are &lt;math&gt;1+\sqrt{2006}(i)&lt;/math&gt; and &lt;math&gt;1-\sqrt{2006}(i)&lt;/math&gt;. The product of these two nonreal roots is (&lt;math&gt;1+\sqrt{2006}(i)&lt;/math&gt;)(&lt;math&gt;1-\sqrt{2006}(i)&lt;/math&gt;) which is equal to &lt;math&gt;1+\sqrt {2006}&lt;/math&gt;. The floor of that value is &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jerry122805