https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Theultimate123&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-21T03:14:19Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_7&diff=109876 2017 AMC 12B Problems/Problem 7 2019-09-19T01:38:27Z <p>Theultimate123: </p> <hr /> <div>==Problem 7==<br /> The functions &lt;math&gt;\sin(x)&lt;/math&gt; and &lt;math&gt;\cos(x)&lt;/math&gt; are periodic with least period &lt;math&gt;2\pi&lt;/math&gt;. What is the least period of the function &lt;math&gt;\cos(\sin(x))&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)} &lt;/math&gt; The function is not periodic.<br /> <br /> ==Solution==<br /> Start by noting that &lt;math&gt;\cos(-x)=\cos(x)&lt;/math&gt;. Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period &lt;math&gt;\pi&lt;/math&gt;, so the answer is, surprisingly, &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Trigonometry Problems]]</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=104586 2019 AIME I Problems/Problem 15 2019-03-16T19:08:20Z <p>Theultimate123: /* Solution 2 */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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;.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\tfrac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\tfrac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\tfrac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\tfrac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we hav e &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=104584 2019 AIME I Problems/Problem 15 2019-03-16T19:07:44Z <p>Theultimate123: </p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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;.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\frac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\frac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\frac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\frac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we hav e &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=104583 2019 AIME I Problems/Problem 15 2019-03-16T19:07:32Z <p>Theultimate123: </p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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;.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt; and &lt;math&gt;XQ=11/2&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Let the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. Then, since &lt;math&gt;RA^2=RB^2&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt; lies on the radical axis of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, which is &lt;math&gt;\overline{PQ}&lt;/math&gt;. It follows that &lt;cmath&gt;-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).&lt;/cmath&gt;<br /> Let &lt;math&gt;Q'&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;. By the Midpoint of Harmonic Bundles Lemma, &lt;cmath&gt;RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,&lt;/cmath&gt;<br /> whence &lt;math&gt;Q=Q'&lt;/math&gt;. Like above, &lt;math&gt;XP=\frac{11-\sqrt{61}}2&lt;/math&gt;. Since &lt;math&gt;XQ=\frac{11}2&lt;/math&gt;, we establish that &lt;math&gt;PQ=\frac{\sqrt{61}}2&lt;/math&gt;, from which &lt;math&gt;PQ^2=\frac{61}4&lt;/math&gt;, and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we hav e &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=104557 2019 AIME I Problems/Problem 7 2019-03-16T00:06:58Z <p>Theultimate123: /* Problem 7 */ formatting</p> <hr /> <div>==Problem 7==<br /> There are positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that satisfy the system of equations &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &amp;= 60\\<br /> \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &amp;= 570.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Let &lt;math&gt;m&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;n&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;y&lt;/math&gt;. Find &lt;math&gt;3m+2n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Add the two equations to get that &lt;math&gt;\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Then, we use the theorem &lt;math&gt;\log a+\log b=\log ab&lt;/math&gt; to get the equation, &lt;math&gt;\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Using the theorem that &lt;math&gt;\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y&lt;/math&gt;, along with the previously mentioned theorem, we can get the equation &lt;math&gt;3\log(xy)=630&lt;/math&gt;.<br /> This can easily be simplified to &lt;math&gt;\log(xy)=210&lt;/math&gt;, or &lt;math&gt;xy = 10^{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;10^{210}&lt;/math&gt; can be factored into &lt;math&gt;2^{210} \cdot 5^{210}&lt;/math&gt;, and &lt;math&gt;m+n&lt;/math&gt; equals to the sum of the exponents of 2 and 5, which is &lt;math&gt;210+210 = 420&lt;/math&gt;.<br /> Multiply by two to get &lt;math&gt;2m +2n&lt;/math&gt;, which is &lt;math&gt;840&lt;/math&gt;.<br /> Then, use the first equation (&lt;math&gt;\log x + 2\log(\gcd(x,y)) = 60&lt;/math&gt;) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the &lt;math&gt;gcd x&lt;/math&gt;. Then, turn the equation into &lt;math&gt;3\log x = 60&lt;/math&gt;, which yields &lt;math&gt;\log x = 20&lt;/math&gt;, or &lt;math&gt;x = 10^{20}&lt;/math&gt;.<br /> Factor this into &lt;math&gt;2^{20} \cdot 5^{20}&lt;/math&gt;, and add the two 20's, resulting in m, which is 40.<br /> Add &lt;math&gt;m&lt;/math&gt; to &lt;math&gt;2m + 2n&lt;/math&gt; (which is 840) to get &lt;math&gt;40+840 = \boxed{880}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Crappier Solution)==<br /> <br /> First simplifying the first and second equations, we get that <br /> <br /> &lt;cmath&gt;\log_{10}(x\cdot\text{gcd}(x,y)^2)=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_{10}(y\cdot\text{lcm}(x,y)^2)=570&lt;/cmath&gt;<br /> <br /> <br /> Thus, when the two equations are added, we have that<br /> &lt;cmath&gt;\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630&lt;/cmath&gt;<br /> When simplified, this equals <br /> &lt;cmath&gt;\log_{10}(x^3y^3)=630&lt;/cmath&gt;<br /> so this means that<br /> &lt;cmath&gt;x^3y^3=10^{630}&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;xy=10^{210}.&lt;/cmath&gt;<br /> <br /> Now, the following cannot be done on a proof contest but let's (intuitively) assume that &lt;math&gt;x&lt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both powers of &lt;math&gt;10&lt;/math&gt;. This means the first equation would simplify to &lt;cmath&gt;x^3=10^{60}&lt;/cmath&gt; and &lt;cmath&gt;y^3=10^{570}.&lt;/cmath&gt; Therefore, &lt;math&gt;x=10^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}&lt;/math&gt; and if we plug these values back, it works! &lt;math&gt;10^{20}&lt;/math&gt; has &lt;math&gt;20\cdot2=40&lt;/math&gt; total factors and &lt;math&gt;10^{190}&lt;/math&gt; has &lt;math&gt;190\cdot2=380&lt;/math&gt; so &lt;cmath&gt;3\cdot 40 + 2\cdot 380 = \boxed{880}.&lt;/cmath&gt;<br /> <br /> Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.<br /> <br /> ==Solution 3 (Easy Solution)==<br /> Let &lt;math&gt;x=10^a&lt;/math&gt; and &lt;math&gt;y=10^b&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;/math&gt;. Then the given equations become &lt;math&gt;3a=60&lt;/math&gt; and &lt;math&gt;3b=570&lt;/math&gt;. Therefore, &lt;math&gt;x=10^{20}=2^{20}\cdot5^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}=2^{190}\cdot5^{190}&lt;/math&gt;. Our answer is &lt;math&gt;3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=104556 2019 AIME I Problems/Problem 7 2019-03-16T00:06:33Z <p>Theultimate123: /* Problem 7 */</p> <hr /> <div>==Problem 7==<br /> There are positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that satisfy the system of equations &lt;cmath&gt;\log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) = 60&lt;/cmath&gt;&lt;cmath&gt;\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.&lt;/cmath&gt; Let &lt;math&gt;m&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;n&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;y&lt;/math&gt;. Find &lt;math&gt;3m+2n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Add the two equations to get that &lt;math&gt;\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Then, we use the theorem &lt;math&gt;\log a+\log b=\log ab&lt;/math&gt; to get the equation, &lt;math&gt;\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Using the theorem that &lt;math&gt;\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y&lt;/math&gt;, along with the previously mentioned theorem, we can get the equation &lt;math&gt;3\log(xy)=630&lt;/math&gt;.<br /> This can easily be simplified to &lt;math&gt;\log(xy)=210&lt;/math&gt;, or &lt;math&gt;xy = 10^{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;10^{210}&lt;/math&gt; can be factored into &lt;math&gt;2^{210} \cdot 5^{210}&lt;/math&gt;, and &lt;math&gt;m+n&lt;/math&gt; equals to the sum of the exponents of 2 and 5, which is &lt;math&gt;210+210 = 420&lt;/math&gt;.<br /> Multiply by two to get &lt;math&gt;2m +2n&lt;/math&gt;, which is &lt;math&gt;840&lt;/math&gt;.<br /> Then, use the first equation (&lt;math&gt;\log x + 2\log(\gcd(x,y)) = 60&lt;/math&gt;) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the &lt;math&gt;gcd x&lt;/math&gt;. Then, turn the equation into &lt;math&gt;3\log x = 60&lt;/math&gt;, which yields &lt;math&gt;\log x = 20&lt;/math&gt;, or &lt;math&gt;x = 10^{20}&lt;/math&gt;.<br /> Factor this into &lt;math&gt;2^{20} \cdot 5^{20}&lt;/math&gt;, and add the two 20's, resulting in m, which is 40.<br /> Add &lt;math&gt;m&lt;/math&gt; to &lt;math&gt;2m + 2n&lt;/math&gt; (which is 840) to get &lt;math&gt;40+840 = \boxed{880}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Crappier Solution)==<br /> <br /> First simplifying the first and second equations, we get that <br /> <br /> &lt;cmath&gt;\log_{10}(x\cdot\text{gcd}(x,y)^2)=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_{10}(y\cdot\text{lcm}(x,y)^2)=570&lt;/cmath&gt;<br /> <br /> <br /> Thus, when the two equations are added, we have that<br /> &lt;cmath&gt;\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630&lt;/cmath&gt;<br /> When simplified, this equals <br /> &lt;cmath&gt;\log_{10}(x^3y^3)=630&lt;/cmath&gt;<br /> so this means that<br /> &lt;cmath&gt;x^3y^3=10^{630}&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;xy=10^{210}.&lt;/cmath&gt;<br /> <br /> Now, the following cannot be done on a proof contest but let's (intuitively) assume that &lt;math&gt;x&lt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both powers of &lt;math&gt;10&lt;/math&gt;. This means the first equation would simplify to &lt;cmath&gt;x^3=10^{60}&lt;/cmath&gt; and &lt;cmath&gt;y^3=10^{570}.&lt;/cmath&gt; Therefore, &lt;math&gt;x=10^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}&lt;/math&gt; and if we plug these values back, it works! &lt;math&gt;10^{20}&lt;/math&gt; has &lt;math&gt;20\cdot2=40&lt;/math&gt; total factors and &lt;math&gt;10^{190}&lt;/math&gt; has &lt;math&gt;190\cdot2=380&lt;/math&gt; so &lt;cmath&gt;3\cdot 40 + 2\cdot 380 = \boxed{880}.&lt;/cmath&gt;<br /> <br /> Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.<br /> <br /> ==Solution 3 (Easy Solution)==<br /> Let &lt;math&gt;x=10^a&lt;/math&gt; and &lt;math&gt;y=10^b&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;/math&gt;. Then the given equations become &lt;math&gt;3a=60&lt;/math&gt; and &lt;math&gt;3b=570&lt;/math&gt;. Therefore, &lt;math&gt;x=10^{20}=2^{20}\cdot5^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}=2^{190}\cdot5^{190}&lt;/math&gt;. Our answer is &lt;math&gt;3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=104555 2019 AIME I Problems/Problem 11 2019-03-16T00:05:22Z <p>Theultimate123: </p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integers lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;BM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> size(8cm);<br /> defaultpen(fontsize(8pt));<br /> pair A, B, C, I, IA, IB, IC;<br /> A=(0, 4sqrt(5));<br /> B=(-1, 0);<br /> C=(1, 0);<br /> I=incenter(A, B, C);<br /> IA=2*circumcenter(I,B,C)-I;<br /> IB=2*circumcenter(I,C,A)-I;<br /> IC=2*circumcenter(I,A,B)-I;<br /> <br /> draw(B -- A -- C); draw(IB -- IC); draw(incircle(A, B, C));<br /> draw(foot(IB, B, C) -- foot(IC, B, C));<br /> draw(circle(IA, length(IA-foot(IA, B, C))));<br /> draw(arc(IB, IB-(4sqrt(5), 0), IB-(0, 4sqrt(5))));<br /> draw(arc(IC, IC-(0, 4sqrt(5)), IC+(4sqrt(5), 0)));<br /> draw(circle(I, 2/sqrt(5)+sqrt(5)));<br /> <br /> dot(&quot;$A$&quot;, A, N);<br /> dot(&quot;$B$&quot;, B, SW);<br /> dot(&quot;$C$&quot;, C, SE);<br /> dot(&quot;$I$&quot;, I, N);<br /> dot(&quot;$I_A$&quot;, IA, S);<br /> dot(&quot;$I_B$&quot;, IB, NE);<br /> dot(&quot;$I_C$&quot;, IC, NW);<br /> &lt;/asy&gt;<br /> First assume that &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;, and scale up later. Notice that &lt;math&gt;\overline{I_BAI_C}\parallel\overline{BC}&lt;/math&gt;. Then, the height from &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so if &lt;math&gt;K=[ABC]&lt;/math&gt;, we know &lt;math&gt;K=\sqrt{x^2-1}&lt;/math&gt;. Then, if &lt;math&gt;r_D&lt;/math&gt; denotes the &lt;math&gt;D&lt;/math&gt;-exradius for &lt;math&gt;D\in\{A,B,C\}&lt;/math&gt; and &lt;math&gt;s=x+1&lt;/math&gt; denotes the semiperimeter, &lt;cmath&gt;r_A=\frac{K}{s-2}=\frac{K}{x-1},\;r_b=r_C=\frac{K}{s-x}=K,\text{ and }r=\frac{K}{s}=\frac{K}{x+1}.&lt;/cmath&gt;Then, if &lt;math&gt;X&lt;/math&gt; denotes the tangency point between the &lt;math&gt;B&lt;/math&gt;-excircle and &lt;math&gt;\overline{BC}&lt;/math&gt;, it is known that &lt;math&gt;BX=s&lt;/math&gt;, so &lt;math&gt;AI_B=s-1=x&lt;/math&gt;. Furthermore, &lt;math&gt;AI=\sqrt{(s-2)^2+r^2}=\sqrt{(x-1)^2+(K/(x+1))^2}&lt;/math&gt;. Then, &lt;cmath&gt;r+2r_A=II_A=II_B-r_B.&lt;/cmath&gt;It follows that<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> II_B&amp;=r+2r_A+r_B\\<br /> \sqrt{AI^2+AI_B^2}&amp;=\frac{K}{x+1}+\frac{2K}{x-1}+K\\<br /> \sqrt{x^2+(x-1)^2+\left(\frac{K}{x+1}\right)^2}&amp;=K\left(\frac1{x+1}+\frac2{x-1}+1\right)\\<br /> \frac{\sqrt{(x^2+(x-1)^2)(x+1)^2+x^2-1}}{x+1}&amp;=K\left(\frac{x^2+3x}{x^2-1}\right)\\<br /> \frac{\sqrt{2x^3(x+1)}}{x+1}&amp;=\frac{x(x+3)}{\sqrt{x^2-1}}\\<br /> 2x(x-1)&amp;=x^2+6x^2+9\\<br /> 0&amp;=x^2-8x-9\\<br /> &amp;=(x+1)(x-9),<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> whence &lt;math&gt;x=9&lt;/math&gt;. Then, since &lt;math&gt;\gcd(2,9,9)=1&lt;/math&gt;, the smallest possible perimeter is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104554 2019 AIME I Problems/Problem 6 2019-03-16T00:04:10Z <p>Theultimate123: </p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution 1 (Trig)==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 3 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Solution 4 (Algebraic Bashing)==<br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt;. We can use the right triangles in the problem to create equations. Let &lt;math&gt;a=NP, b=PK, c=NO, d=OM, e=OP, f=PC, and g=NC.&lt;/math&gt; We are trying to find &lt;math&gt;d.&lt;/math&gt; We can find &lt;math&gt;7&lt;/math&gt; equations. They are<br /> &lt;cmath&gt;4225+d^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;4225+d^2+16d+64=a^2+2ab+b^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2=64,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2+2ef+f^2=784,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2+2ef+f^2=g^2,&lt;/cmath&gt;<br /> and &lt;cmath&gt;g^2+784=a^2+2ab+b^2.&lt;/cmath&gt;<br /> We can subtract the fifth equation from the sixth equation to get &lt;math&gt;a^2-b^2=g^2-784.&lt;/math&gt; We can subtract the fourth equation from the third equation to get &lt;math&gt;a^2-b^2=c^2-64.&lt;/math&gt; Combining these equations gives &lt;math&gt;c^2-64=g^2-784&lt;/math&gt; so &lt;math&gt;g^2=c^2+720.&lt;/math&gt; Substituting this into the seventh equation gives &lt;math&gt;c^2+1504=a^2+2ab+b^2.&lt;/math&gt; Substituting this into the second equation gives &lt;math&gt;4225+d^2+16d+64=c^2+1504&lt;/math&gt;. Subtracting the first equation from this gives &lt;math&gt;16d+64=1504.&lt;/math&gt; Solving this equation, we find that &lt;math&gt;d=\boxed{090}.&lt;/math&gt;<br /> (Solution by DottedCaculator)<br /> <br /> ==Solution 5 (5-second PoP)==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair K, L, M, NN, X, O;<br /> K=(-sqrt(98^2+65^2)/2, 0);<br /> NN=(sqrt(98^2+65^2)/2, 0);<br /> L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));<br /> M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));<br /> X=foot(L, K, NN);<br /> O=extension(L, X, K, M);<br /> draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));<br /> draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);<br /> <br /> draw(rightanglemark(K, L, NN, 100));<br /> draw(rightanglemark(K, M, NN, 100));<br /> draw(rightanglemark(L, X, NN, 100));<br /> dot(&quot;$K$&quot;, K, SW);<br /> dot(&quot;$L$&quot;, L, unit(L));<br /> dot(&quot;$M$&quot;, M, unit(M));<br /> dot(&quot;$N$&quot;, NN, SE);<br /> dot(&quot;$X$&quot;, X, S);<br /> &lt;/asy&gt;<br /> Notice that &lt;math&gt;KLMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{KN}&lt;/math&gt; and &lt;math&gt;XOMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{ON}&lt;/math&gt;. Furthermore, &lt;math&gt;(XLN)&lt;/math&gt; is tangent to &lt;math&gt;\overline{KL}&lt;/math&gt;. Then, &lt;cmath&gt;KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,&lt;/cmath&gt;and &lt;math&gt;MO=KM-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_10&diff=104553 2019 AIME I Problems/Problem 10 2019-03-16T00:02:55Z <p>Theultimate123: </p> <hr /> <div>The 2019 AIME I takes place on March 13, 2019.<br /> <br /> ==Problem 10==<br /> For distinct complex numbers &lt;math&gt;z_1,z_2,\dots,z_{673}&lt;/math&gt;, the polynomial<br /> &lt;cmath&gt; (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 &lt;/cmath&gt;can be expressed as &lt;math&gt;x^{2019} + 20x^{2018} + 19x^{2017}+g(x)&lt;/math&gt;, where &lt;math&gt;g(x)&lt;/math&gt; is a polynomial with complex coefficients and with degree at most &lt;math&gt;2016&lt;/math&gt;. The value of<br /> &lt;cmath&gt; \left| \sum_{1 \le j &lt;k \le 673} z_jz_k \right| &lt;/cmath&gt;can be expressed in the form &lt;math&gt;\tfrac{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;.<br /> <br /> ==Solution 1==<br /> In order to begin this problem, we must first understand what it is asking for. The notation <br /> &lt;cmath&gt; \left| \sum_{1 \le j &lt;k \le 673} z_jz_k \right| &lt;/cmath&gt;<br /> simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or<br /> &lt;cmath&gt;(z_1z_2+z_1z_3+ \dots + z_1z_{672}+z_1z_{673})+(z_2z_3+z_2z_4+ \dots +z_2z_{673}) + (z_3z_4+z_3z_5+ \dots +z_3z_{673}) + \dots +z_{672}z_{673}.&lt;/cmath&gt; Call this sum &lt;math&gt;S&lt;/math&gt;. <br /> <br /> Now we can begin the problem. Rewrite the polynomial as &lt;math&gt;P=(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2) \dots (x-z_{673})(x-z_{673})(x-z_{673})&lt;/math&gt;. Then we have that the roots of &lt;math&gt;P&lt;/math&gt; are &lt;math&gt;z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}&lt;/math&gt;. <br /> <br /> By Vieta's formulas, we have that the sum of the roots of &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(-1)^1 * \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})&lt;/math&gt;. Thus, &lt;math&gt;z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.&lt;/math&gt; <br /> <br /> Similarly, we also have that the the sum of the roots of &lt;math&gt;P&lt;/math&gt; taken two at a time is &lt;math&gt;(-1)^2 * \dfrac{19}{1} = 19.&lt;/math&gt; This is equal to &lt;math&gt;z_1^2+z_1^2+z_1^2+z_1z_2+z_1z_2+z_1z_2+ \dots = \\ 3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673}) = 3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9S.&lt;/math&gt; <br /> <br /> Now we need to find and expression for &lt;math&gt;z_1^2+z_2^2+ \dots + z_{673}^2&lt;/math&gt; in terms of &lt;math&gt;S&lt;/math&gt;. We note that &lt;math&gt;(z_1+z_2+z_3+ \dots +z_{673})^2= (-20/3)^2=\dfrac{400}{9} \\ =(z_1^2+z_2^2+ \dots + z_{673}^2)+2(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673})=(z_1^2+z_2^2+ \dots + z_{673}^2)+2S.&lt;/math&gt; <br /> Thus, &lt;math&gt;z_1^2+z_2^2+ \dots + z_{673}^2= \dfrac{400}{9} -2S&lt;/math&gt;. <br /> <br /> Plugging this into our other Vieta equation, we have &lt;math&gt; 3 \left( \dfrac{400}{9} -2S \right) +9S = 19&lt;/math&gt;. This gives &lt;math&gt;S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}&lt;/math&gt;. Since 343 is relatively prime to 9, &lt;math&gt;m+n = 343+9 = \fbox{352}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> This is a quick fake solve using &lt;math&gt;z_q = 0&lt;/math&gt; where &lt;math&gt;3 \le q \le 673&lt;/math&gt; and only &lt;math&gt;z_1,z_2 \neq 0&lt;/math&gt; .<br /> <br /> By Vieta's, &lt;cmath&gt;3q_1+3q_2=-20&lt;/cmath&gt; and &lt;cmath&gt;3q_1^2+3q_2^2+9q_1q_2 = 19.&lt;/cmath&gt;<br /> Rearranging gives &lt;math&gt;q_1 + q_2 = \dfrac{-20}{3}&lt;/math&gt; and &lt;math&gt;3(q_1^2 + 2q_1q_2 + q_2^2) + 3q_1q_2 = 19&lt;/math&gt; giving &lt;math&gt; 3(q_1 + q_2)^2 + 3q_1q_2 =\dfrac{19}{3}&lt;/math&gt;.<br /> <br /> Substituting gives &lt;math&gt;3(\dfrac{400}{9}) + 3q_1q_2 = 19&lt;/math&gt; which simplifies to &lt;math&gt;\dfrac{400}{3} + 3q_1q_2 = \dfrac{57}{3}&lt;/math&gt;<br /> &lt;math&gt;3q_1q_2 = \dfrac{-343}{3}&lt;/math&gt;, &lt;math&gt;q_1q_2 = \dfrac{-343}{9}&lt;/math&gt;, &lt;math&gt;|\dfrac{-343}{9}|=\dfrac{343}{9}&lt;/math&gt;, &lt;math&gt;m+n = 343+9 = \fbox{352}.&lt;/math&gt;<br /> <br /> ~Ish_Sahh<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;x=\sum_{1\le j&lt;k\le 673} z_jz_k&lt;/math&gt;. By Vieta's, &lt;cmath&gt;3\sum_{i=1}^{673}z_i=-20\implies \sum_{i=1}^{673}z_i=-\frac{20}3.&lt;/cmath&gt;Then, consider the &lt;math&gt;19x^{2017}&lt;/math&gt; term. To produce the product of two roots, the two roots can either be either &lt;math&gt;(z_i,z_i)&lt;/math&gt; for some &lt;math&gt;i&lt;/math&gt;, or &lt;math&gt;(z_j,z_k)&lt;/math&gt; for some &lt;math&gt;j&lt;k&lt;/math&gt;. In the former case, this can happen in &lt;math&gt;\tbinom 32=3&lt;/math&gt; ways, and in the latter case, this can happen in &lt;math&gt;3^2=9&lt;/math&gt; ways. Hence,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 19=3\sum_{i=1}^{673} z_i^2+9\sum_{1\le j&lt;k\le 673} z_jz_k=3\left(\left(-\frac{20}3\right)^2-2x\right)+9x&amp;=\frac{400}3+3x\\<br /> \implies x&amp;=-\frac{343}9,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> and the requested sum is &lt;math&gt;343+9=\boxed{352}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=104551 2019 AIME I Problems/Problem 12 2019-03-16T00:01:44Z <p>Theultimate123: Note: I put my solution as Solution 1 as I think it is more elegant.</p> <hr /> <div>==Problem 12==<br /> Given &lt;math&gt;f(z) = z^2-19z&lt;/math&gt;, there are complex numbers &lt;math&gt;z&lt;/math&gt; with the property that &lt;math&gt;z&lt;/math&gt;, &lt;math&gt;f(z)&lt;/math&gt;, and &lt;math&gt;f(f(z))&lt;/math&gt; are the vertices of a right triangle in the complex plane with a right angle at &lt;math&gt;f(z)&lt;/math&gt;. There are positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; such that one such value of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;m+\sqrt{n}+11i&lt;/math&gt;. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Notice that we must have &lt;cmath&gt;\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in\mathbb I.&lt;/cmath&gt;However, &lt;math&gt;f(t)=t(t-20)&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{f(f(z))-f(z)}{f(z)-z}&amp;=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\<br /> &amp;=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\<br /> &amp;=(z-19)(z+1)\\<br /> &amp;=(z-9)^2-100.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Then, the real part of &lt;math&gt;(z-9)^2&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\text{Im}(z-9)=\text{Im}(z)=11&lt;/math&gt;, let &lt;math&gt;z-9=a+11i&lt;/math&gt;. Then, &lt;cmath&gt;100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.&lt;/cmath&gt;It follows that &lt;math&gt;z=9+\sqrt{221}+11i&lt;/math&gt;, and the requested sum is &lt;math&gt;9+221=\boxed{230}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> We will use the fact that segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; are perpendicular in the complex plane if and only if &lt;math&gt;\frac{a-b}{b-c}\in i\mathbb{R}&lt;/math&gt;. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. <br /> <br /> Now to apply this: <br /> &lt;cmath&gt;\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{z^2-20z}{z^4-38z^3+341z^2+380z}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{z(z-20)}{z(z+1)(z-19)(z-20)}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{(z+1)(z-19)}\in i\mathbb{R}&lt;/cmath&gt;<br /> <br /> The factorization of the nasty denominator above is made easier with the intuition that &lt;math&gt;(z-20)&lt;/math&gt; must be a divisor for the problem to lead anywhere. Now we know &lt;math&gt;(z+1)(z-19)\in i\mathbb{R}&lt;/math&gt; so using the fact that the imaginary part of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;11i&lt;/math&gt; and calling the real part r, <br /> <br /> &lt;cmath&gt;(r+1+11i)(r-19+11i)\in i\mathbb{R}&lt;/cmath&gt;<br /> &lt;cmath&gt;r^2-18r-140=0&lt;/cmath&gt;<br /> <br /> solving the above quadratic yields &lt;math&gt;r=9+\sqrt{221}&lt;/math&gt; so our answer is &lt;math&gt;9+221=\boxed{230}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_13&diff=104550 2019 AIME I Problems/Problem 13 2019-03-16T00:00:14Z <p>Theultimate123: Note: I put my solution as Solution 1 as I feel like it has a diagram and is formatted well.</p> <hr /> <div>==Problem 13==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and &lt;math&gt;CA=6&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on ray &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;AB&lt;AD&lt;AE&lt;/math&gt;. The point &lt;math&gt;F \neq C&lt;/math&gt; is a point of intersection of the circumcircles of &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle EBC&lt;/math&gt; satisfying &lt;math&gt;DF=2&lt;/math&gt; and &lt;math&gt;EF=7&lt;/math&gt;. Then &lt;math&gt;BE&lt;/math&gt; can be expressed as &lt;math&gt;\tfrac{a+b\sqrt{c}}{d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are positive integers such that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are relatively prime, and &lt;math&gt;c&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> size(10cm);<br /> pair A, B, C, D, EE, F, X;<br /> B=dir(270-aCos(9/16));<br /> C=dir(270+aCos(9/16));<br /> A=intersectionpoint(circle((0, 0), 1), (B+0.01*(1, 3sqrt(7))) -- (B+100*(1, 3sqrt(7))));<br /> D=B-5/16*(sqrt(2)+1)*(A-B);<br /> EE=B-(5+21*sqrt(2))/16*(A-B);<br /> F=intersectionpoints(circumcircle(A, C, D), circumcircle(B, C, EE));<br /> X=extension(A, B, C, F);<br /> <br /> draw(B -- C -- A -- EE -- F -- C); draw(D -- F);<br /> draw(circumcircle(A, C, D)); draw(circumcircle(C, EE, F));<br /> <br /> dot(&quot;$A$&quot;, A, N);<br /> dot(&quot;$B$&quot;, B, NW);<br /> dot(&quot;$C$&quot;, C, E);<br /> dot(&quot;$D$&quot;, D, SW);<br /> dot(&quot;$E$&quot;, EE, SW);<br /> dot(&quot;$F$&quot;, F, W);<br /> &lt;/asy&gt;<br /> Notice that &lt;cmath&gt;\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.&lt;/cmath&gt;By the Law of Cosines, &lt;cmath&gt;\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.&lt;/cmath&gt;Then, &lt;cmath&gt;DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.&lt;/cmath&gt;Let &lt;math&gt;X=\overline{AB}\cap\overline{CF}&lt;/math&gt;, &lt;math&gt;a=XB&lt;/math&gt;, and &lt;math&gt;b=XD&lt;/math&gt;. Then, &lt;cmath&gt;XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.&lt;/cmath&gt;However, since &lt;math&gt;\triangle XFD\sim\triangle XAC&lt;/math&gt;, &lt;math&gt;XF=\tfrac{4+a}3&lt;/math&gt;, but since &lt;math&gt;\triangle XFE\sim\triangle XBC&lt;/math&gt;, &lt;cmath&gt;\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,&lt;/cmath&gt;and the requested sum is &lt;math&gt;5+21+2+4=\boxed{032}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Define &lt;math&gt;\omega_1&lt;/math&gt; to be the circumcircle of &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; to be the circumcircle of &lt;math&gt;\triangle EBC&lt;/math&gt;.<br /> <br /> Because of exterior angles, <br /> <br /> &lt;math&gt;\angle ACB = \angle CBE - \angle CAD&lt;/math&gt;<br /> <br /> But &lt;math&gt;\angle CBE = \angle CFE&lt;/math&gt; because &lt;math&gt;CBFE&lt;/math&gt; is cyclic. In addition, &lt;math&gt;\angle CAD = \angle CFD&lt;/math&gt; because &lt;math&gt;CAFD&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle ACB = \angle CFE - \angle CFD&lt;/math&gt;. But &lt;math&gt;\angle CFE - \angle CFD = \angle DFE&lt;/math&gt;, so &lt;math&gt;\angle ACB = \angle DFE&lt;/math&gt;. Using Law of Cosines on &lt;math&gt;\triangle ABC&lt;/math&gt;, we can figure out that &lt;math&gt;\cos(\angle ACB) = \frac{3}{4}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = \angle DFE&lt;/math&gt;, &lt;math&gt;\cos(\angle DFE) = \frac{3}{4}&lt;/math&gt;. We are given that &lt;math&gt;DF = 2&lt;/math&gt; and &lt;math&gt;FE = 7&lt;/math&gt;, so we can use Law of Cosines on &lt;math&gt;\triangle DEF&lt;/math&gt; to find that &lt;math&gt;DE = 4\sqrt{2}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the intersection of segment &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{CF}&lt;/math&gt;. Using Power of a Point with respect to &lt;math&gt;G&lt;/math&gt; within &lt;math&gt;\omega_1&lt;/math&gt;, we find that &lt;math&gt;AG \cdot GD = CG \cdot GF&lt;/math&gt;. We can also apply Power of a Point with respect to &lt;math&gt;G&lt;/math&gt; within &lt;math&gt;\omega_2&lt;/math&gt; to find that &lt;math&gt;CG \cdot GF = BG \cdot GE&lt;/math&gt;. Therefore, &lt;math&gt;AG \cdot GD = BG \cdot GE&lt;/math&gt;.<br /> <br /> &lt;math&gt;AG \cdot GD = BG \cdot GE&lt;/math&gt;<br /> <br /> &lt;math&gt;(AB + BG) \cdot GD = BG \cdot (GD + DE)&lt;/math&gt;<br /> <br /> &lt;math&gt;AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE&lt;/math&gt;<br /> <br /> &lt;math&gt;AB \cdot GD = BG \cdot DE&lt;/math&gt;<br /> <br /> &lt;math&gt;4 \cdot GD = BG \cdot 4\sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;GD = BG \cdot \sqrt{2}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\triangle GAC&lt;/math&gt; is similar to &lt;math&gt;\triangle GFD&lt;/math&gt;. &lt;math&gt;GF = \frac{BG + 4}{3}&lt;/math&gt;. Also note that &lt;math&gt;\triangle GBC&lt;/math&gt; is similar to &lt;math&gt;\triangle GFE&lt;/math&gt;, which gives us &lt;math&gt;GF = \frac{7 \cdot BG}{5}&lt;/math&gt;. Solving this system of linear equations, we get &lt;math&gt;BG = \frac{5}{4}&lt;/math&gt;. Now, we can solve for &lt;math&gt;BE&lt;/math&gt;, which is equal to &lt;math&gt;BG(\sqrt{2} + 1) + 4\sqrt{2}&lt;/math&gt;. This simplifies to &lt;math&gt;\frac{5 + 21\sqrt{2}}{4}&lt;/math&gt;, which means our answer is &lt;math&gt;\boxed{032}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Construct &lt;math&gt;FC&lt;/math&gt; and let &lt;math&gt;FC\cap AE=K&lt;/math&gt;. Let &lt;math&gt;FK=x&lt;/math&gt;. Using &lt;math&gt;\triangle FKE\sim \triangle BKC&lt;/math&gt;, &lt;cmath&gt;BK=\frac{5}{7}x&lt;/cmath&gt; Using &lt;math&gt;\triangle FDK\sim ACK&lt;/math&gt;, it can be found taht &lt;cmath&gt;3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}&lt;/cmath&gt; This also means that &lt;math&gt;BK=\frac{21}{4}-4=\frac{5}{4}&lt;/math&gt;. It suffices to find &lt;math&gt;KE&lt;/math&gt;. It is easy to see the following: &lt;cmath&gt;180-\angle ABC=\angle KBC=\angle KFE&lt;/cmath&gt; Using reverse Law of Cosines on &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}&lt;/math&gt;. Using Law of Cosines on &lt;math&gt;\triangle EFK&lt;/math&gt; gives &lt;math&gt;KE=\frac{21\sqrt 2}{4}&lt;/math&gt;, so &lt;math&gt;BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}&lt;/math&gt;.<br /> -franchester<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=104549 2019 AIME I Problems/Problem 15 2019-03-15T23:57:59Z <p>Theultimate123: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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;.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;$O$&quot;, O, S);<br /> dot(&quot;$A$&quot;, A, A);<br /> dot(&quot;$B$&quot;, B, B);<br /> dot(&quot;$P$&quot;, P, dir(70));<br /> dot(&quot;$Q$&quot;, Q, dir(200));<br /> dot(&quot;$O_1$&quot;, O1, SW);<br /> dot(&quot;$O_2$&quot;, O2, SE);<br /> dot(&quot;$X$&quot;, X, X);<br /> dot(&quot;$Y$&quot;, Y, Y);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we hav e &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=104548 2019 AIME I Problems/Problem 15 2019-03-15T23:57:30Z <p>Theultimate123: Note: I put my solution as Solution 1 as I feel like it is better organized and clearer than the other one.</p> <hr /> <div>==Problem 15==<br /> <br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a chord of a circle &lt;math&gt;\omega&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be a point on the chord &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and is internally tangent to &lt;math&gt;\omega&lt;/math&gt;. Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Line &lt;math&gt;PQ&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Assume that &lt;math&gt;AP=5&lt;/math&gt;, &lt;math&gt;PB=3&lt;/math&gt;, &lt;math&gt;XY=11&lt;/math&gt;, and &lt;math&gt;PQ^2 = \tfrac{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;.<br /> <br /> ==Solution 1==<br /> <br /> [asy]<br /> size(8cm);<br /> pair O, A, B, P, O1, O2, Q, X, Y;<br /> O=(0, 0);<br /> A=dir(140); B=dir(40);<br /> P=(3A+5B)/8;<br /> O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br /> O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br /> Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)));<br /> X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br /> Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br /> <br /> draw(circle(O, 1));<br /> draw(circle(O1, length(A-O1)));<br /> draw(circle(O2, length(B-O2)));<br /> draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br /> <br /> dot(&quot;&lt;math&gt;O&lt;/math&gt;&quot;, O, S);<br /> dot(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, A, A);<br /> dot(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, B, B);<br /> dot(&quot;&lt;math&gt;P&lt;/math&gt;&quot;, P, dir(70));<br /> dot(&quot;&lt;math&gt;Q&lt;/math&gt;&quot;, Q, dir(200));<br /> dot(&quot;&lt;math&gt;O_1&lt;/math&gt;&quot;, O1, SW);<br /> dot(&quot;&lt;math&gt;O_2&lt;/math&gt;&quot;, O2, SE);<br /> dot(&quot;&lt;math&gt;X&lt;/math&gt;&quot;, X, X);<br /> dot(&quot;&lt;math&gt;Y&lt;/math&gt;&quot;, Y, Y);<br /> [/asy]<br /> Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;, respectively. There is a homothety at &lt;math&gt;A&lt;/math&gt; sending &lt;math&gt;\omega&lt;/math&gt; to &lt;math&gt;\omega_1&lt;/math&gt; that sends &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;O_1&lt;/math&gt;, so &lt;math&gt;\overline{OO_2}\parallel\overline{O_1P}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{OO_1}\parallel\overline{O_2P}&lt;/math&gt;, so &lt;math&gt;OO_1PO_2&lt;/math&gt; is a parallelogram. Moreover, &lt;cmath&gt;\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,&lt;/cmath&gt;whence &lt;math&gt;OO_1O_2Q&lt;/math&gt; is cyclic. However, &lt;cmath&gt;OO_1=O_2P=O_2Q,&lt;/cmath&gt;so &lt;math&gt;OO_1O_2Q&lt;/math&gt; is an isosceles trapezoid. Since &lt;math&gt;\overline{O_1O_2}\perp\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{OQ}\perp\overline{XY}&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{XY}&lt;/math&gt;.<br /> <br /> By Power of a Point, &lt;math&gt;PX\cdot PY=PA\cdot PB=15&lt;/math&gt;. Since &lt;math&gt;PX+PY=XY=11&lt;/math&gt;, &lt;cmath&gt;XP=\frac{11-\sqrt{61}}2\implies PQ=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,&lt;/cmath&gt;and the requested sum is &lt;math&gt;61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 2==<br /> <br /> Firstly we need to notice that &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;. Assume the center of circle &lt;math&gt;w, w_1, w_2&lt;/math&gt; are &lt;math&gt;O, O_1, O_2&lt;/math&gt;, respectively. Then &lt;math&gt;A, O_2, O&lt;/math&gt; are collinear and &lt;math&gt;O, O_1, B&lt;/math&gt; are collinear. Link &lt;math&gt;O_1P, O_2P, O_1Q, O_2Q&lt;/math&gt;. Notice that, &lt;math&gt;\angle B=\angle A=\angle APO_2=\angle BPO_1&lt;/math&gt;. As a result, &lt;math&gt;PO_1\parallel O_2O&lt;/math&gt; and &lt;math&gt;QO_1\parallel O_2P&lt;/math&gt;. So we have parallelogram &lt;math&gt;PO_2O_1O&lt;/math&gt;. So &lt;math&gt;\angle O_2PO_1=\angle O&lt;/math&gt; Notice that, &lt;math&gt;O_1O_2\bot PQ&lt;/math&gt; and &lt;math&gt;O_1O_2&lt;/math&gt; divide &lt;math&gt;PQ&lt;/math&gt; into two equal length pieces, So we have &lt;math&gt;\angle O_2PO_1=\angle O_2QO_1=\angle O&lt;/math&gt;. As a result, &lt;math&gt;O_2, Q, O, O_1,&lt;/math&gt; lie on one circle. So &lt;math&gt;\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P&lt;/math&gt;. Notice that &lt;math&gt;\angle O_1PQ+\angle O_2O_1P=90^{\circ}&lt;/math&gt;, we have &lt;math&gt;\angle OQP=90^{\circ}&lt;/math&gt;. As a result, &lt;math&gt;OQ\bot PQ&lt;/math&gt;. So &lt;math&gt;Q&lt;/math&gt; is the middle point of &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> Back to our problem. Assume &lt;math&gt;XP=x&lt;/math&gt;, &lt;math&gt;PY=y&lt;/math&gt; and &lt;math&gt;x&lt;y&lt;/math&gt;. Then we have &lt;math&gt;AP\cdot PB=XP\cdot PY&lt;/math&gt;, that is, &lt;math&gt;xy=15&lt;/math&gt;. Also, &lt;math&gt;XP+PY=x+y=XY=11&lt;/math&gt;. Solve these above, we have &lt;math&gt;x=\frac{11-\sqrt{61}}{2}=XP&lt;/math&gt;. As a result, we hav e &lt;math&gt;PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}&lt;/math&gt;. So, we have &lt;math&gt;PQ^2=\frac{61}{4}&lt;/math&gt;. As a result, our answer is &lt;math&gt;m+n=61+4=\boxed{065}&lt;/math&gt;.<br /> <br /> Solution By BladeRunnerAUG (Fanyuchen20020715).<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=104022 2014 AIME II Problems/Problem 10 2019-03-04T04:56:02Z <p>Theultimate123: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)== <br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Notice that &lt;cmath&gt;\frac1{w+z} = \frac{w+z}{wz} \implies 0 = w^2 + wz + z^2 = \frac{w^3-z^3}{w-z}.&lt;/cmath&gt;<br /> Hence, &lt;math&gt;w=ze^{2\pi i/3},ze^{4\pi i/3}&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; is an equilateral triangle with circumradius &lt;math&gt;2014&lt;/math&gt;. Then, &lt;cmath&gt;[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,&lt;/cmath&gt;<br /> and the answer is &lt;math&gt;3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=104021 2014 AIME II Problems/Problem 10 2019-03-04T04:55:06Z <p>Theultimate123: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)== <br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Notice that<br /> \begin{align*}<br /> \frac1{w+z} &amp;= \frac{w+z}{wz}\\<br /> \implies wz &amp;= {(w+z)}^2\\<br /> \implies w^2 + wz + z^2 &amp;= 0\\<br /> \implies \frac{w^3-z^3}{w-z} &amp;= 0\\<br /> \end{align*}<br /> Hence, &lt;math&gt;w=ze^{2\pi i/3},ze^{4\pi i/3}&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; is an equilateral triangle with circumradius &lt;math&gt;2014&lt;/math&gt;. Then, &lt;cmath&gt;[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,&lt;/cmath&gt;<br /> and the answer is &lt;math&gt;3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=104020 2014 AIME II Problems/Problem 10 2019-03-04T04:54:20Z <p>Theultimate123: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)== <br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Our equation can be simplified like the following.<br /> \begin{align*}<br /> \frac{1}{w+z} &amp;= \frac{w+z}{wz}&lt;math&gt;\\<br /> \implies wz &amp;= {(w+z)}^2\\<br /> \implies w^2 + wz + z^2 &amp;= 0\\<br /> \implies \frac{w^3-z^3}{w-z} &amp;= 0\\<br /> \end{align*}<br /> Hence, &lt;/math&gt;w=ze^{2\pi i/3},ze^{4\pi i/3}&lt;math&gt;, and &lt;/math&gt;P&lt;math&gt; is an equilateral triangle with circumradius &lt;/math&gt;2014&lt;math&gt;. Then, &lt;cmath&gt;[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,&lt;/cmath&gt;<br /> and the answer is &lt;/math&gt;3\cdot 1007^2\equiv 3\cdot 7^2\equiv\boxed{147}\pmod{1000}$.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_4&diff=100592 1990 AHSME Problems/Problem 4 2019-01-19T04:35:20Z <p>Theultimate123: /* Problem */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> draw((0,0)--(16,0)--(21,5*sqrt(3))--(5,5*sqrt(3))--cycle,dot);<br /> draw((5,5*sqrt(3))--(1,5*sqrt(3))--(16,0),dot);<br /> MP(&quot;A&quot;,(0,0),S);MP(&quot;B&quot;,(16,0),S);MP(&quot;C&quot;,(21,5sqrt(3)),NE);MP(&quot;D&quot;,(5,5sqrt(3)),N);MP(&quot;E&quot;,(1,5sqrt(3)),N);<br /> MP(&quot;16&quot;,(8,0),S);MP(&quot;10&quot;,(18.5,5sqrt(3)/2),E);MP(&quot;4&quot;,(3,5sqrt(3)),N);<br /> dot((4,4sqrt(3)));<br /> MP(&quot;F&quot;,(4,4sqrt(3)),dir(210));<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a parallelogram with &lt;math&gt;\angle{ABC}=120^\circ, AB=16&lt;/math&gt; and &lt;math&gt;BC=10.&lt;/math&gt; Extend &lt;math&gt;\overline{CD}&lt;/math&gt; through &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt; so that &lt;math&gt;DE=4.&lt;/math&gt; If &lt;math&gt;\overline{BE}&lt;/math&gt; intersects &lt;math&gt;\overline{AD}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt;, then &lt;math&gt;FD&lt;/math&gt; is closest to<br /> <br /> &lt;math&gt;\text{(A) } 1\quad<br /> \text{(B) } 2\quad<br /> \text{(C) } 3\quad<br /> \text{(D) } 4\quad<br /> \text{(E) } 5&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;DFE&lt;/math&gt; and &lt;math&gt;ADB&lt;/math&gt; are similar triangles, so &lt;math&gt;FD&lt;/math&gt; is one quarter the length of the corresponding side &lt;math&gt;AF&lt;/math&gt;.<br /> Thus it is one fifth of the length of &lt;math&gt;AD&lt;/math&gt;, which means &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=3|num-a=5}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&diff=97256 2010 USAMO Problems/Problem 1 2018-08-15T22:13:16Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;AXYZB&lt;/math&gt; be a convex pentagon inscribed in a semicircle of diameter<br /> &lt;math&gt;AB&lt;/math&gt;. Denote by &lt;math&gt;P, Q, R, S&lt;/math&gt; the feet of the perpendiculars from &lt;math&gt;Y&lt;/math&gt; onto<br /> lines &lt;math&gt;AX, BX, AZ, BZ&lt;/math&gt;, respectively. Prove that the acute angle<br /> formed by lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; is half the size of &lt;math&gt;\angle XOZ&lt;/math&gt;, where<br /> &lt;math&gt;O&lt;/math&gt; is the midpoint of segment &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\alpha = \angle BAZ&lt;/math&gt;, &lt;math&gt;\beta = \angle ABX&lt;/math&gt;.<br /> Since &lt;math&gt;XY&lt;/math&gt; is a chord of the circle with diameter &lt;math&gt;AB&lt;/math&gt;,<br /> &lt;math&gt;\angle XAY = \angle XBY = \gamma&lt;/math&gt;. From the chord &lt;math&gt;YZ&lt;/math&gt;,<br /> we conclude &lt;math&gt;\angle YAZ = \angle YBZ = \delta&lt;/math&gt;.<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import olympiad;<br /> <br /> // Scale<br /> unitsize(1inch);<br /> real r = 1.75;<br /> <br /> // Semi-circle: centre O, radius r, diameter A--B.<br /> pair O = (0,0); dot(O); label(&quot;$O$&quot;, O, plain.S);<br /> pair A = r * plain.W; dot(A); label(&quot;$A$&quot;, A, unit(A));<br /> pair B = r * plain.E; dot(B); label(&quot;$B$&quot;, B, unit(B));<br /> draw(arc(O, r, 0, 180)--cycle);<br /> <br /> // points X, Y, Z<br /> real alpha = 22.5;<br /> real beta = 15;<br /> real delta = 30;<br /> pair X = r * dir(180 - 2*beta); dot(X); label(&quot;$X$&quot;, X, unit(X));<br /> pair Y = r * dir(2*(alpha + delta)); dot(Y); label(&quot;$Y$&quot;, Y, unit(Y));<br /> pair Z = r * dir(2*alpha); dot(Z); label(&quot;$Z$&quot;, Z, unit(Z));<br /> <br /> // Feet of perpendiculars from Y<br /> pair P = foot(Y, A, X); dot(P); label(&quot;$P$&quot;, P, unit(P-Y)); dot(P);<br /> pair Q = foot(Y, B, X); dot(P); label(&quot;$Q$&quot;, Q, unit(A-Q)); dot(Q);<br /> pair R = foot(Y, B, Z); dot(R); label(&quot;$S$&quot;, R, unit(R-Y)); dot(R);<br /> pair S = foot(Y, A, Z); dot(S); label(&quot;$R$&quot;, S, unit(B-S)); dot(S);<br /> pair T = foot(Y, A, B); dot(T); label(&quot;$T$&quot;, T, unit(T-Y)); dot(T);<br /> <br /> // Segments<br /> draw(B--X); draw(B--Y); draw(B--R);<br /> draw(A--Z); draw(A--Y); draw(A--P);<br /> draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);<br /> draw(R--T); draw(P--T);<br /> <br /> // Right angles<br /> draw(rightanglemark(A, X, B, 3));<br /> draw(rightanglemark(A, Y, B, 3));<br /> draw(rightanglemark(A, Z, B, 3));<br /> draw(rightanglemark(A, P, Y, 3));<br /> draw(rightanglemark(Y, R, B, 3));<br /> draw(rightanglemark(Y, S, A, 3));<br /> draw(rightanglemark(B, Q, Y, 3));<br /> <br /> // Acute angles<br /> import markers;<br /> void langle(pair A, pair B, pair C, string l=&quot;&quot;, real r=40, int n=1, int nm = 0)<br /> {<br /> string sl = &quot;$\scriptstyle{&quot; + l + &quot;}$&quot;;<br /> marker m = (nm &gt; 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;<br /> markangle(Label(sl), radius=r, n=n, A, B, C, m);<br /> }<br /> langle(B, A, Z, &quot;\alpha&quot; );<br /> langle(X, B, A, &quot;\beta&quot;, n=2);<br /> langle(Y, A, X, &quot;\gamma&quot;, nm=1);<br /> langle(Y, B, X, &quot;\gamma&quot;, nm=1);<br /> langle(Z, A, Y, &quot;\delta&quot;, nm=2);<br /> langle(Z, B, Y, &quot;\delta&quot;, nm=2);<br /> langle(R, S, Y, &quot;\alpha+\delta&quot;, r=23);<br /> langle(Y, Q, P, &quot;\beta+\gamma&quot;, r=23);<br /> langle(R, T, P, &quot;\chi&quot;, r=15);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> Triangles &lt;math&gt;BQY&lt;/math&gt; and &lt;math&gt;APY&lt;/math&gt; are both right-triangles, and share the<br /> angle &lt;math&gt;\gamma&lt;/math&gt;, therefore they are similar, and so the ratio &lt;math&gt;PY :<br /> YQ = AY : YB&lt;/math&gt;. Now by [[Thales' theorem]] the angles &lt;math&gt;\angle AXB =<br /> \angle AYB = \angle AZB&lt;/math&gt; are all right-angles. Also, &lt;math&gt;\angle PYQ&lt;/math&gt;,<br /> being the fourth angle in a quadrilateral with 3 right-angles is<br /> again a right-angle. Therefore &lt;math&gt;\triangle PYQ \sim \triangle AYB&lt;/math&gt; and<br /> &lt;math&gt;\angle YQP = \angle YBA = \gamma + \beta&lt;/math&gt;.<br /> Similarly, &lt;math&gt;RY : YS = AY : YB&lt;/math&gt;, and so &lt;math&gt;\angle YRS = \angle YAB = \alpha + \delta&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;RY&lt;/math&gt; is perpendicular to &lt;math&gt;AZ&lt;/math&gt; so the direction &lt;math&gt;RY&lt;/math&gt; is &lt;math&gt;\alpha&lt;/math&gt; counterclockwise from the vertical, and since &lt;math&gt;\angle YRS = \alpha + \delta&lt;/math&gt; we see that &lt;math&gt;SR&lt;/math&gt; is &lt;math&gt;\delta&lt;/math&gt; clockwise from the vertical. (Draw an actual vertical line segment if necessary.)<br /> <br /> Similarly, &lt;math&gt;QY&lt;/math&gt; is perpendicular to &lt;math&gt;BX&lt;/math&gt; so the direction &lt;math&gt;QY&lt;/math&gt; is &lt;math&gt;\beta&lt;/math&gt; clockwise from the vertical, and since &lt;math&gt;\angle YQP&lt;/math&gt; is &lt;math&gt;\gamma + \beta&lt;/math&gt; we see that &lt;math&gt;QY&lt;/math&gt; is &lt;math&gt;\gamma&lt;/math&gt; counterclockwise from the vertical.<br /> <br /> Therefore the lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; intersect at an angle &lt;math&gt;\chi = \gamma<br /> + \delta&lt;/math&gt;. Now by the central angle theorem &lt;math&gt;2\gamma = \angle XOY&lt;/math&gt;<br /> and &lt;math&gt;2\delta = \angle YOZ&lt;/math&gt;, and so &lt;math&gt;2(\gamma + \delta) = \angle XOZ&lt;/math&gt;,<br /> and we are done.<br /> <br /> ''Note that &lt;math&gt;RTQY&lt;/math&gt; is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?''<br /> <br /> ===Footnote===<br /> We can prove a bit more. Namely, the extensions of the segments<br /> &lt;math&gt;RS&lt;/math&gt; and &lt;math&gt;PQ&lt;/math&gt; meet at a point on the diameter &lt;math&gt;AB&lt;/math&gt; that is vertically<br /> below the point &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;YS = AY \sin(\delta)&lt;/math&gt; and is inclined &lt;math&gt;\alpha&lt;/math&gt; counterclockwise<br /> from the vertical, the point &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;AY \sin(\delta) \sin(\alpha)&lt;/math&gt;<br /> horizontally to the right of &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;AS = AY \cos(\delta)&lt;/math&gt;, so &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;AS \sin(\alpha) = AY<br /> \cos(\delta)\sin(\alpha)&lt;/math&gt; vertically above the diameter &lt;math&gt;AB&lt;/math&gt;. Also,<br /> the segment &lt;math&gt;SR&lt;/math&gt; is inclined &lt;math&gt;\delta&lt;/math&gt; clockwise from the vertical,<br /> so if we extend it down from &lt;math&gt;S&lt;/math&gt; towards the diameter &lt;math&gt;AB&lt;/math&gt; it will<br /> meet the diameter at a point which is<br /> &lt;math&gt;AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)&lt;/math&gt;<br /> horizontally to the left of &lt;math&gt;S&lt;/math&gt;. This places the intersection point<br /> of &lt;math&gt;RS&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; vertically below &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Similarly, and by symmetry the intersection point of &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;<br /> is directly below &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AB&lt;/math&gt;, so the lines through &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt;<br /> meet at a point &lt;math&gt;T&lt;/math&gt; on the diameter that is vertically below &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> ===Footnote to the Footnote===<br /> The Footnote's claim is more easily proved as follows.<br /> <br /> Note that because &lt;math&gt;\angle{QPY}&lt;/math&gt; and &lt;math&gt;\angle{YAB}&lt;/math&gt; are both complementary to &lt;math&gt;\beta + \gamma&lt;/math&gt;, they must be equal. Now, let &lt;math&gt;PQ&lt;/math&gt; intersect diameter &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;T'&lt;/math&gt;. Then &lt;math&gt;PYT'A&lt;/math&gt; is cyclic and so &lt;math&gt;\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ&lt;/math&gt;. Hence &lt;math&gt;T'YSB&lt;/math&gt; is cyclic as well, and so we deduce that &lt;math&gt;\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.&lt;/math&gt; Hence &lt;math&gt;S, R, T'&lt;/math&gt; are collinear and so &lt;math&gt;T = T'&lt;/math&gt;. This proves the Footnote.<br /> <br /> ===Footnote to the Footnote to the Footnote===<br /> The Footnote's claim can be proved even more easily as follows.<br /> <br /> Drop an altitude from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; at point &lt;math&gt;T&lt;/math&gt;. Notice that &lt;math&gt;P, Q, T&lt;/math&gt; are collinear because they form the Simson line of &lt;math&gt;\triangle AXB&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt;. Also notice that &lt;math&gt;P, Q, T&lt;/math&gt; are collinear because they form the Simson line of &lt;math&gt;\triangle AZB&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt;. Since &lt;math&gt;T&lt;/math&gt; is at the diameter &lt;math&gt;AB&lt;/math&gt;, lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; must intersect at the diameter.<br /> <br /> ===Another footnote===<br /> There is another, more simpler solution using Simson lines. Can you find it?<br /> <br /> ==Operation Diagram==<br /> Of course, as with any geometry problem, DRAW A HUGE DIAGRAM spanning at least one page. And label all your right angles, noting &lt;math&gt;rectangles PXQY&lt;/math&gt; and &lt;math&gt;YSZR&lt;/math&gt;. It looks like there are a couple of key angles we need to diagram. Let's take &lt;math&gt;\angle{ZAB} = \alpha, \angle{XBA} = \beta, \angle{YAZ} = \angle{YBZ} = \delta&lt;/math&gt;. From there &lt;math&gt;\angle{XOZ}=180^\circ - \angle{XOA}-\angle{ZOB}=180-2(\beta + \alpha)&lt;/math&gt;.<br /> <br /> Move on to the part about the intersection of &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt;. Call the intersection &lt;math&gt;J&lt;/math&gt;. Note that by Simson Lines from point &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle{ABX}&lt;/math&gt; and &lt;math&gt;\triangle{AZB}&lt;/math&gt;, &lt;math&gt;YJ&lt;/math&gt; is perpendicular to &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;AB&lt;/math&gt;. Immediately note that we are trying to show that &lt;math&gt;\angle{PJS} = 90 - \beta - \alpha&lt;/math&gt;. <br /> <br /> It suffices to show that referencing quadrilateral &lt;math&gt;QR~J&lt;/math&gt;, where &lt;math&gt;~&lt;/math&gt; represents the intersection of &lt;math&gt;XB, AZ&lt;/math&gt;, we have reflex &lt;math&gt;\angle{Q~R} + \angle{BQJ} + \angle{ARJ} = 270 + \alpha + \beta&lt;/math&gt;. Note that the reflex angle is &lt;math&gt;180^\circ + \angle{A~X} = 180^\circ + (90^\circ - \angle{XA*}) = 270^\circ - ((90 - \beta) - \alpha) = 180 ^\circ + \alpha + \beta&lt;/math&gt;, therefore it suffices to show that &lt;math&gt;\angle{BQJ} + \angle{ARJ} = 90^\circ&lt;/math&gt;. To make this proof more accessible, note that via (cyclic) rectangles &lt;math&gt;PXQY&lt;/math&gt; and &lt;math&gt;YSZR&lt;/math&gt;, it suffices to prove &lt;math&gt;\angle{YPJ} + \angle{YSJ} = 90^\circ&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;\angle{YPJ} = \angle{YPQ} = \angle{YXQ} = \angle{YXB} = \angle{YAB} = \alpha + \delta&lt;/math&gt;.<br /> Note &lt;math&gt;\angle{YSJ} = \angle{YSR} = \angle{YZR} = \angle{YZA} = \angle{YBA} = \angle{YBX} + \angle{XBA} = ((90^\circ - \alpha) - \delta - \beta) + \beta = 90^\circ - \alpha - \delta&lt;/math&gt;, which completes the proof.<br /> <br /> ===Footnote to Operation Diagram===<br /> For reference/feasibility records: took expiLnCalc ~56 minutes (consecutively).<br /> During the problem expiLnCalc realized that the inclusion of &lt;math&gt;\delta&lt;/math&gt; was necessary when trying to show that &lt;math&gt;\angle{YSJ}+\angle{YPJ}=90^\circ&lt;/math&gt;. Don't be afraid to attempt several different strategies, and always be humble!<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> currentpicture=new picture;<br /> size(12cm);<br /> pair O, A, B, X, Y, Z, P, Q, R, SS, T;<br /> O=(0, 0);<br /> A=(-1, 0);<br /> B=(1, 0);<br /> X=(Cos(144), Sin(144));<br /> Y=(Cos(105), Sin(105));<br /> Z=(Cos(27), Sin(27));<br /> P=foot(Y, A, X);<br /> Q=foot(Y, B, X);<br /> R=foot(Y, A, Z);<br /> SS=foot(Y, B, Z);<br /> T=foot(Y, A, B);<br /> dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T);<br /> draw(arc(O, 1, 0, 180));<br /> draw(circumcircle(T, A, Y), dotted);<br /> draw(circumcircle(T, B, Y), dotted);<br /> draw(A -- B);<br /> draw(Z -- O -- X -- A -- Z -- B -- X);<br /> draw(A -- Y -- B);<br /> draw(P -- T -- SS);<br /> draw(P -- Y -- Q); draw(R -- Y -- SS);<br /> draw(X -- P); draw(Z -- SS);<br /> draw(Y -- T);<br /> draw(rightanglemark(Y, T, B, 1.25));<br /> draw(rightanglemark(Y, P, A, 1.25));<br /> draw(rightanglemark(Y, Q, X, 1.25));<br /> draw(rightanglemark(Y, R, Z, 1.25));<br /> draw(rightanglemark(Y, SS, B, 1.25));<br /> draw(rightanglemark(A, X, B, 1.25));<br /> draw(rightanglemark(A, Y, B, 1.25));<br /> draw(rightanglemark(A, Z, B, 1.25));<br /> label(&quot;$O$&quot;, O, S);<br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$X$&quot;, X, (X-B)/length(X-B));<br /> label(&quot;$Y$&quot;, Y, Y);<br /> label(&quot;$Z$&quot;, Z, (Z-A)/length(Z-A));<br /> label(&quot;$P$&quot;, P, (P-T)/length(P-T));<br /> label(&quot;$Q$&quot;, Q, SW);<br /> label(&quot;$R$&quot;, R, SE);<br /> label(&quot;$S$&quot;, SS, (SS-T)/length(SS-T));<br /> label(&quot;$T$&quot;, T, S);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;T&lt;/math&gt; be the projection of &lt;math&gt;Y&lt;/math&gt; onto &lt;math&gt;\overline{AB}&lt;/math&gt;. Notice that &lt;math&gt;T&lt;/math&gt; lies on the Simson Line &lt;math&gt;\overline{PQ}&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle AXB&lt;/math&gt;, and the Simson Line &lt;math&gt;\overline{RS}&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle AZB&lt;/math&gt;. Hence, &lt;math&gt;T=\overline{PQ}\cap\overline{RS}&lt;/math&gt;, so it suffices to show that &lt;math&gt;\angle PTS=\tfrac{1}{2}\angle XOZ&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;TAPY&lt;/math&gt; and &lt;math&gt;TBSY&lt;/math&gt; are cyclic quadrilaterals, &lt;cmath&gt;\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ=\frac{1}{2}\angle XOY+\frac{1}{2}\angle YOZ=\frac{1}{2}\angle XOZ,&lt;/cmath&gt;<br /> as required. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=2010|before=First problem|num-a=2}}<br /> {{USAJMO newbox|year=2010|num-b=2|num-a=4}}<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&diff=97255 2010 USAMO Problems/Problem 1 2018-08-15T22:10:24Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;AXYZB&lt;/math&gt; be a convex pentagon inscribed in a semicircle of diameter<br /> &lt;math&gt;AB&lt;/math&gt;. Denote by &lt;math&gt;P, Q, R, S&lt;/math&gt; the feet of the perpendiculars from &lt;math&gt;Y&lt;/math&gt; onto<br /> lines &lt;math&gt;AX, BX, AZ, BZ&lt;/math&gt;, respectively. Prove that the acute angle<br /> formed by lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; is half the size of &lt;math&gt;\angle XOZ&lt;/math&gt;, where<br /> &lt;math&gt;O&lt;/math&gt; is the midpoint of segment &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\alpha = \angle BAZ&lt;/math&gt;, &lt;math&gt;\beta = \angle ABX&lt;/math&gt;.<br /> Since &lt;math&gt;XY&lt;/math&gt; is a chord of the circle with diameter &lt;math&gt;AB&lt;/math&gt;,<br /> &lt;math&gt;\angle XAY = \angle XBY = \gamma&lt;/math&gt;. From the chord &lt;math&gt;YZ&lt;/math&gt;,<br /> we conclude &lt;math&gt;\angle YAZ = \angle YBZ = \delta&lt;/math&gt;.<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import olympiad;<br /> <br /> // Scale<br /> unitsize(1inch);<br /> real r = 1.75;<br /> <br /> // Semi-circle: centre O, radius r, diameter A--B.<br /> pair O = (0,0); dot(O); label(&quot;$O$&quot;, O, plain.S);<br /> pair A = r * plain.W; dot(A); label(&quot;$A$&quot;, A, unit(A));<br /> pair B = r * plain.E; dot(B); label(&quot;$B$&quot;, B, unit(B));<br /> draw(arc(O, r, 0, 180)--cycle);<br /> <br /> // points X, Y, Z<br /> real alpha = 22.5;<br /> real beta = 15;<br /> real delta = 30;<br /> pair X = r * dir(180 - 2*beta); dot(X); label(&quot;$X$&quot;, X, unit(X));<br /> pair Y = r * dir(2*(alpha + delta)); dot(Y); label(&quot;$Y$&quot;, Y, unit(Y));<br /> pair Z = r * dir(2*alpha); dot(Z); label(&quot;$Z$&quot;, Z, unit(Z));<br /> <br /> // Feet of perpendiculars from Y<br /> pair P = foot(Y, A, X); dot(P); label(&quot;$P$&quot;, P, unit(P-Y)); dot(P);<br /> pair Q = foot(Y, B, X); dot(P); label(&quot;$Q$&quot;, Q, unit(A-Q)); dot(Q);<br /> pair R = foot(Y, B, Z); dot(R); label(&quot;$S$&quot;, R, unit(R-Y)); dot(R);<br /> pair S = foot(Y, A, Z); dot(S); label(&quot;$R$&quot;, S, unit(B-S)); dot(S);<br /> pair T = foot(Y, A, B); dot(T); label(&quot;$T$&quot;, T, unit(T-Y)); dot(T);<br /> <br /> // Segments<br /> draw(B--X); draw(B--Y); draw(B--R);<br /> draw(A--Z); draw(A--Y); draw(A--P);<br /> draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);<br /> draw(R--T); draw(P--T);<br /> <br /> // Right angles<br /> draw(rightanglemark(A, X, B, 3));<br /> draw(rightanglemark(A, Y, B, 3));<br /> draw(rightanglemark(A, Z, B, 3));<br /> draw(rightanglemark(A, P, Y, 3));<br /> draw(rightanglemark(Y, R, B, 3));<br /> draw(rightanglemark(Y, S, A, 3));<br /> draw(rightanglemark(B, Q, Y, 3));<br /> <br /> // Acute angles<br /> import markers;<br /> void langle(pair A, pair B, pair C, string l=&quot;&quot;, real r=40, int n=1, int nm = 0)<br /> {<br /> string sl = &quot;$\scriptstyle{&quot; + l + &quot;}&quot;;<br /> marker m = (nm &gt; 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;<br /> markangle(Label(sl), radius=r, n=n, A, B, C, m);<br /> }<br /> langle(B, A, Z, &quot;\alpha&quot; );<br /> langle(X, B, A, &quot;\beta&quot;, n=2);<br /> langle(Y, A, X, &quot;\gamma&quot;, nm=1);<br /> langle(Y, B, X, &quot;\gamma&quot;, nm=1);<br /> langle(Z, A, Y, &quot;\delta&quot;, nm=2);<br /> langle(Z, B, Y, &quot;\delta&quot;, nm=2);<br /> langle(R, S, Y, &quot;\alpha+\delta&quot;, r=23);<br /> langle(Y, Q, P, &quot;\beta+\gamma&quot;, r=23);<br /> langle(R, T, P, &quot;\chi&quot;, r=15);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> Triangles &lt;math&gt;BQY&lt;/math&gt; and &lt;math&gt;APY&lt;/math&gt; are both right-triangles, and share the<br /> angle &lt;math&gt;\gamma&lt;/math&gt;, therefore they are similar, and so the ratio &lt;math&gt;PY :<br /> YQ = AY : YB&lt;/math&gt;. Now by [[Thales' theorem]] the angles &lt;math&gt;\angle AXB =<br /> \angle AYB = \angle AZB&lt;/math&gt; are all right-angles. Also, &lt;math&gt;\angle PYQ&lt;/math&gt;,<br /> being the fourth angle in a quadrilateral with 3 right-angles is<br /> again a right-angle. Therefore &lt;math&gt;\triangle PYQ \sim \triangle AYB&lt;/math&gt; and<br /> &lt;math&gt;\angle YQP = \angle YBA = \gamma + \beta&lt;/math&gt;.<br /> Similarly, &lt;math&gt;RY : YS = AY : YB&lt;/math&gt;, and so &lt;math&gt;\angle YRS = \angle YAB = \alpha + \delta&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;RY&lt;/math&gt; is perpendicular to &lt;math&gt;AZ&lt;/math&gt; so the direction &lt;math&gt;RY&lt;/math&gt; is &lt;math&gt;\alpha&lt;/math&gt; counterclockwise from the vertical, and since &lt;math&gt;\angle YRS = \alpha + \delta&lt;/math&gt; we see that &lt;math&gt;SR&lt;/math&gt; is &lt;math&gt;\delta&lt;/math&gt; clockwise from the vertical. (Draw an actual vertical line segment if necessary.)<br /> <br /> Similarly, &lt;math&gt;QY&lt;/math&gt; is perpendicular to &lt;math&gt;BX&lt;/math&gt; so the direction &lt;math&gt;QY&lt;/math&gt; is &lt;math&gt;\beta&lt;/math&gt; clockwise from the vertical, and since &lt;math&gt;\angle YQP&lt;/math&gt; is &lt;math&gt;\gamma + \beta&lt;/math&gt; we see that &lt;math&gt;QY&lt;/math&gt; is &lt;math&gt;\gamma&lt;/math&gt; counterclockwise from the vertical.<br /> <br /> Therefore the lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; intersect at an angle &lt;math&gt;\chi = \gamma<br /> + \delta&lt;/math&gt;. Now by the central angle theorem &lt;math&gt;2\gamma = \angle XOY&lt;/math&gt;<br /> and &lt;math&gt;2\delta = \angle YOZ&lt;/math&gt;, and so &lt;math&gt;2(\gamma + \delta) = \angle XOZ&lt;/math&gt;,<br /> and we are done.<br /> <br /> ''Note that &lt;math&gt;RTQY&lt;/math&gt; is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?''<br /> <br /> ===Footnote===<br /> We can prove a bit more. Namely, the extensions of the segments<br /> &lt;math&gt;RS&lt;/math&gt; and &lt;math&gt;PQ&lt;/math&gt; meet at a point on the diameter &lt;math&gt;AB&lt;/math&gt; that is vertically<br /> below the point &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;YS = AY \sin(\delta)&lt;/math&gt; and is inclined &lt;math&gt;\alpha&lt;/math&gt; counterclockwise<br /> from the vertical, the point &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;AY \sin(\delta) \sin(\alpha)&lt;/math&gt;<br /> horizontally to the right of &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;AS = AY \cos(\delta)&lt;/math&gt;, so &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;AS \sin(\alpha) = AY<br /> \cos(\delta)\sin(\alpha)&lt;/math&gt; vertically above the diameter &lt;math&gt;AB&lt;/math&gt;. Also,<br /> the segment &lt;math&gt;SR&lt;/math&gt; is inclined &lt;math&gt;\delta&lt;/math&gt; clockwise from the vertical,<br /> so if we extend it down from &lt;math&gt;S&lt;/math&gt; towards the diameter &lt;math&gt;AB&lt;/math&gt; it will<br /> meet the diameter at a point which is<br /> &lt;math&gt;AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)&lt;/math&gt;<br /> horizontally to the left of &lt;math&gt;S&lt;/math&gt;. This places the intersection point<br /> of &lt;math&gt;RS&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; vertically below &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> Similarly, and by symmetry the intersection point of &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;<br /> is directly below &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AB&lt;/math&gt;, so the lines through &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt;<br /> meet at a point &lt;math&gt;T&lt;/math&gt; on the diameter that is vertically below &lt;math&gt;Y&lt;/math&gt;.<br /> <br /> ===Footnote to the Footnote===<br /> The Footnote's claim is more easily proved as follows.<br /> <br /> Note that because &lt;math&gt;\angle{QPY}&lt;/math&gt; and &lt;math&gt;\angle{YAB}&lt;/math&gt; are both complementary to &lt;math&gt;\beta + \gamma&lt;/math&gt;, they must be equal. Now, let &lt;math&gt;PQ&lt;/math&gt; intersect diameter &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;T'&lt;/math&gt;. Then &lt;math&gt;PYT'A&lt;/math&gt; is cyclic and so &lt;math&gt;\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ&lt;/math&gt;. Hence &lt;math&gt;T'YSB&lt;/math&gt; is cyclic as well, and so we deduce that &lt;math&gt;\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.&lt;/math&gt; Hence &lt;math&gt;S, R, T'&lt;/math&gt; are collinear and so &lt;math&gt;T = T'&lt;/math&gt;. This proves the Footnote.<br /> <br /> ===Footnote to the Footnote to the Footnote===<br /> The Footnote's claim can be proved even more easily as follows.<br /> <br /> Drop an altitude from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; at point &lt;math&gt;T&lt;/math&gt;. Notice that &lt;math&gt;P, Q, T&lt;/math&gt; are collinear because they form the Simson line of &lt;math&gt;\triangle AXB&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt;. Also notice that &lt;math&gt;P, Q, T&lt;/math&gt; are collinear because they form the Simson line of &lt;math&gt;\triangle AZB&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt;. Since &lt;math&gt;T&lt;/math&gt; is at the diameter &lt;math&gt;AB&lt;/math&gt;, lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; must intersect at the diameter.<br /> <br /> ===Another footnote===<br /> There is another, more simpler solution using Simson lines. Can you find it?<br /> <br /> ==Operation Diagram==<br /> Of course, as with any geometry problem, DRAW A HUGE DIAGRAM spanning at least one page. And label all your right angles, noting &lt;math&gt;rectangles PXQY&lt;/math&gt; and &lt;math&gt;YSZR&lt;/math&gt;. It looks like there are a couple of key angles we need to diagram. Let's take &lt;math&gt;\angle{ZAB} = \alpha, \angle{XBA} = \beta, \angle{YAZ} = \angle{YBZ} = \delta&lt;/math&gt;. From there &lt;math&gt;\angle{XOZ}=180^\circ - \angle{XOA}-\angle{ZOB}=180-2(\beta + \alpha)&lt;/math&gt;.<br /> <br /> Move on to the part about the intersection of &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt;. Call the intersection &lt;math&gt;J&lt;/math&gt;. Note that by Simson Lines from point &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle{ABX}&lt;/math&gt; and &lt;math&gt;\triangle{AZB}&lt;/math&gt;, &lt;math&gt;YJ&lt;/math&gt; is perpendicular to &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;AB&lt;/math&gt;. Immediately note that we are trying to show that &lt;math&gt;\angle{PJS} = 90 - \beta - \alpha&lt;/math&gt;. <br /> <br /> It suffices to show that referencing quadrilateral &lt;math&gt;QR~J&lt;/math&gt;, where &lt;math&gt;~&lt;/math&gt; represents the intersection of &lt;math&gt;XB, AZ&lt;/math&gt;, we have reflex &lt;math&gt;\angle{Q~R} + \angle{BQJ} + \angle{ARJ} = 270 + \alpha + \beta&lt;/math&gt;. Note that the reflex angle is &lt;math&gt;180^\circ + \angle{A~X} = 180^\circ + (90^\circ - \angle{XA*}) = 270^\circ - ((90 - \beta) - \alpha) = 180 ^\circ + \alpha + \beta&lt;/math&gt;, therefore it suffices to show that &lt;math&gt;\angle{BQJ} + \angle{ARJ} = 90^\circ&lt;/math&gt;. To make this proof more accessible, note that via (cyclic) rectangles &lt;math&gt;PXQY&lt;/math&gt; and &lt;math&gt;YSZR&lt;/math&gt;, it suffices to prove &lt;math&gt;\angle{YPJ} + \angle{YSJ} = 90^\circ&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;\angle{YPJ} = \angle{YPQ} = \angle{YXQ} = \angle{YXB} = \angle{YAB} = \alpha + \delta&lt;/math&gt;.<br /> Note &lt;math&gt;\angle{YSJ} = \angle{YSR} = \angle{YZR} = \angle{YZA} = \angle{YBA} = \angle{YBX} + \angle{XBA} = ((90^\circ - \alpha) - \delta - \beta) + \beta = 90^\circ - \alpha - \delta&lt;/math&gt;, which completes the proof.<br /> <br /> ===Footnote to Operation Diagram===<br /> For reference/feasibility records: took expiLnCalc ~56 minutes (consecutively).<br /> During the problem expiLnCalc realized that the inclusion of &lt;math&gt;\delta&lt;/math&gt; was necessary when trying to show that &lt;math&gt;\angle{YSJ}+\angle{YPJ}=90^\circ&lt;/math&gt;. Don't be afraid to attempt several different strategies, and always be humble!<br /> <br /> ==Solution 2==<br /> [asy]<br /> currentpicture=new picture;<br /> size(12cm);<br /> pair O, A, B, X, Y, Z, P, Q, R, SS, T;<br /> O=(0, 0);<br /> A=(-1, 0);<br /> B=(1, 0);<br /> X=(Cos(144), Sin(144));<br /> Y=(Cos(105), Sin(105));<br /> Z=(Cos(27), Sin(27));<br /> P=foot(Y, A, X);<br /> Q=foot(Y, B, X);<br /> R=foot(Y, A, Z);<br /> SS=foot(Y, B, Z);<br /> T=foot(Y, A, B);<br /> dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T);<br /> draw(arc(O, 1, 0, 180));<br /> draw(circumcircle(T, A, Y), dotted);<br /> draw(circumcircle(T, B, Y), dotted);<br /> draw(A -- B);<br /> draw(Z -- O -- X -- A -- Z -- B -- X);<br /> draw(A -- Y -- B);<br /> draw(P -- T -- SS);<br /> draw(P -- Y -- Q); draw(R -- Y -- SS);<br /> draw(X -- P); draw(Z -- SS);<br /> draw(Y -- T);<br /> draw(rightanglemark(Y, T, B, 1.25));<br /> draw(rightanglemark(Y, P, A, 1.25));<br /> draw(rightanglemark(Y, Q, X, 1.25));<br /> draw(rightanglemark(Y, R, Z, 1.25));<br /> draw(rightanglemark(Y, SS, B, 1.25));<br /> draw(rightanglemark(A, X, B, 1.25));<br /> draw(rightanglemark(A, Y, B, 1.25));<br /> draw(rightanglemark(A, Z, B, 1.25));<br /> label(&quot;&lt;math&gt;O&lt;/math&gt;&quot;, O, S);<br /> label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, A, SW);<br /> label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, B, SE);<br /> label(&quot;&lt;math&gt;X&lt;/math&gt;&quot;, X, (X-B)/length(X-B));<br /> label(&quot;&lt;math&gt;Y&lt;/math&gt;&quot;, Y, Y);<br /> label(&quot;&lt;math&gt;Z&lt;/math&gt;&quot;, Z, (Z-A)/length(Z-A));<br /> label(&quot;&lt;math&gt;P&lt;/math&gt;&quot;, P, (P-T)/length(P-T));<br /> label(&quot;&lt;math&gt;Q&lt;/math&gt;&quot;, Q, SW);<br /> label(&quot;&lt;math&gt;R&lt;/math&gt;&quot;, R, SE);<br /> label(&quot;&lt;math&gt;S&lt;/math&gt;&quot;, SS, (SS-T)/length(SS-T));<br /> label(&quot;&lt;math&gt;T&lt;/math&gt;&quot;, T, S);<br /> [/asy]<br /> Let &lt;math&gt;T&lt;/math&gt; be the projection of &lt;math&gt;Y&lt;/math&gt; onto &lt;math&gt;\seg{AB}&lt;/math&gt;. Notice that &lt;math&gt;T&lt;/math&gt; lies on the Simson Line &lt;math&gt;\seg{PQ}&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle AXB&lt;/math&gt;, and the Simson Line &lt;math&gt;\seg{RS}&lt;/math&gt; from &lt;math&gt;Y&lt;/math&gt; to &lt;math&gt;\triangle AZB&lt;/math&gt;. Hence, &lt;math&gt;T=\seg{PQ}\cap\seg{RS}&lt;/math&gt;, so it suffices to show that &lt;math&gt;\angle PTS=\frac{1}{2}\angle XOZ&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;TAPY&lt;/math&gt; and &lt;math&gt;TBSY&lt;/math&gt; are cyclic quadrilaterals, &lt;cmath&gt;\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ=\frac{1}{2}\angle XOY+\frac{1}{2}\angle YOZ=\frac{1}{2}\angle XOZ,&lt;/cmath&gt;<br /> as required. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=2010|before=First problem|num-a=2}}<br /> {{USAJMO newbox|year=2010|num-b=2|num-a=4}}<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=96050 2014 AIME II Problems/Problem 10 2018-07-09T06:01:27Z <p>Theultimate123: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)== <br /> <br /> *Commenter's note: this solution made me lose brain cells, as it is so wordy for something so simple.<br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Our equation can be simplified like the following.<br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> We recognize this as the Law of Cosines with angle &lt;math&gt;120&lt;/math&gt; degrees.<br /> Our polygon is an equilateral triangle, say &lt;math&gt;ABC&lt;/math&gt;, with center &lt;math&gt;O&lt;/math&gt; at the origin and &lt;math&gt;AO=BO=CO=2014&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;3*[ABO]=3*(1007*1007\sqrt{3})=3*1007^2*\sqrt{3}=3042147\sqrt{3}&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{147}&lt;/math&gt;.<br /> <br /> -Solution by TheUltimate123<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_1&diff=94022 2018 USAJMO Problems/Problem 1 2018-04-18T23:24:02Z <p>Theultimate123: </p> <hr /> <div>== Problem ==<br /> <br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> <br /> -no two consecutive digits are equal, and<br /> <br /> -the last digit is a prime.<br /> <br /> ==Solution 1==<br /> The answer is &lt;math&gt;\boxed{\frac{2}{5}\left(9^n+(-1)^{n+1}\right)}&lt;/math&gt;.<br /> <br /> Suppose &lt;math&gt;a_n&lt;/math&gt; denotes the number of &lt;math&gt;n&lt;/math&gt;-digit numbers that satisfy the condition. We claim &lt;math&gt;a_n=4\cdot 9^{n-1}-a_{n-1}&lt;/math&gt;, with &lt;math&gt;a_1=4&lt;/math&gt;.<br /> [i]Proof.[/i] It is trivial to show that &lt;math&gt;a_1=4&lt;/math&gt;. Now, we can do casework on whether or not the tens digit of the &lt;math&gt;n&lt;/math&gt;-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in &lt;math&gt;a_{n-1}&lt;/math&gt; ways and choose the units digit in &lt;math&gt;3&lt;/math&gt; ways, since it must be prime and not equal to the tens digit. Therefore, there are &lt;math&gt;3a_{n-1}&lt;/math&gt; ways in this case.<br /> <br /> If the tens digit is not prime, we can use complementary counting. First we consider the number of &lt;math&gt;(n-1)&lt;/math&gt;-digit integers that do not have consecutive digits. There are &lt;math&gt;9&lt;/math&gt; ways to choose the first digit and &lt;math&gt;9&lt;/math&gt; ways to choose the remaining digits. Thus, there are &lt;math&gt;9^{n-1}&lt;/math&gt; integers that satisfy this. Therefore, the number of those &lt;math&gt;(n-1)&lt;/math&gt;-digit integers whose units digit is not prime is &lt;math&gt;9^{n-1}-a_{n-1}&lt;/math&gt;. It is easy to see that there are &lt;math&gt;4&lt;/math&gt; ways to choose the units digit, so there are &lt;math&gt;4\left(9^{n-1}-a_{n-1}\right)&lt;/math&gt; numbers in this case. It follows that &lt;cmath&gt;a_n=3a_{n-1}+4\left(9^{n-1}-a_{n-1}\right)=4\cdot 9^{n-1}-a_{n-1},&lt;/cmath&gt;<br /> and our claim has been proven.<br /> <br /> Then, we can use induction to show that &lt;math&gt;a_n=\frac{2}{5}\left(9^n+(-1)^{n+1}\right)&lt;/math&gt;. It is easy to see that our base case is true, as &lt;math&gt;a_1=4&lt;/math&gt;. Then, &lt;cmath&gt;a_{n+1}=4\cdot 9^n-a_n=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1},&lt;/cmath&gt;<br /> which is equal to &lt;cmath&gt;a_{n+1}=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right),&lt;/cmath&gt;<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|beforetext=|before=First Problem|num-a=2}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_1&diff=94021 2018 USAJMO Problems/Problem 1 2018-04-18T23:21:16Z <p>Theultimate123: Created page with &quot;== Problem == For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions: [list] [*] no...&quot;</p> <hr /> <div>== Problem ==<br /> <br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, find the number of &lt;math&gt;n&lt;/math&gt;-digit positive integers that satisfy both of the following conditions:<br /> [list]<br /> [*] no two consecutive digits are equal, and<br /> [*] the last digit is a prime.<br /> [/list]<br /> <br /> ==Solution 1==<br /> The answer is &lt;math&gt;\boxed{\frac{2}{5}\left(9^n+(-1)^{n+1}\right)}&lt;/math&gt;.<br /> <br /> Suppose &lt;math&gt;a_n&lt;/math&gt; denotes the number of &lt;math&gt;n&lt;/math&gt;-digit numbers that satisfy the condition. We claim &lt;math&gt;a_n=4\cdot 9^{n-1}-a_{n-1}&lt;/math&gt;, with &lt;math&gt;a_1=4&lt;/math&gt;.<br /> [i]Proof.[/i] It is trivial to show that &lt;math&gt;a_1=4&lt;/math&gt;. Now, we can do casework on whether or not the tens digit of the &lt;math&gt;n&lt;/math&gt;-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in &lt;math&gt;a_{n-1}&lt;/math&gt; ways and choose the units digit in &lt;math&gt;3&lt;/math&gt; ways, since it must be prime and not equal to the tens digit. Therefore, there are &lt;math&gt;3a_{n-1}&lt;/math&gt; ways in this case.<br /> <br /> If the tens digit is not prime, we can use complementary counting. First we consider the number of &lt;math&gt;(n-1)&lt;/math&gt;-digit integers that do not have consecutive digits. There are &lt;math&gt;9&lt;/math&gt; ways to choose the first digit and &lt;math&gt;9&lt;/math&gt; ways to choose the remaining digits. Thus, there are &lt;math&gt;9^{n-1}&lt;/math&gt; integers that satisfy this. Therefore, the number of those &lt;math&gt;(n-1)&lt;/math&gt;-digit integers whose units digit is not prime is &lt;math&gt;9^{n-1}-a_{n-1}&lt;/math&gt;. It is easy to see that there are &lt;math&gt;4&lt;/math&gt; ways to choose the units digit, so there are &lt;math&gt;4\left(9^{n-1}-a_{n-1}\right)&lt;/math&gt; numbers in this case. It follows that &lt;cmath&gt;a_n=3a_{n-1}+4\left(9^{n-1}-a_{n-1}\right)=4\cdot 9^{n-1}-a_{n-1},&lt;/cmath&gt;<br /> and our claim has been proven.<br /> <br /> Then, we can use induction to show that &lt;math&gt;a_n=\frac{2}{5}\left(9^n+(-1)^{n+1}\right)&lt;/math&gt;. It is easy to see that our base case is true, as &lt;math&gt;a_1=4&lt;/math&gt;. Then,<br /> \begin{align*}<br /> a_{n+1}&amp;=4\cdot 9^n-a_n\\<br /> &amp;=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)\\<br /> &amp;=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1}\\<br /> &amp;=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}\\<br /> &amp;=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}\\<br /> &amp;=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right),<br /> \end{align*}<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Solution by TheUltimate123.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|beforetext=|before=First Problem|num-a=2}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO&diff=94020 2018 USAJMO 2018-04-18T23:19:12Z <p>Theultimate123: </p> <hr /> <div>The test was held on April 18th and 19th, 2018. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br /> <br /> [[2018 USAJMO Problems]]<br /> * [[2018 USAJMO Problems/Problem 1]]<br /> * [[2018 USAJMO Problems/Problem 2]]<br /> * [[2018 USAJMO Problems/Problem 3]]<br /> * [[2018 USAJMO Problems/Problem 4]]<br /> * [[2018 USAJMO Problems/Problem 5]]<br /> * [[2018 USAJMO Problems/Problem 6]]<br /> <br /> == See Also ==<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]<br /> * [[USAJMO]]<br /> <br /> {{USAJMO newbox|year= 2018 |before=[[2017 USAJMO]]|after=[[2019 USAJMO]]}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=USAJMO_Problems_and_Solutions&diff=94019 USAJMO Problems and Solutions 2018-04-18T23:18:18Z <p>Theultimate123: </p> <hr /> <div>[[USAJMO]] problems and solutions.<br /> <br /> *[[2018 USAJMO]]<br /> *[[2017 USAJMO]]<br /> *[[2016 USAJMO]]<br /> *[[2015 USAJMO]]<br /> *[[2014 USAJMO]]<br /> *[[2013 USAJMO]]<br /> *[[2012 USAJMO]]<br /> *[[2011 USAJMO]]<br /> *[[2010 USAJMO]]</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=93093 2018 AIME I Problems/Problem 4 2018-03-09T23:48:16Z <p>Theultimate123: /* Problem 4 */</p> <hr /> <div>==Problem 4==<br /> In &lt;math&gt;\triangle ABC, AB = AC = 10&lt;/math&gt; and &lt;math&gt;BC = 12&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD = DE = EC&lt;/math&gt;. Then &lt;math&gt;AD&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{p}{q}&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 /> ==Solution 1 (No Trig)==<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import cse5;<br /> unitsize(10mm);<br /> pathpen=black;<br /> dotfactor=3;<br /> <br /> pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br /> pair[] dotted = {A,B,C,D,E,F,G};<br /> <br /> D(A--B);<br /> D(C--B);<br /> D(A--C);<br /> D(D--E);<br /> pathpen=dashed;<br /> D(B--F);<br /> D(D--G);<br /> <br /> dot(dotted);<br /> label(&quot;A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,E,NE);<br /> label(&quot;$F$&quot;,F,NE);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$x$&quot;,H,NW);<br /> label(&quot;$x$&quot;,I,NW);<br /> label(&quot;$x&quot;,J,NE);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> We draw the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; to get point &lt;math&gt;F&lt;/math&gt;. We notice that the triangle's height from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt; is 8 because it is a &lt;math&gt;3-4-5&lt;/math&gt; Right Triangle. To find the length of &lt;math&gt;\overline{BF}&lt;/math&gt;, we let &lt;math&gt;h&lt;/math&gt; represent &lt;math&gt;\overline{BF}&lt;/math&gt; and set up an equation by finding two ways to express the area. The equation is &lt;math&gt;(8)(12)=(10)(h)&lt;/math&gt;, which leaves us with &lt;math&gt;h=9.6&lt;/math&gt;. We then solve for the length &lt;math&gt;\overline{AF}&lt;/math&gt;, which is done through pythagorean theorm and get &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;2.8&lt;/math&gt;. We can now see that &lt;math&gt;\triangle ABF&lt;/math&gt; is a &lt;math&gt;7-24-25&lt;/math&gt; Right Triangle. Thus, we set &lt;math&gt;\overline{AG}&lt;/math&gt; as &lt;math&gt;5-&lt;/math&gt;&lt;math&gt;\tfrac{x}{2}&lt;/math&gt;, and yield that &lt;math&gt;\overline{AD}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;(&lt;/math&gt; &lt;math&gt;5-&lt;/math&gt; &lt;math&gt;\tfrac{x}{2}&lt;/math&gt; &lt;math&gt;)&lt;/math&gt; &lt;math&gt;(&lt;/math&gt; &lt;math&gt;\tfrac{25}{7}&lt;/math&gt; &lt;math&gt;)&lt;/math&gt;. Now, we can see &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;= 10-x&lt;/math&gt;, so we have &lt;math&gt;\overline{AB}&lt;/math&gt; &lt;math&gt;=10=&lt;/math&gt; &lt;math&gt;10-x+&lt;/math&gt; &lt;math&gt;(&lt;/math&gt; &lt;math&gt;5-&lt;/math&gt; &lt;math&gt;\tfrac{x}{2}&lt;/math&gt; &lt;math&gt;)&lt;/math&gt; &lt;math&gt;(&lt;/math&gt; &lt;math&gt;\tfrac{25}{7}&lt;/math&gt; &lt;math&gt;)&lt;/math&gt;. Solving this equation, we yield &lt;math&gt;39x=250&lt;/math&gt;, or &lt;math&gt;x=&lt;/math&gt; &lt;math&gt;\tfrac{250}{39}&lt;/math&gt;. Thus, our final answer is &lt;math&gt;250+39=\boxed{289}&lt;/math&gt;.<br /> ~bluebacon008<br /> <br /> ==Solution 2 (Coordinates)==<br /> Let &lt;math&gt;B = (0, 0)&lt;/math&gt;, &lt;math&gt;C = (12, 0)&lt;/math&gt;, and &lt;math&gt;A = (6, 8)&lt;/math&gt;. Then, let &lt;math&gt;x&lt;/math&gt; be in the interval &lt;math&gt;0&lt;x&lt;2&lt;/math&gt; and parametrically define &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; as &lt;math&gt;(6-3x, 8-4x)&lt;/math&gt; and &lt;math&gt;(12-3x, 4x)&lt;/math&gt; respectively. Note that &lt;math&gt;AD = 5x&lt;/math&gt;, so &lt;math&gt;DE = 5x&lt;/math&gt;. This means that<br /> &lt;cmath&gt;\begin{align*}<br /> \sqrt{36+(8x-8)^2} &amp;= 5x\\<br /> 36+(8x-8)^2 &amp;= 25x^2\\<br /> 64x^2-128x+100 &amp;= 25x^2\\<br /> 39x^2-128x+100 &amp;= 0\\<br /> x &amp;= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br /> x &amp;= \dfrac{100}{78}, 2\\<br /> \end{align*}&lt;/cmath&gt;<br /> However, since &lt;math&gt;2&lt;/math&gt; is extraneous by definition, &lt;math&gt;x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}&lt;/math&gt; ~ mathwiz0803<br /> <br /> ==Solution 3 (Law of Cosines)==<br /> As shown in the diagram, let &lt;math&gt;x&lt;/math&gt; denote &lt;math&gt;\overline{AD}&lt;/math&gt;. Let us denote the foot of the altitude of &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt; as &lt;math&gt;F&lt;/math&gt;. Note that &lt;math&gt;\overline{AE}&lt;/math&gt; can be expressed as &lt;math&gt;10-x&lt;/math&gt; and &lt;math&gt;\triangle{ABF}&lt;/math&gt; is a &lt;math&gt;6-8-10&lt;/math&gt; triangle . Therefore, &lt;math&gt;\sin(\angle{BAF})=\frac{3}{5}&lt;/math&gt; and &lt;math&gt;\cos(\angle{BAF})=\frac{4}{5}&lt;/math&gt;. Before we can proceed with the Law of Cosines, we must determine &lt;math&gt;\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}&lt;/math&gt;. Using LOC, we can write the following statement:<br /> &lt;cmath&gt;(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies&lt;/cmath&gt;<br /> &lt;cmath&gt;0=(10-x)^2-\frac{14x}{25}(10-x)\implies&lt;/cmath&gt;<br /> &lt;cmath&gt;0=10-x-\frac{14x}{25}\implies&lt;/cmath&gt;<br /> &lt;cmath&gt;10=\frac{39x}{25}\implies x=\frac{250}{39}&lt;/cmath&gt;<br /> Thus, the desired answer is &lt;math&gt;\boxed{289}&lt;/math&gt; ~ blitzkrieg21<br /> <br /> ==Solution 4==<br /> In isosceles triangle, draw the altitude from &lt;math&gt;D&lt;/math&gt; onto &lt;math&gt;\overline{AD}&lt;/math&gt;. Let the point of intersection be &lt;math&gt;X&lt;/math&gt;. Clearly, &lt;math&gt;AE=10-AD&lt;/math&gt;, and hence &lt;math&gt;AX=\frac{10-AD}{2}&lt;/math&gt;.<br /> <br /> Now, we recognise that the perpendicular from &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;\overline{AD}&lt;/math&gt; gives us two &lt;math&gt;6&lt;/math&gt;-&lt;math&gt;8&lt;/math&gt;-&lt;math&gt;10&lt;/math&gt; triangles. So, we calculate &lt;math&gt;\sin \angle ABC=\frac{4}{5}&lt;/math&gt; and &lt;math&gt;\cos \angle ABC=\frac{3}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle BAC = 180-2\cdot\angle ABC&lt;/math&gt;. And hence,<br /> <br /> &lt;math&gt;\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br /> = -\cos (2\cdot\angle ABC)<br /> = \sin^2 \angle ABC - \cos^2 \angle ABC<br /> = \frac{16}{25}-\frac{9}{25}=\frac{7}{25}&lt;/math&gt;<br /> <br /> Inspecting &lt;math&gt;\triangle ADX&lt;/math&gt; gives us &lt;math&gt;\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}&lt;/math&gt;<br /> Solving the equation &lt;math&gt;\frac{10-x}{2x}=\frac{7}{25}&lt;/math&gt; gives &lt;math&gt;x= \frac{250}{39} \implies\boxed{289}&lt;/math&gt;<br /> <br /> ~novus677<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_3&diff=93092 2018 AIME I Problems/Problem 3 2018-03-09T23:48:04Z <p>Theultimate123: /* Question */</p> <hr /> <div>==Question==<br /> Kathy has &lt;math&gt;5&lt;/math&gt; red cards and &lt;math&gt;5&lt;/math&gt; green cards. She shuffles the &lt;math&gt;10&lt;/math&gt; cards and lays out &lt;math&gt;5&lt;/math&gt; of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is &lt;math&gt; \frac{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;.<br /> <br /> ==Solution 1==<br /> <br /> We have &lt;math&gt;2+4\cdot 2&lt;/math&gt; cases total.<br /> <br /> The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.<br /> <br /> Obviously the denominator is &lt;math&gt;10\cdot 9\cdot 8\cdot 7\cdot 6&lt;/math&gt;, since we are choosing a card without replacement.<br /> <br /> Then, we have for the numerator for the two of all red and green: &lt;cmath&gt;5\cdot 4\cdot 3\cdot 2\cdot 1.&lt;/cmath&gt;<br /> <br /> For the 4 and 1, we have: &lt;cmath&gt;5\cdot 4\cdot 3\cdot 2\cdot 5.&lt;/cmath&gt;<br /> <br /> For the 3 and 2, we have: &lt;cmath&gt;5\cdot 4\cdot 3\cdot 5\cdot 4.&lt;/cmath&gt;<br /> <br /> For the 2 and 3, we have: &lt;cmath&gt;5\cdot 4\cdot 5\cdot 4\cdot 3.&lt;/cmath&gt;<br /> <br /> For the 1 and 4, we have: &lt;cmath&gt;5\cdot 5\cdot 4\cdot 3\cdot 2.&lt;/cmath&gt;<br /> <br /> Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: &lt;math&gt;\dfrac{31}{126}&lt;/math&gt;<br /> <br /> Thus the answer is &lt;math&gt;31 + 126 = \boxed{157}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Our probability will be &lt;math&gt;\dfrac{\text{number of &quot;happy&quot; configurations of cards}}{\text{total number of configurations of cards}}.&lt;/math&gt;<br /> <br /> First of all, we have &lt;math&gt;10&lt;/math&gt; choices for the first card, &lt;math&gt;9&lt;/math&gt; choices for the second card, &lt;math&gt;8&lt;/math&gt; choices for the third card, &lt;math&gt;7&lt;/math&gt; choices for the fourth card, and &lt;math&gt;6&lt;/math&gt; choices for the last card. This gives a total of &lt;math&gt;10\cdot 9\cdot 8\cdot 7\cdot 6&lt;/math&gt; possible ways for five cards to be chosen.<br /> <br /> Finding the number of configurations that make Kathy happy is a more difficult task, however, and we will resort to casework to do it.<br /> <br /> First, let's look at the appearances of the &quot;happy configurations&quot; that Kathy likes. Based on the premise of the problem, we can realize that there are ten cases for the appearance of the configurations: &lt;cmath&gt;\text{RRRRR},&lt;/cmath&gt; &lt;cmath&gt;\text{GGGGG},&lt;/cmath&gt; &lt;cmath&gt;\text{RRRRG},&lt;/cmath&gt; &lt;cmath&gt;\text{GGGGR},&lt;/cmath&gt; &lt;cmath&gt;\text{RRRGG},&lt;/cmath&gt; &lt;cmath&gt;\text{GGGRR},&lt;/cmath&gt; &lt;cmath&gt;\text{RRGGG},&lt;/cmath&gt; &lt;cmath&gt;\text{GGRRR},&lt;/cmath&gt; &lt;cmath&gt;\text{RGGGG},&lt;/cmath&gt; &lt;cmath&gt;\text{GRRRR}.&lt;/cmath&gt; But this doesn't mean there are 10 &quot;happy configurations&quot; in total-- remember that we've been treating these cards as distinguishable, so we must continue to do so.<br /> <br /> To lighten the load of 10 cases on the human brain, we can note that in the eyes of what we will soon do, &lt;math&gt;RRRRR&lt;/math&gt; and &lt;math&gt;GGGGG&lt;/math&gt; are effectively equivalent, and therefore may be treated in the same case. We will have to multiply by 2 at the end, though.<br /> <br /> Similarly, we can equate &lt;math&gt;RRRRG,&lt;/math&gt; &lt;math&gt;GGGGR,&lt;/math&gt; &lt;math&gt;RGGGG,&lt;/math&gt; and &lt;math&gt;GRRRR,&lt;/math&gt; as well as &lt;math&gt;RRRGG,&lt;/math&gt; &lt;math&gt;GGGRR,&lt;/math&gt; &lt;math&gt;RRGGG,&lt;/math&gt; and &lt;math&gt;GGRRR,&lt;/math&gt; so that we just have three cases. We can approach each of these cases with constructive counting.<br /> <br /> Case 1: &lt;math&gt;RRRRR&lt;/math&gt;-type.<br /> <br /> For this case, there are &lt;math&gt;5&lt;/math&gt; available choices for the first card, &lt;math&gt;4&lt;/math&gt; available choices for the second card, &lt;math&gt;3&lt;/math&gt; for the third card, &lt;math&gt;2&lt;/math&gt; for the fourth card, and &lt;math&gt;1&lt;/math&gt; for the last card. This leads to a total of &lt;math&gt;5\cdot 4\cdot 3\cdot 2\cdot 1=120&lt;/math&gt; configurations for this case. There are &lt;math&gt;2&lt;/math&gt; cases of this type.<br /> <br /> Case 2: &lt;math&gt;RRRRG&lt;/math&gt;-type.<br /> <br /> For this case, there are &lt;math&gt;5&lt;/math&gt; available choices for the first card, &lt;math&gt;4&lt;/math&gt; available choices for the second card, &lt;math&gt;3&lt;/math&gt; for the third card, &lt;math&gt;2&lt;/math&gt; for the fourth card, and &lt;math&gt;5&lt;/math&gt; choices for the last card (not &lt;math&gt;1&lt;/math&gt;, because we're doing a new color). This leads to a total of &lt;math&gt;5\cdot 4\cdot 3\cdot 2\cdot 5=600&lt;/math&gt; configurations for this case. There are &lt;math&gt;4&lt;/math&gt; cases of this type.<br /> <br /> Case 3: &lt;math&gt;RRRGG&lt;/math&gt;-type.<br /> <br /> For this case, there are &lt;math&gt;5&lt;/math&gt; available choices for the first card, &lt;math&gt;4&lt;/math&gt; available choices for the second card, &lt;math&gt;3&lt;/math&gt; for the third card, &lt;math&gt;5&lt;/math&gt; for the fourth card, and &lt;math&gt;4&lt;/math&gt; choices for the last card. This leads to a total of &lt;math&gt;5\cdot 4\cdot 3\cdot 5\cdot 4=1200&lt;/math&gt; configurations for this case. There are &lt;math&gt;4&lt;/math&gt; cases of this type.<br /> <br /> Adding the cases up gives &lt;math&gt;2\cdot 120+4\cdot 600+4\cdot 1200=7440&lt;/math&gt; &quot;happy&quot; configurations in total.<br /> <br /> This means that the probability that Kathy is happy will be &lt;math&gt;\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},&lt;/math&gt; which simplifies to &lt;math&gt;\dfrac{31}{126}.&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;31+126=\boxed{157.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_8&diff=93091 2018 AIME I Problems/Problem 8 2018-03-09T23:47:26Z <p>Theultimate123: </p> <hr /> <div>Let &lt;math&gt;ABCDEF&lt;/math&gt; be an equiangular hexagon such that &lt;math&gt;AB=6, BC=8, CD=10&lt;/math&gt;, and &lt;math&gt;DE=12&lt;/math&gt;. Denote by &lt;math&gt;d&lt;/math&gt; the diameter of the largest circle that fits inside the hexagon. Find &lt;math&gt;d^2&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> [[image:2018_AIME_I-8.png|center|500px]]<br /> - cooljoseph<br /> <br /> First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that &lt;math&gt;EF=2, FA=16&lt;/math&gt;. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length &lt;math&gt;6+8+10=24&lt;/math&gt;. Then, if you drew it to scale, notice that the &quot;widest&quot; this circle can be according to &lt;math&gt;AF, CD&lt;/math&gt; is &lt;math&gt;7\sqrt{3}&lt;/math&gt;. And it will be obvious that the sides won't be inside the circle, so our answer is &lt;math&gt;\boxed{147}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==Solution 2==<br /> <br /> Like solution 1, draw out the large equilateral triangle with side length &lt;math&gt;24&lt;/math&gt;. Let the tangent point of the circle at &lt;math&gt;\overline{CD}&lt;/math&gt; be G and the tangent point of the circle at &lt;math&gt;\overline{AF}&lt;/math&gt; be H. Clearly, GH is the diameter of our circle, and is also perpendicular to &lt;math&gt;\overline{CD}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt;.<br /> <br /> The equilateral triangle of side length &lt;math&gt;10&lt;/math&gt; is similar to our large equilateral triangle of &lt;math&gt;24&lt;/math&gt;. And the height of the former equilateral triangle is &lt;math&gt;\sqrt{10^2-5^2}=5\sqrt{3}&lt;/math&gt;. By our similarity condition,<br /> &lt;math&gt;\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}&lt;/math&gt;<br /> <br /> Solving this equation gives &lt;math&gt;d=7\sqrt{3}&lt;/math&gt;, and &lt;math&gt;d^2=\boxed{147}&lt;/math&gt;<br /> <br /> ~novus677<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems&diff=93089 2018 AIME I Problems 2018-03-09T23:46:50Z <p>Theultimate123: /* Problem 4 */ extra )</p> <hr /> <div>{{AIME Problems|year=2018|n=I}}<br /> <br /> ==Problem 1==<br /> Let &lt;math&gt;S&lt;/math&gt; be the number of ordered pairs of integers &lt;math&gt;(a,b)&lt;/math&gt; with &lt;math&gt;1 \leq a \leq 100&lt;/math&gt; and &lt;math&gt;b \geq 0&lt;/math&gt; such that the polynomial &lt;math&gt;x^2+ax+b&lt;/math&gt; can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> The number &lt;math&gt;n&lt;/math&gt; can be written in base &lt;math&gt;14&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{b}\text{ }\underline{c}&lt;/math&gt;, can be written in base &lt;math&gt;15&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{b}&lt;/math&gt;, and can be written in base &lt;math&gt;6&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }&lt;/math&gt;, where &lt;math&gt;a &gt; 0&lt;/math&gt;. Find the base-&lt;math&gt;10&lt;/math&gt; representation of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Kathy has &lt;math&gt;5&lt;/math&gt; red cards and &lt;math&gt;5&lt;/math&gt; green cards. She shuffles the &lt;math&gt;10&lt;/math&gt; cards and lays out &lt;math&gt;5&lt;/math&gt; of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is &lt;math&gt; \frac{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;.<br /> <br /> [[2018 AIME I Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> In &lt;math&gt;\triangle ABC, AB = AC = 10&lt;/math&gt; and &lt;math&gt;BC = 12&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD = DE = EC&lt;/math&gt;. Then &lt;math&gt;AD&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{p}{q}&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 /> [[2018 AIME I Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \log_2(2x+y) = \log_4(x^2+xy+7y^2) &lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that<br /> &lt;cmath&gt; \log_3(3x+y) = \log_9(3x^2+4xy+Ky^2). &lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of complex numbers &lt;math&gt;z&lt;/math&gt; with the properties that &lt;math&gt;|z|=1&lt;/math&gt; and &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is a real number. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> A right hexagonal prism has height &lt;math&gt;2&lt;/math&gt;. The bases are regular hexagons with side length &lt;math&gt;1&lt;/math&gt;. Any &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;12&lt;/math&gt; vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).<br /> <br /> [[2018 AIME I Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be an equiangular hexagon such that &lt;math&gt;AB=6, BC=8, CD=10&lt;/math&gt;, and &lt;math&gt;DE=12&lt;/math&gt;. Denote &lt;math&gt;d&lt;/math&gt; the diameter of the largest circle that fits inside the hexagon. Find &lt;math&gt;d^2&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> Find the number of four-element subsets of &lt;math&gt;\{1,2,3,4,\dots, 20\}&lt;/math&gt; with the property that two distinct elements of a subset have a sum of &lt;math&gt;16&lt;/math&gt;, and two distinct elements of a subset have a sum of &lt;math&gt;24&lt;/math&gt;. For example, &lt;math&gt;\{3,5,13,19\}&lt;/math&gt; and &lt;math&gt;\{6,10,20,18\}&lt;/math&gt; are two such subsets.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point &lt;math&gt;A&lt;/math&gt;. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path &lt;math&gt;AJABCHCHIJA&lt;/math&gt;, which has &lt;math&gt;10&lt;/math&gt; steps. Let &lt;math&gt;n&lt;/math&gt; be the number of paths with &lt;math&gt;15&lt;/math&gt; steps that begin and end at point &lt;math&gt;A&lt;/math&gt;. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> <br /> draw(unitcircle);<br /> draw(scale(2) * unitcircle);<br /> for(int d = 90; d &lt; 360 + 90; d += 72){<br /> draw(2 * dir(d) -- dir(d));<br /> }<br /> <br /> dot(1 * dir( 90), linewidth(5));<br /> dot(1 * dir(162), linewidth(5));<br /> dot(1 * dir(234), linewidth(5));<br /> dot(1 * dir(306), linewidth(5));<br /> dot(1 * dir(378), linewidth(5));<br /> dot(2 * dir(378), linewidth(5));<br /> dot(2 * dir(306), linewidth(5));<br /> dot(2 * dir(234), linewidth(5));<br /> dot(2 * dir(162), linewidth(5));<br /> dot(2 * dir( 90), linewidth(5));<br /> <br /> label(&quot;A$&quot;, 1 * dir( 90), -dir( 90));<br /> label(&quot;$B$&quot;, 1 * dir(162), -dir(162));<br /> label(&quot;$C$&quot;, 1 * dir(234), -dir(234));<br /> label(&quot;$D$&quot;, 1 * dir(306), -dir(306));<br /> label(&quot;$E$&quot;, 1 * dir(378), -dir(378));<br /> label(&quot;$F$&quot;, 2 * dir(378), dir(378));<br /> label(&quot;$G$&quot;, 2 * dir(306), dir(306));<br /> label(&quot;$H$&quot;, 2 * dir(234), dir(234));<br /> label(&quot;$I$&quot;, 2 * dir(162), dir(162));<br /> label(&quot;$J$&quot;, 2 * dir( 90), dir( 90));<br /> &lt;/asy&gt;<br /> <br /> [[2018 AIME I Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;3^n&lt;/math&gt; is written in base &lt;math&gt;143&lt;/math&gt;, its two right-most digits in base &lt;math&gt;143&lt;/math&gt; are &lt;math&gt;01&lt;/math&gt;.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{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&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have side lengths &lt;math&gt;AB=30&lt;/math&gt;, &lt;math&gt;BC=32&lt;/math&gt;, and &lt;math&gt;AC=34&lt;/math&gt;. Point &lt;math&gt;X&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt;, and points &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; are the incenters of &lt;math&gt;\triangle ABX&lt;/math&gt; and &lt;math&gt;\triangle ACX&lt;/math&gt;, respectively. Find the minimum possible area of &lt;math&gt;\triangle AI_1I_2&lt;/math&gt; as &lt;math&gt;X&lt;/math&gt; varies along &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;SP_1P_2P_3EP_4P_5&lt;/math&gt; be a heptagon. A frog starts jumping at vertex &lt;math&gt;S&lt;/math&gt;. From any vertex of the heptagon except &lt;math&gt;E&lt;/math&gt;, the frog may jump to either of the two adjacent vertices. When it reaches vertex &lt;math&gt;E&lt;/math&gt;, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than &lt;math&gt;12&lt;/math&gt; jumps that end at &lt;math&gt;E&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, &lt;math&gt;A,\text{ }B,\text{ }C&lt;/math&gt;, which can each be inscribed in a circle with radius &lt;math&gt;1&lt;/math&gt;. Let &lt;math&gt;\varphi_A&lt;/math&gt; denote the measure of the acute angle made by the diagonals of quadrilateral &lt;math&gt;A&lt;/math&gt;, and define &lt;math&gt;\varphi_B&lt;/math&gt; and &lt;math&gt;\varphi_C&lt;/math&gt; similarly. Suppose that &lt;math&gt;\sin\varphi_A=\frac{2}{3}&lt;/math&gt;, &lt;math&gt;\sin\varphi_B=\frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\sin\varphi_C=\frac{6}{7}&lt;/math&gt;. All three quadrilaterals have the same area &lt;math&gt;K&lt;/math&gt;, which can be written in the form &lt;math&gt;\dfrac{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;.<br /> <br /> [[2018 AIME I Problems/Problem 15 | Solution]]<br /> <br /> {{AIME box|year=2018|n=I|before=[[2017 AIME II]]|after=[[2018 AIME II]]}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems&diff=93088 2018 AIME I Problems 2018-03-09T23:46:15Z <p>Theultimate123: /* Problem 3 */ there was no latex around RRGGG, etc.</p> <hr /> <div>{{AIME Problems|year=2018|n=I}}<br /> <br /> ==Problem 1==<br /> Let &lt;math&gt;S&lt;/math&gt; be the number of ordered pairs of integers &lt;math&gt;(a,b)&lt;/math&gt; with &lt;math&gt;1 \leq a \leq 100&lt;/math&gt; and &lt;math&gt;b \geq 0&lt;/math&gt; such that the polynomial &lt;math&gt;x^2+ax+b&lt;/math&gt; can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> The number &lt;math&gt;n&lt;/math&gt; can be written in base &lt;math&gt;14&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{b}\text{ }\underline{c}&lt;/math&gt;, can be written in base &lt;math&gt;15&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{b}&lt;/math&gt;, and can be written in base &lt;math&gt;6&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }&lt;/math&gt;, where &lt;math&gt;a &gt; 0&lt;/math&gt;. Find the base-&lt;math&gt;10&lt;/math&gt; representation of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Kathy has &lt;math&gt;5&lt;/math&gt; red cards and &lt;math&gt;5&lt;/math&gt; green cards. She shuffles the &lt;math&gt;10&lt;/math&gt; cards and lays out &lt;math&gt;5&lt;/math&gt; of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is &lt;math&gt; \frac{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;.<br /> <br /> [[2018 AIME I Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> In &lt;math&gt;\triangle ABC, AB = AC = 10&lt;/math&gt; and &lt;math&gt;BC = 12&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;\overline{AC}&lt;/math&gt;) so that &lt;math&gt;AD = DE = EC&lt;/math&gt;. Then &lt;math&gt;AD&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{p}{q}&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 /> [[2018 AIME I Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \log_2(2x+y) = \log_4(x^2+xy+7y^2) &lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that<br /> &lt;cmath&gt; \log_3(3x+y) = \log_9(3x^2+4xy+Ky^2). &lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of complex numbers &lt;math&gt;z&lt;/math&gt; with the properties that &lt;math&gt;|z|=1&lt;/math&gt; and &lt;math&gt;z^{6!}-z^{5!}&lt;/math&gt; is a real number. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> A right hexagonal prism has height &lt;math&gt;2&lt;/math&gt;. The bases are regular hexagons with side length &lt;math&gt;1&lt;/math&gt;. Any &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;12&lt;/math&gt; vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).<br /> <br /> [[2018 AIME I Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be an equiangular hexagon such that &lt;math&gt;AB=6, BC=8, CD=10&lt;/math&gt;, and &lt;math&gt;DE=12&lt;/math&gt;. Denote &lt;math&gt;d&lt;/math&gt; the diameter of the largest circle that fits inside the hexagon. Find &lt;math&gt;d^2&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> Find the number of four-element subsets of &lt;math&gt;\{1,2,3,4,\dots, 20\}&lt;/math&gt; with the property that two distinct elements of a subset have a sum of &lt;math&gt;16&lt;/math&gt;, and two distinct elements of a subset have a sum of &lt;math&gt;24&lt;/math&gt;. For example, &lt;math&gt;\{3,5,13,19\}&lt;/math&gt; and &lt;math&gt;\{6,10,20,18\}&lt;/math&gt; are two such subsets.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point &lt;math&gt;A&lt;/math&gt;. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path &lt;math&gt;AJABCHCHIJA&lt;/math&gt;, which has &lt;math&gt;10&lt;/math&gt; steps. Let &lt;math&gt;n&lt;/math&gt; be the number of paths with &lt;math&gt;15&lt;/math&gt; steps that begin and end at point &lt;math&gt;A&lt;/math&gt;. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> <br /> draw(unitcircle);<br /> draw(scale(2) * unitcircle);<br /> for(int d = 90; d &lt; 360 + 90; d += 72){<br /> draw(2 * dir(d) -- dir(d));<br /> }<br /> <br /> dot(1 * dir( 90), linewidth(5));<br /> dot(1 * dir(162), linewidth(5));<br /> dot(1 * dir(234), linewidth(5));<br /> dot(1 * dir(306), linewidth(5));<br /> dot(1 * dir(378), linewidth(5));<br /> dot(2 * dir(378), linewidth(5));<br /> dot(2 * dir(306), linewidth(5));<br /> dot(2 * dir(234), linewidth(5));<br /> dot(2 * dir(162), linewidth(5));<br /> dot(2 * dir( 90), linewidth(5));<br /> <br /> label(&quot;$A$&quot;, 1 * dir( 90), -dir( 90));<br /> label(&quot;$B$&quot;, 1 * dir(162), -dir(162));<br /> label(&quot;$C$&quot;, 1 * dir(234), -dir(234));<br /> label(&quot;$D$&quot;, 1 * dir(306), -dir(306));<br /> label(&quot;$E$&quot;, 1 * dir(378), -dir(378));<br /> label(&quot;$F$&quot;, 2 * dir(378), dir(378));<br /> label(&quot;$G$&quot;, 2 * dir(306), dir(306));<br /> label(&quot;$H$&quot;, 2 * dir(234), dir(234));<br /> label(&quot;$I$&quot;, 2 * dir(162), dir(162));<br /> label(&quot;$J$&quot;, 2 * dir( 90), dir( 90));<br /> &lt;/asy&gt;<br /> <br /> [[2018 AIME I Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;3^n&lt;/math&gt; is written in base &lt;math&gt;143&lt;/math&gt;, its two right-most digits in base &lt;math&gt;143&lt;/math&gt; are &lt;math&gt;01&lt;/math&gt;.<br /> <br /> <br /> [[2018 AIME I Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{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&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have side lengths &lt;math&gt;AB=30&lt;/math&gt;, &lt;math&gt;BC=32&lt;/math&gt;, and &lt;math&gt;AC=34&lt;/math&gt;. Point &lt;math&gt;X&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt;, and points &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; are the incenters of &lt;math&gt;\triangle ABX&lt;/math&gt; and &lt;math&gt;\triangle ACX&lt;/math&gt;, respectively. Find the minimum possible area of &lt;math&gt;\triangle AI_1I_2&lt;/math&gt; as &lt;math&gt;X&lt;/math&gt; varies along &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;SP_1P_2P_3EP_4P_5&lt;/math&gt; be a heptagon. A frog starts jumping at vertex &lt;math&gt;S&lt;/math&gt;. From any vertex of the heptagon except &lt;math&gt;E&lt;/math&gt;, the frog may jump to either of the two adjacent vertices. When it reaches vertex &lt;math&gt;E&lt;/math&gt;, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than &lt;math&gt;12&lt;/math&gt; jumps that end at &lt;math&gt;E&lt;/math&gt;.<br /> <br /> [[2018 AIME I Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, &lt;math&gt;A,\text{ }B,\text{ }C&lt;/math&gt;, which can each be inscribed in a circle with radius &lt;math&gt;1&lt;/math&gt;. Let &lt;math&gt;\varphi_A&lt;/math&gt; denote the measure of the acute angle made by the diagonals of quadrilateral &lt;math&gt;A&lt;/math&gt;, and define &lt;math&gt;\varphi_B&lt;/math&gt; and &lt;math&gt;\varphi_C&lt;/math&gt; similarly. Suppose that &lt;math&gt;\sin\varphi_A=\frac{2}{3}&lt;/math&gt;, &lt;math&gt;\sin\varphi_B=\frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\sin\varphi_C=\frac{6}{7}&lt;/math&gt;. All three quadrilaterals have the same area &lt;math&gt;K&lt;/math&gt;, which can be written in the form &lt;math&gt;\dfrac{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;.<br /> <br /> [[2018 AIME I Problems/Problem 15 | Solution]]<br /> <br /> {{AIME box|year=2018|n=I|before=[[2017 AIME II]]|after=[[2018 AIME II]]}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_12&diff=92723 2017 AIME II Problems/Problem 12 2018-03-05T03:07:12Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Circle &lt;math&gt;C_0&lt;/math&gt; has radius &lt;math&gt;1&lt;/math&gt;, and the point &lt;math&gt;A_0&lt;/math&gt; is a point on the circle. Circle &lt;math&gt;C_1&lt;/math&gt; has radius &lt;math&gt;r&lt;1&lt;/math&gt; and is internally tangent to &lt;math&gt;C_0&lt;/math&gt; at point &lt;math&gt;A_0&lt;/math&gt;. Point &lt;math&gt;A_1&lt;/math&gt; lies on circle &lt;math&gt;C_1&lt;/math&gt; so that &lt;math&gt;A_1&lt;/math&gt; is located &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from &lt;math&gt;A_0&lt;/math&gt; on &lt;math&gt;C_1&lt;/math&gt;. Circle &lt;math&gt;C_2&lt;/math&gt; has radius &lt;math&gt;r^2&lt;/math&gt; and is internally tangent to &lt;math&gt;C_1&lt;/math&gt; at point &lt;math&gt;A_1&lt;/math&gt;. In this way a sequence of circles &lt;math&gt;C_1,C_2,C_3,\ldots&lt;/math&gt; and a sequence of points on the circles &lt;math&gt;A_1,A_2,A_3,\ldots&lt;/math&gt; are constructed, where circle &lt;math&gt;C_n&lt;/math&gt; has radius &lt;math&gt;r^n&lt;/math&gt; and is internally tangent to circle &lt;math&gt;C_{n-1}&lt;/math&gt; at point &lt;math&gt;A_{n-1}&lt;/math&gt;, and point &lt;math&gt;A_n&lt;/math&gt; lies on &lt;math&gt;C_n&lt;/math&gt; &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from point &lt;math&gt;A_{n-1}&lt;/math&gt;, as shown in the figure below. There is one point &lt;math&gt;B&lt;/math&gt; inside all of these circles. When &lt;math&gt;r = \frac{11}{60}&lt;/math&gt;, the distance from the center &lt;math&gt;C_0&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{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;.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0),125));<br /> draw(Circle((25,0),100));<br /> draw(Circle((25,20),80));<br /> draw(Circle((9,20),64));<br /> dot((125,0));<br /> label(&quot;$A_0$&quot;,(125,0),E);<br /> dot((25,100));<br /> label(&quot;$A_1$&quot;,(25,100),SE);<br /> dot((-55,20));<br /> label(&quot;$A_2$&quot;,(-55,20),E);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Impose a coordinate system and let the center of &lt;math&gt;C_0&lt;/math&gt; be &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;A_0&lt;/math&gt; be &lt;math&gt;(1,0)&lt;/math&gt;. Therefore &lt;math&gt;A_1=(1-r,r)&lt;/math&gt;, &lt;math&gt;A_2=(1-r-r^2,r-r^2)&lt;/math&gt;, &lt;math&gt;A_3=(1-r-r^2+r^3,r-r^2-r^3)&lt;/math&gt;, &lt;math&gt;A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)&lt;/math&gt;, and so on, where the signs alternate in groups of &lt;math&gt;2&lt;/math&gt;. The limit of all these points is point &lt;math&gt;B&lt;/math&gt;. Using the geometric series formula on &lt;math&gt;B&lt;/math&gt; and reducing the expression, we get &lt;math&gt;B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)&lt;/math&gt;. The distance from &lt;math&gt;B&lt;/math&gt; to the origin is &lt;math&gt;\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.&lt;/math&gt; Let &lt;math&gt;r=\frac{11}{60}&lt;/math&gt;, and the distance from the origin is &lt;math&gt;\frac{49}{61}&lt;/math&gt;. &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let the center of circle &lt;math&gt;C_i&lt;/math&gt; be &lt;math&gt;O_i&lt;/math&gt;. Note that &lt;math&gt;O_0BO_1&lt;/math&gt; is a right triangle, with right angle at &lt;math&gt;B&lt;/math&gt;. Also, &lt;math&gt;O_1B=\frac{11}{60}O_0B&lt;/math&gt;, or &lt;math&gt;O_0B = \frac{60}{61}O_0O_1&lt;/math&gt;. It is clear that &lt;math&gt;O_0O_1=1-r=\frac{49}{60}&lt;/math&gt;, so &lt;math&gt;O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}&lt;/math&gt;. Our answer is &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;<br /> <br /> -william122<br /> <br /> ==Solution 3==<br /> Note that there is an invariance, Consider the entire figure &lt;math&gt;\mathcal{F}&lt;/math&gt;. Perform a &lt;math&gt;90^\circ&lt;/math&gt; counterclockwise rotation, then scale by &lt;math&gt;r&lt;/math&gt; with respect to &lt;math&gt;(1, 0)&lt;/math&gt;. It is easy to see that the new figure &lt;math&gt;\mathcal{F}' \cup S^1 = \mathcal{F}&lt;/math&gt;, so &lt;math&gt;B&lt;/math&gt; is invariant.<br /> <br /> Using the invariance, Let &lt;math&gt;B = (x,y)&lt;/math&gt;. Then rotating and scaling, &lt;math&gt;B = (1-r(1+y), rx)&lt;/math&gt;. Equating, we find &lt;math&gt;x = \frac{1-r}{r^2+1}, y = \frac{r-r^2}{r^2+1}&lt;/math&gt;. The distance is thus &lt;math&gt;\frac{49}{61}&lt;/math&gt;. Our answer is &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;<br /> <br /> -Isogonal<br /> <br /> ==Solution 4==<br /> Using the invariance again as in Solution 3, assume &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt; away from the origin. The locus of possible points is a circle with radius &lt;math&gt;d&lt;/math&gt;. Consider the following diagram.<br /> &lt;asy&gt;<br /> size(7cm);<br /> draw(circle((0,0), 49/61));<br /> draw((0,0)--(0.790110185, 0.144853534));<br /> draw((0,0)--(-0.144853534, 0.790110185));<br /> draw((-0.144853534, 0.790110185)--(1,0));<br /> draw((0,0)--(1,0));<br /> draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));<br /> <br /> label(&quot;$O$&quot;, (0,0), SW);<br /> label(&quot;$(1,0)$&quot;, (1,0), E);<br /> label(&quot;$B$&quot;, (0.790110185, 0.144853534), NE);<br /> label(&quot;$B'$&quot;, (-0.144853534, 0.790110185), N);<br /> label(&quot;$d$&quot;, (0.5 * 49/61, 0), S);<br /> <br /> &lt;/asy&gt;<br /> <br /> Let the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;(1,0)&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. As &lt;math&gt;B&lt;/math&gt; is invariant, &lt;math&gt;x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}&lt;/math&gt;. Then by Power of a Point, &lt;math&gt;x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)&lt;/math&gt;. Solving, &lt;math&gt;d = \frac{49}{61}&lt;/math&gt;. Our answer is &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;<br /> <br /> -Isogonal<br /> <br /> ==Solution 5==<br /> <br /> We place this on the complex plane with &lt;math&gt;C_0=0&lt;/math&gt;. Notice that &lt;math&gt;B&lt;/math&gt; is the center of &lt;math&gt;C_k&lt;/math&gt; as &lt;math&gt;k&lt;/math&gt; approaches &lt;math&gt;\infty&lt;/math&gt;. We have &lt;math&gt;C_1=C_0+1-\frac{11}{60}=C_0+\frac{49}{60}&lt;/math&gt;. Similarly, &lt;math&gt;C_2=C_1+\frac{11}{60}\cdot\frac{49}{60}i&lt;/math&gt;, &lt;math&gt;C_3=C_2+\left(\frac{11}{60}\right)^2\cdot\frac{49}{60}i^2&lt;/math&gt;, and so on. Therefore, &lt;math&gt;B=\frac{49}{60}+\frac{49}{60}\cdot\frac{11}{60}i+\frac{49}{60}\cdot\left(\frac{11}{60}i\right)^2+\ldots&lt;/math&gt;<br /> <br /> It follows that &lt;math&gt;B=\frac{\frac{49}{60}}{1-\frac{11}{60}i}=\frac{49}{60-11i}&lt;/math&gt;. We seek &lt;math&gt;|B|&lt;/math&gt;, which is &lt;math&gt;\frac{49}{|60-11i|}=\frac{49}{61}&lt;/math&gt;, and the answer is &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> -TheUltimate123<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_8&diff=92650 2017 AIME II Problems/Problem 8 2018-03-03T06:49:07Z <p>Theultimate123: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;2017&lt;/math&gt; such that &lt;cmath&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}&lt;/cmath&gt; is an integer.<br /> <br /> ==Solution 1 (Not Rigorous)==<br /> Writing the last two terms with a common denominator, we have &lt;math&gt;\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}&lt;/math&gt; By inspection. this yields that &lt;math&gt;n \equiv 0, 24 \pmod{30}&lt;/math&gt;. Therefore, we get the final answer of &lt;math&gt;67 + 67 = \boxed{134}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Taking out the &lt;math&gt;1+n&lt;/math&gt; part of the expression and writing the remaining terms under a common denominator, we get &lt;math&gt;\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)&lt;/math&gt;. Therefore the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; must equal &lt;math&gt;720m&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;.<br /> Taking both sides mod &lt;math&gt;2&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{2}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be even. If &lt;math&gt;n&lt;/math&gt; is even, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;2a&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;2a&lt;/math&gt; in the expression, &lt;math&gt;64a^6+192a^5+480a^4+960a^3+1440a^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;16&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is even, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;3&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{3}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be a multiple of &lt;math&gt;3&lt;/math&gt;. If &lt;math&gt;n&lt;/math&gt; is a multiple of three, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;3b&lt;/math&gt; where &lt;math&gt;b&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;3b&lt;/math&gt; in the expression, &lt;math&gt;729b^6+1458b^5+2430b^4+3240b^3+3240b^2&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisibly by &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, the result is &lt;math&gt;n^6+n^5 \equiv 0 \pmod{5}&lt;/math&gt;. The only values of &lt;math&gt;n (\text{mod }5)&lt;/math&gt; that satisfy the equation are &lt;math&gt;n\equiv0(\text{mod }5)&lt;/math&gt; and &lt;math&gt;n\equiv4(\text{mod }5)&lt;/math&gt;. Therefore if &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; will be a multiple of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The only way to get the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; to be divisible by &lt;math&gt;720=16 \cdot 9 \cdot 5&lt;/math&gt; is to have &lt;math&gt;n \equiv 0 \pmod{2}&lt;/math&gt;, &lt;math&gt;n \equiv 0 \pmod{3}&lt;/math&gt;, and &lt;math&gt;n \equiv 0 \text{ or } 4 \pmod{5}&lt;/math&gt;. By the Chinese Remainder Theorem or simple guessing and checking, we see &lt;math&gt;n\equiv0,24 \pmod{30}&lt;/math&gt;. Because no numbers between &lt;math&gt;2011&lt;/math&gt; and &lt;math&gt;2017&lt;/math&gt; are equivalent to &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;24&lt;/math&gt; mod &lt;math&gt;30&lt;/math&gt;, the answer is &lt;math&gt;\frac{2010}{30}\times2=\boxed{134}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Note that &lt;math&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}&lt;/math&gt; will have a denominator that divides &lt;math&gt;5!&lt;/math&gt;. Therefore, for the expression to be an integer, &lt;math&gt;\frac{n^6}{6!}&lt;/math&gt; must have a denominator that divides &lt;math&gt;5!&lt;/math&gt;. Thus, &lt;math&gt;6\mid n^6&lt;/math&gt;, and &lt;math&gt;6\mid n&lt;/math&gt;. Let &lt;math&gt;n=6m&lt;/math&gt;. Substituting gives &lt;math&gt;1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}&lt;/math&gt;. Note that the first &lt;math&gt;5&lt;/math&gt; terms are integers, so it suffices for &lt;math&gt;\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}&lt;/math&gt; to be an integer. This simplifies to &lt;math&gt;\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)&lt;/math&gt;. It follows that &lt;math&gt;5\mid m^5(m+1)&lt;/math&gt;. Therefore, &lt;math&gt;m&lt;/math&gt; is either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; modulo &lt;math&gt;5&lt;/math&gt;. However, we seek the number of &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n=6m&lt;/math&gt;. By CRT, &lt;math&gt;n&lt;/math&gt; is either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;24&lt;/math&gt; modulo &lt;math&gt;30&lt;/math&gt;, and the answer is &lt;math&gt;67+67=\boxed{134}&lt;/math&gt;.<br /> <br /> -TheUltimate123<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_8&diff=92649 2017 AIME II Problems/Problem 8 2018-03-03T06:47:57Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;2017&lt;/math&gt; such that &lt;cmath&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}&lt;/cmath&gt; is an integer.<br /> <br /> ==Solution 1 (Not Rigorous)==<br /> Writing the last two terms with a common denominator, we have &lt;math&gt;\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}&lt;/math&gt; By inspection. this yields that &lt;math&gt;n \equiv 0, 24 \pmod{30}&lt;/math&gt;. Therefore, we get the final answer of &lt;math&gt;67 + 67 = \boxed{134}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Taking out the &lt;math&gt;1+n&lt;/math&gt; part of the expression and writing the remaining terms under a common denominator, we get &lt;math&gt;\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)&lt;/math&gt;. Therefore the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; must equal &lt;math&gt;720m&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;.<br /> Taking both sides mod &lt;math&gt;2&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{2}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be even. If &lt;math&gt;n&lt;/math&gt; is even, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;2a&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;2a&lt;/math&gt; in the expression, &lt;math&gt;64a^6+192a^5+480a^4+960a^3+1440a^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;16&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is even, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;3&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{3}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be a multiple of &lt;math&gt;3&lt;/math&gt;. If &lt;math&gt;n&lt;/math&gt; is a multiple of three, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;3b&lt;/math&gt; where &lt;math&gt;b&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;3b&lt;/math&gt; in the expression, &lt;math&gt;729b^6+1458b^5+2430b^4+3240b^3+3240b^2&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisibly by &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, the result is &lt;math&gt;n^6+n^5 \equiv 0 \pmod{5}&lt;/math&gt;. The only values of &lt;math&gt;n (\text{mod }5)&lt;/math&gt; that satisfy the equation are &lt;math&gt;n\equiv0(\text{mod }5)&lt;/math&gt; and &lt;math&gt;n\equiv4(\text{mod }5)&lt;/math&gt;. Therefore if &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; will be a multiple of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The only way to get the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; to be divisible by &lt;math&gt;720=16 \cdot 9 \cdot 5&lt;/math&gt; is to have &lt;math&gt;n \equiv 0 \pmod{2}&lt;/math&gt;, &lt;math&gt;n \equiv 0 \pmod{3}&lt;/math&gt;, and &lt;math&gt;n \equiv 0 \text{ or } 4 \pmod{5}&lt;/math&gt;. By the Chinese Remainder Theorem or simple guessing and checking, we see &lt;math&gt;n\equiv0,24 \pmod{30}&lt;/math&gt;. Because no numbers between &lt;math&gt;2011&lt;/math&gt; and &lt;math&gt;2017&lt;/math&gt; are equivalent to &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;24&lt;/math&gt; mod &lt;math&gt;30&lt;/math&gt;, the answer is &lt;math&gt;\frac{2010}{30}\times2=\boxed{134}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Note that &lt;math&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}&lt;/math&gt; will have a denominator that divides &lt;math&gt;5!&lt;/math&gt;. Therefore, for the expression to be an integer, &lt;math&gt;\frac{n^6}{6!}&lt;/math&gt; must have a denominator that divides &lt;math&gt;5!&lt;/math&gt;. Thus, &lt;math&gt;6\mid n^6&lt;/math&gt;, and &lt;math&gt;6\mid n&lt;/math&gt;. Let &lt;math&gt;n=6m&lt;/math&gt;. Substituting gives &lt;math&gt;1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}&lt;/math&gt;. Note that the first &lt;math&gt;5&lt;/math&gt; terms are integers, so it suffices for &lt;math&gt;\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}&lt;/math&gt; to be an integer. This simplifies to &lt;math&gt;\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)&lt;/math&gt;. It follows that &lt;math&gt;5\mid m^5(m+1)&lt;/math&gt;. Therefore, &lt;math&gt;m&lt;/math&gt; is either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; modulo &lt;math&gt;5&lt;/math&gt;. However, we seek the number of &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n=6m&lt;/math&gt;. By CRT, &lt;math&gt;n&lt;/math&gt; is either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;24&lt;/math&gt; modulo &lt;math&gt;30&lt;/math&gt;, and the answer is &lt;math&gt;67+67=\boxed{134}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_13&diff=92265 2013 AIME II Problems/Problem 13 2018-02-23T06:53:15Z <p>Theultimate123: /* Solution */</p> <hr /> <div>==Problem 13==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AC = BC&lt;/math&gt;, and point &lt;math&gt;D&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;CD = 3\cdot BD&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. Given that &lt;math&gt;CE = \sqrt{7}&lt;/math&gt; and &lt;math&gt;BE = 3&lt;/math&gt;, the area of &lt;math&gt;\triangle ABC&lt;/math&gt; can be expressed in the form &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> After drawing the figure, we suppose &lt;math&gt;BD=a&lt;/math&gt;, so that &lt;math&gt;CD=3a&lt;/math&gt;, &lt;math&gt;AC=4a&lt;/math&gt;, and &lt;math&gt;AE=ED=b&lt;/math&gt;.<br /> <br /> Using Law of Cosines for &lt;math&gt;\triangle AEC&lt;/math&gt; and &lt;math&gt;\triangle CED&lt;/math&gt;,we get<br /> <br /> &lt;cmath&gt;b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+7+2\sqrt{7}\cdot \cos(\angle CED)=16a^2\qquad (2)&lt;/cmath&gt;<br /> So, &lt;math&gt;(1)+(2)&lt;/math&gt;, we get&lt;cmath&gt;2b^2+14=25a^2. \qquad (3)&lt;/cmath&gt;<br /> <br /> Using Law of Cosines in &lt;math&gt;\triangle ACD&lt;/math&gt;, we get<br /> <br /> &lt;cmath&gt;b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2&lt;/cmath&gt;<br /> <br /> So, &lt;cmath&gt;\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)&lt;/cmath&gt;<br /> <br /> Using Law of Cosines in &lt;math&gt;\triangle EDC&lt;/math&gt; and &lt;math&gt;\triangle EDB&lt;/math&gt;, we get<br /> <br /> &lt;cmath&gt;b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;b^2+a^2+2\cdot a\cdot b\cdot \cos(\angle ADC)=9.\qquad (6)&lt;/cmath&gt;<br /> <br /> &lt;math&gt;(5)+(6)&lt;/math&gt;, and according to &lt;math&gt;(4)&lt;/math&gt;, we can get <br /> &lt;cmath&gt;37a^2+2b^2=48. \qquad (7)&lt;/cmath&gt;<br /> <br /> Using &lt;math&gt;(3)&lt;/math&gt; and &lt;math&gt;(7)&lt;/math&gt;, we can solve &lt;math&gt;a=1&lt;/math&gt; and &lt;math&gt;b=\frac{\sqrt{22}}{2}&lt;/math&gt;.<br /> <br /> Finally, we use Law of Cosines for &lt;math&gt;\triangle ADB&lt;/math&gt;, <br /> <br /> &lt;cmath&gt;4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2&lt;/cmath&gt;<br /> <br /> then &lt;math&gt;AB=2\sqrt{7}&lt;/math&gt;, so the height of this &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\sqrt{4^2-(\sqrt{7})^2}=3&lt;/math&gt;.<br /> <br /> Then the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;3\sqrt{7}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{010}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;X&lt;/math&gt; be the foot of the altitude from &lt;math&gt;C&lt;/math&gt; with other points labelled as shown below.<br /> &lt;asy&gt;<br /> size(200);<br /> pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;<br /> draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);<br /> label(&quot;$A$&quot;,A,SW);label(&quot;$B$&quot;,B,SE);label(&quot;$C$&quot;,C,N);label(&quot;$D$&quot;,D,NE);label(&quot;$L$&quot;,L,NW);label(&quot;$M$&quot;,M,S);<br /> pair X=foot(C,A,B), Y=foot(L,A,B);<br /> pair EE=D/2;<br /> label(&quot;$X$&quot;,X,S);label(&quot;$E$&quot;,EE,NW);label(&quot;$Y$&quot;,Y,S);<br /> draw(C--X^^L--Y,dotted);<br /> draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L));<br /> &lt;/asy&gt;<br /> Now we proceed using [[mass points]]. To balance along the segment &lt;math&gt;BC&lt;/math&gt;, we assign &lt;math&gt;B&lt;/math&gt; a mass of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;. Therefore, &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;. As &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;, we must assign &lt;math&gt;A&lt;/math&gt; a mass of &lt;math&gt;4&lt;/math&gt; as well. This gives &lt;math&gt;L&lt;/math&gt; a mass of &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; a mass of &lt;math&gt;7&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;AB=b&lt;/math&gt; be the base of the triangle, and let &lt;math&gt;CX=h&lt;/math&gt; be the height. Then as &lt;math&gt;AM:MB=3:4&lt;/math&gt;, and as &lt;math&gt;AX=\frac{b}{2}&lt;/math&gt;, we know that &lt;cmath&gt;MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.&lt;/cmath&gt; Also, as &lt;math&gt;CE:EM=7:1&lt;/math&gt;, we know that &lt;math&gt;EM=\frac{1}{\sqrt{7}}&lt;/math&gt;. Therefore, by the Pythagorean Theorem on &lt;math&gt;\triangle {XCM}&lt;/math&gt;, we know that &lt;cmath&gt;\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.&lt;/cmath&gt;<br /> <br /> Also, as &lt;math&gt;LE:BE=5:3&lt;/math&gt;, we know that &lt;math&gt;BL=\frac{8}{5}\cdot 3=\frac{24}{5}&lt;/math&gt;. Furthermore, as &lt;math&gt;\triangle YLA\sim \triangle XCA&lt;/math&gt;, and as &lt;math&gt;AL:LC=1:4&lt;/math&gt;, we know that &lt;math&gt;LY=\frac{h}{5}&lt;/math&gt; and &lt;math&gt;AY=\frac{b}{10}&lt;/math&gt;, so &lt;math&gt;YB=\frac{9b}{10}&lt;/math&gt;. Therefore, by the Pythagorean Theorem on &lt;math&gt;\triangle BLY&lt;/math&gt;, we get &lt;cmath&gt;\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.&lt;/cmath&gt;<br /> Solving this system of equations yields &lt;math&gt;b=2\sqrt{7}&lt;/math&gt; and &lt;math&gt;h=3&lt;/math&gt;. Therefore, the area of the triangle is &lt;math&gt;3\sqrt{7}&lt;/math&gt;, giving us an answer of &lt;math&gt;\boxed{010}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively.<br /> Then &lt;math&gt;D = (\frac{3a}{4}, \frac{h}{4})&lt;/math&gt; and &lt;math&gt;E = (-\frac{a}{8},\frac{h}{8}).&lt;/math&gt; <br /> &lt;math&gt;EC^2 = 7&lt;/math&gt; implies &lt;math&gt;a^2 + 49h^2 = 448&lt;/math&gt;; &lt;math&gt;EB^2 = 9&lt;/math&gt; implies &lt;math&gt;81a^2 + h^2 = 576.&lt;/math&gt;<br /> Solve this system of equations simultaneously, &lt;math&gt;a=\sqrt{7}&lt;/math&gt; and &lt;math&gt;h=3&lt;/math&gt;. <br /> Area of the triangle is ah = &lt;math&gt;3\sqrt{7}&lt;/math&gt;, giving us an answer of &lt;math&gt;\boxed{010}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> &lt;asy&gt;<br /> size(200);<br /> pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;<br /> draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);<br /> label(&quot;$A$&quot;,A,SW);label(&quot;$B$&quot;,B,SE);label(&quot;$C$&quot;,C,N);label(&quot;$D$&quot;,D,NE);label(&quot;$L$&quot;,L,NW);label(&quot;$M$&quot;,M,S);<br /> pair EE=D/2;<br /> label(&quot;$\sqrt{7}$&quot;, C--EE, W); label(&quot;$x$&quot;, D--B, E); label(&quot;$3x$&quot;, C--D, E); label(&quot;$l$&quot;, EE--D, N); label(&quot;$3$&quot;, EE--B, N);<br /> label(&quot;$E&quot;,EE,NW);<br /> &lt;/asy&gt;<br /> (Thanks to writer of Solution 2)<br /> <br /> Let &lt;math&gt;BD = x&lt;/math&gt;. Then &lt;math&gt;CD = 3x&lt;/math&gt; and &lt;math&gt;AC = 4x&lt;/math&gt;. Also, let &lt;math&gt;AE = ED = l&lt;/math&gt;. Using Stewart's Theorem on &lt;math&gt;\bigtriangleup CEB&lt;/math&gt; gives us the equation &lt;math&gt;(x)(3x)(4x) + (4x)(l^2) = 27x + 7x&lt;/math&gt; or, after simplifying, &lt;math&gt;4l^2 = 34 - 12x^2&lt;/math&gt;. We use Stewart's again on &lt;math&gt;\bigtriangleup CAD&lt;/math&gt;: &lt;math&gt;(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)&lt;/math&gt;, which becomes &lt;math&gt;2l^2 = 25x^2 - 14&lt;/math&gt;. Substituting &lt;math&gt;2l^2 = 17 - 6x^2&lt;/math&gt;, we see that &lt;math&gt;31x^2 = 31&lt;/math&gt;, or &lt;math&gt;x = 1&lt;/math&gt;. Then &lt;math&gt;l^2 = \frac{11}{2}&lt;/math&gt;.<br /> <br /> We now use Law of Cosines on &lt;math&gt;\bigtriangleup CAD&lt;/math&gt;. &lt;math&gt;(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\cos C&lt;/math&gt;. Plugging in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;22 = 16 + 9 - 2(4)(3)\cos C&lt;/math&gt;, so &lt;math&gt;\cos C = \frac{1}{8}&lt;/math&gt;. Using the Pythagorean trig identity &lt;math&gt;\sin^2 + \cos^2 = 1&lt;/math&gt;, &lt;math&gt;\sin^2 C = 1 - \frac{1}{64}&lt;/math&gt;, so &lt;math&gt;\sin C = \frac{3\sqrt{7}}{8}&lt;/math&gt;. <br /> <br /> &lt;math&gt;[ABC] = \frac{1}{2} AC \cdot BC \sin C = (\frac{1}{2})(4)(4)(\frac{3\sqrt{7}}{8}) = 3\sqrt{7}&lt;/math&gt;, and our answer is &lt;math&gt;3 + 7 = \boxed{010}&lt;/math&gt;.<br /> <br /> ===Solution 5 (Barycentric Coordinates)===<br /> Let ABC be the reference triangle, with &lt;math&gt;A=(1,0,0)&lt;/math&gt;, &lt;math&gt;B=(0,1,0)&lt;/math&gt;, and &lt;math&gt;C=(0,0,1)&lt;/math&gt;. We can easily calculate &lt;math&gt;D=(0,\frac{3}{4},\frac{1}{4})&lt;/math&gt; and subsequently &lt;math&gt;E=(\frac{1}{2},\frac{3}{8},\frac{1}{8})&lt;/math&gt;. Using distance formula on &lt;math&gt;\overline{EC}=(\frac{1}{2},\frac{3}{8},-\frac{7}{8})&lt;/math&gt; and &lt;math&gt;\overline{EB}=(\frac{1}{2},-\frac{5}{8},\frac{1}{8})&lt;/math&gt; gives <br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \begin{cases}<br /> 7&amp;=|EC|^2=-a^2 \cdot \frac{3}{8} \cdot (-\frac{7}{8})-b^2 \cdot \frac{1}{2} \cdot (-\frac{7}{8})-c^2 \cdot \frac{1}{2} \cdot \frac{3}{8} \\<br /> 9&amp;=|EB|^2=-a^2 \cdot (-\frac{5}{8}) \cdot \frac{1}{8}-b^2 \cdot \frac{1}{2} \cdot \frac{1}{8}-c^2 \cdot \frac{1}{2} \cdot (-\frac{5}{8}) \\<br /> \end{cases}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> But we know that &lt;math&gt;a=b&lt;/math&gt;, so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \begin{cases}<br /> 7\cdot 64&amp;=3\cdot 7\cdot a^2+b^2\cdot 4\cdot 7-c^2\cdot 4\cdot 3\\<br /> 9\cdot 64&amp;=5a^2-4b^2+4\cdot 5\cdot c^2 \\<br /> \end{cases}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \begin{cases}<br /> 7\cdot 64&amp;=49a^2-12c^2 \\<br /> 9\cdot 64&amp;=a^2+20c^2 \\<br /> \end{cases}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \begin{cases}<br /> 5\cdot 7\cdot 64&amp;=245a^2-60c^2 \\<br /> 3\cdot 9\cdot 64&amp;=3a^2+60c^2 \\<br /> \end{cases}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Then we add the equations to get<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 62\cdot 64&amp;=248a^2 \\<br /> a^2 &amp;=16 \\<br /> a &amp;=4 \\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Then plugging gives &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=2\sqrt{7}&lt;/math&gt;. Then the height from &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;, and the area is &lt;math&gt;3\sqrt{7}&lt;/math&gt; and our answer is &lt;math&gt;\boxed{010}&lt;/math&gt;.<br /> <br /> ===Solution 6 (Desperate for points)===<br /> <br /> Note that &lt;math&gt;\triangle BEC&lt;/math&gt; uniquely determines &lt;math&gt;\triangle ABC&lt;/math&gt;. Assume WTMLOG (without too much loss of generality) that &lt;math&gt;\angle BEC=90^\circ&lt;/math&gt;. Then, the answer is &lt;math&gt;2\cdot \frac{1}{2}\cdot 3\cdot \sqrt{7}=3\sqrt{7}\rightarrow\boxed{010}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_14&diff=92063 2011 AIME II Problems/Problem 14 2018-02-19T20:15:23Z <p>Theultimate123: /* Solution 3 (2-sec solve) */</p> <hr /> <div>==Problem 14==<br /> There are &lt;math&gt;N&lt;/math&gt; [[permutation]]s &lt;math&gt;(a_{1}, a_{2}, ... , a_{30})&lt;/math&gt; of &lt;math&gt;1, 2, \ldots, 30&lt;/math&gt; such that for &lt;math&gt;m \in \left\{{2, 3, 5}\right\}&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;a_{n+m} - a_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;1 \leq n &lt; n+m \leq 30&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> __TOC__<br /> ==Solutions==<br /> ===Solution 1===<br /> Be wary of &quot;position&quot; versus &quot;number&quot; in the solution!<br /> <br /> Each POSITION in the 30-position permutation is uniquely defined by an ordered triple &lt;math&gt;(i, j, k)&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt;th position is defined by this ordered triple where &lt;math&gt;i&lt;/math&gt; is &lt;math&gt;n \mod 2&lt;/math&gt;, &lt;math&gt;j&lt;/math&gt; is &lt;math&gt;n \mod 3&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;n \mod 5&lt;/math&gt;. There are 2 choices for &lt;math&gt;i&lt;/math&gt;, 3 for &lt;math&gt;j&lt;/math&gt;, and 5 for &lt;math&gt;k&lt;/math&gt;, yielding &lt;math&gt;2 \cdot 3 \cdot 5=30&lt;/math&gt; possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if i is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If j is the same, then the two numbers must be equivalent &lt;math&gt;\mod 3&lt;/math&gt;, and if &lt;math&gt;k&lt;/math&gt; is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have &lt;math&gt;1 \mod 2&lt;/math&gt;! It's that the POSITION which NUMBER 1 occupies has &lt;math&gt;1 \mod 2&lt;/math&gt;!<br /> <br /> The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.<br /> <br /> The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.<br /> <br /> Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of &lt;math&gt;1\cdot 1 \cdot 3=3&lt;/math&gt; choices for the placement of 3.<br /> <br /> As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of &lt;math&gt;1 \cdot 1 \cdot 2=2&lt;/math&gt; possible placements for 4.<br /> <br /> 5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.<br /> <br /> All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed. <br /> <br /> Thus, &lt;math&gt;N = 30 \cdot 8 \cdot 3 \cdot 2=30 \cdot 48=1440&lt;/math&gt;. Thus, the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\boxed{440}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> We observe that the condition on the permutations means that two numbers with indices congruent &lt;math&gt;\mod m&lt;/math&gt; are themselves congruent &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m \in \{ 2,3,5\}.&lt;/math&gt; Furthermore, suppose that &lt;math&gt;a_n \equiv k \mod m.&lt;/math&gt; Then, there are &lt;math&gt;30/m&lt;/math&gt; indices congruent to &lt;math&gt;n \mod m,&lt;/math&gt; and &lt;math&gt;30/m&lt;/math&gt; numbers congruent to &lt;math&gt;k \mod m,&lt;/math&gt; because 2, 3, and 5 are all factors of 30. Therefore, since every index congruent to &lt;math&gt;n&lt;/math&gt; must contain a number congruent to &lt;math&gt;k,&lt;/math&gt; and no number can appear twice in the permutation, only the indices congruent to &lt;math&gt;n&lt;/math&gt; contain numbers congruent to &lt;math&gt;k.&lt;/math&gt; In other words, &lt;math&gt;a_i \equiv a_j \mod m \iff i \equiv j \mod m.&lt;/math&gt; But it is not necessary that &lt;math&gt;\textcolor{red}{(a_i\equiv i)\cup (a_j\equiv j)\mod m}&lt;/math&gt;. In fact, if that were the case, there would only be one way to assign the indices, since &lt;math&gt;2,3,5&lt;/math&gt; are relatively prime to each other and &lt;math&gt;30=\text{lcm}(2,3,5)&lt;/math&gt;: &lt;math&gt;\{a_1,a_2,\dots a_{30}\}\in\{1,2,\dots 30\}\text{ }respectively&lt;/math&gt;.<br /> <br /> This tells us that in a valid permutation, the congruence classes &lt;math&gt;\mod m&lt;/math&gt; are simply swapped around, and if the set &lt;math&gt;S&lt;/math&gt; is a congruence class &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m = &lt;/math&gt; 2, 3, or 5, the set &lt;math&gt;\{ a_i \vert i \in S \}&lt;/math&gt; is still a congruence class &lt;math&gt;\mod m.&lt;/math&gt; Clearly, each valid permutation of the numbers 1 through 30 corresponds to exactly one permutation of the congruence classes modulo 2, 3, and 5. Also, if we choose some permutations of the congruence classes modulo 2, 3, and 5, they correspond to exactly one valid permutation of the numbers 1 through 30. This can be shown as follows: First of all, the choice of permutations of the congruence classes gives us every number in the permutation modulo 2, 3, and 5, so by the Chinese Remainder Theorem, it gives us every number &lt;math&gt;\mod 2\cdot 3\cdot 5 = 30.&lt;/math&gt; Because the numbers must be between 1 and 30 inclusive, it thus uniquely determines what number goes in each index. Furthermore, two different indices cannot contain the same number. We will prove this by contradiction, calling the indices &lt;math&gt;a_i&lt;/math&gt; and &lt;math&gt;a_j&lt;/math&gt; for &lt;math&gt;i \neq j.&lt;/math&gt; If &lt;math&gt;a_i=a_j,&lt;/math&gt; then they must have the same residues modulo 2, 3, and 5, and so &lt;math&gt;i \equiv j&lt;/math&gt; modulo 2, 3, and 5. Again using the Chinese Remainder Theorem, we conclude that &lt;math&gt;i \equiv j \mod 30,&lt;/math&gt; so because &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are both between 1 and 30 inclusive, &lt;math&gt;i = j,&lt;/math&gt; giving us a contradiction. Therefore, every choice of permutations of the congruence classes modulo 2, 3, and 5 corresponds to exactly one valid permutation of the numbers 1 through 30.<br /> <br /> In other words, each set of assignment from &lt;math&gt;a_j\rightarrow j\mod (2,3,5)&lt;/math&gt; determines a unique string of &lt;math&gt;30&lt;/math&gt; numbers. For example:<br /> <br /> &lt;math&gt;\left[(0,1)\rightarrow (1,0)\right]\cap\left[(0,1,2)\rightarrow (0,2,1)\right]\cap\left[(0,1,2,3,4)\rightarrow (4,2,3,0,1)\right]&lt;/math&gt;:<br /> &lt;cmath&gt;\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \hline<br /> 2&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0\\ \hline<br /> 3&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1\\ \hline<br /> 5&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1\\ \hline\hline<br /> 30&amp;9&amp;2&amp;13&amp;0&amp;11&amp;4&amp;27&amp;8&amp;25&amp;6&amp;29&amp;22&amp;3&amp;20&amp;1&amp;24&amp;17&amp;28&amp;15&amp;26&amp;19&amp;12&amp;23&amp;10&amp;21&amp;14&amp;7&amp;18&amp;5&amp;16\\ \hline<br /> \end{array} &lt;/cmath&gt;<br /> <br /> We have now established a bijection between valid permutations of the numbers 1 through 30 and permutations of the congruence classes modulo 2, 3, and 5, so &lt;math&gt;N&lt;/math&gt; is equal to the number of permutations of congruence classes. There are always &lt;math&gt;m&lt;/math&gt; congruence classes &lt;math&gt;\mod m,&lt;/math&gt; so &lt;math&gt;N = 2! \cdot 3! \cdot 5! = 2 \cdot 6 \cdot 120 = 1440 \equiv \framebox[1.3\width]{440} \mod 1000.&lt;/math&gt;<br /> <br /> ===Solution 3 (2-sec solve)===<br /> Note that &lt;math&gt;30=2\cdot 3\cdot 5&lt;/math&gt;. Since &lt;math&gt;\gcd(2, 3, 5)=1&lt;/math&gt;, by CRT, for each value &lt;math&gt;k=0\ldots 29&lt;/math&gt; modulo &lt;math&gt;30&lt;/math&gt; there exists a unique ordered triple of values &lt;math&gt;(a, b, c)&lt;/math&gt; such that &lt;math&gt;k\equiv a\pmod{2}&lt;/math&gt;, &lt;math&gt;k\equiv b\pmod{3}&lt;/math&gt;, and &lt;math&gt;k\equiv c\pmod{5}&lt;/math&gt;. Therefore, we can independantly assign the residues modulo &lt;math&gt;2, 3, 5&lt;/math&gt;, so &lt;math&gt;N=2!\cdot 3!\cdot 5!=1440&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> -TheUltimate123<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=13 | num-a=15}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_14&diff=92062 2011 AIME II Problems/Problem 14 2018-02-19T20:14:28Z <p>Theultimate123: /* Solution 3 (2-sec solve) */</p> <hr /> <div>==Problem 14==<br /> There are &lt;math&gt;N&lt;/math&gt; [[permutation]]s &lt;math&gt;(a_{1}, a_{2}, ... , a_{30})&lt;/math&gt; of &lt;math&gt;1, 2, \ldots, 30&lt;/math&gt; such that for &lt;math&gt;m \in \left\{{2, 3, 5}\right\}&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;a_{n+m} - a_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;1 \leq n &lt; n+m \leq 30&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> __TOC__<br /> ==Solutions==<br /> ===Solution 1===<br /> Be wary of &quot;position&quot; versus &quot;number&quot; in the solution!<br /> <br /> Each POSITION in the 30-position permutation is uniquely defined by an ordered triple &lt;math&gt;(i, j, k)&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt;th position is defined by this ordered triple where &lt;math&gt;i&lt;/math&gt; is &lt;math&gt;n \mod 2&lt;/math&gt;, &lt;math&gt;j&lt;/math&gt; is &lt;math&gt;n \mod 3&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;n \mod 5&lt;/math&gt;. There are 2 choices for &lt;math&gt;i&lt;/math&gt;, 3 for &lt;math&gt;j&lt;/math&gt;, and 5 for &lt;math&gt;k&lt;/math&gt;, yielding &lt;math&gt;2 \cdot 3 \cdot 5=30&lt;/math&gt; possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if i is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If j is the same, then the two numbers must be equivalent &lt;math&gt;\mod 3&lt;/math&gt;, and if &lt;math&gt;k&lt;/math&gt; is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have &lt;math&gt;1 \mod 2&lt;/math&gt;! It's that the POSITION which NUMBER 1 occupies has &lt;math&gt;1 \mod 2&lt;/math&gt;!<br /> <br /> The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.<br /> <br /> The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.<br /> <br /> Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of &lt;math&gt;1\cdot 1 \cdot 3=3&lt;/math&gt; choices for the placement of 3.<br /> <br /> As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of &lt;math&gt;1 \cdot 1 \cdot 2=2&lt;/math&gt; possible placements for 4.<br /> <br /> 5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.<br /> <br /> All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed. <br /> <br /> Thus, &lt;math&gt;N = 30 \cdot 8 \cdot 3 \cdot 2=30 \cdot 48=1440&lt;/math&gt;. Thus, the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\boxed{440}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> We observe that the condition on the permutations means that two numbers with indices congruent &lt;math&gt;\mod m&lt;/math&gt; are themselves congruent &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m \in \{ 2,3,5\}.&lt;/math&gt; Furthermore, suppose that &lt;math&gt;a_n \equiv k \mod m.&lt;/math&gt; Then, there are &lt;math&gt;30/m&lt;/math&gt; indices congruent to &lt;math&gt;n \mod m,&lt;/math&gt; and &lt;math&gt;30/m&lt;/math&gt; numbers congruent to &lt;math&gt;k \mod m,&lt;/math&gt; because 2, 3, and 5 are all factors of 30. Therefore, since every index congruent to &lt;math&gt;n&lt;/math&gt; must contain a number congruent to &lt;math&gt;k,&lt;/math&gt; and no number can appear twice in the permutation, only the indices congruent to &lt;math&gt;n&lt;/math&gt; contain numbers congruent to &lt;math&gt;k.&lt;/math&gt; In other words, &lt;math&gt;a_i \equiv a_j \mod m \iff i \equiv j \mod m.&lt;/math&gt; But it is not necessary that &lt;math&gt;\textcolor{red}{(a_i\equiv i)\cup (a_j\equiv j)\mod m}&lt;/math&gt;. In fact, if that were the case, there would only be one way to assign the indices, since &lt;math&gt;2,3,5&lt;/math&gt; are relatively prime to each other and &lt;math&gt;30=\text{lcm}(2,3,5)&lt;/math&gt;: &lt;math&gt;\{a_1,a_2,\dots a_{30}\}\in\{1,2,\dots 30\}\text{ }respectively&lt;/math&gt;.<br /> <br /> This tells us that in a valid permutation, the congruence classes &lt;math&gt;\mod m&lt;/math&gt; are simply swapped around, and if the set &lt;math&gt;S&lt;/math&gt; is a congruence class &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m = &lt;/math&gt; 2, 3, or 5, the set &lt;math&gt;\{ a_i \vert i \in S \}&lt;/math&gt; is still a congruence class &lt;math&gt;\mod m.&lt;/math&gt; Clearly, each valid permutation of the numbers 1 through 30 corresponds to exactly one permutation of the congruence classes modulo 2, 3, and 5. Also, if we choose some permutations of the congruence classes modulo 2, 3, and 5, they correspond to exactly one valid permutation of the numbers 1 through 30. This can be shown as follows: First of all, the choice of permutations of the congruence classes gives us every number in the permutation modulo 2, 3, and 5, so by the Chinese Remainder Theorem, it gives us every number &lt;math&gt;\mod 2\cdot 3\cdot 5 = 30.&lt;/math&gt; Because the numbers must be between 1 and 30 inclusive, it thus uniquely determines what number goes in each index. Furthermore, two different indices cannot contain the same number. We will prove this by contradiction, calling the indices &lt;math&gt;a_i&lt;/math&gt; and &lt;math&gt;a_j&lt;/math&gt; for &lt;math&gt;i \neq j.&lt;/math&gt; If &lt;math&gt;a_i=a_j,&lt;/math&gt; then they must have the same residues modulo 2, 3, and 5, and so &lt;math&gt;i \equiv j&lt;/math&gt; modulo 2, 3, and 5. Again using the Chinese Remainder Theorem, we conclude that &lt;math&gt;i \equiv j \mod 30,&lt;/math&gt; so because &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are both between 1 and 30 inclusive, &lt;math&gt;i = j,&lt;/math&gt; giving us a contradiction. Therefore, every choice of permutations of the congruence classes modulo 2, 3, and 5 corresponds to exactly one valid permutation of the numbers 1 through 30.<br /> <br /> In other words, each set of assignment from &lt;math&gt;a_j\rightarrow j\mod (2,3,5)&lt;/math&gt; determines a unique string of &lt;math&gt;30&lt;/math&gt; numbers. For example:<br /> <br /> &lt;math&gt;\left[(0,1)\rightarrow (1,0)\right]\cap\left[(0,1,2)\rightarrow (0,2,1)\right]\cap\left[(0,1,2,3,4)\rightarrow (4,2,3,0,1)\right]&lt;/math&gt;:<br /> &lt;cmath&gt;\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \hline<br /> 2&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0\\ \hline<br /> 3&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1\\ \hline<br /> 5&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1\\ \hline\hline<br /> 30&amp;9&amp;2&amp;13&amp;0&amp;11&amp;4&amp;27&amp;8&amp;25&amp;6&amp;29&amp;22&amp;3&amp;20&amp;1&amp;24&amp;17&amp;28&amp;15&amp;26&amp;19&amp;12&amp;23&amp;10&amp;21&amp;14&amp;7&amp;18&amp;5&amp;16\\ \hline<br /> \end{array} &lt;/cmath&gt;<br /> <br /> We have now established a bijection between valid permutations of the numbers 1 through 30 and permutations of the congruence classes modulo 2, 3, and 5, so &lt;math&gt;N&lt;/math&gt; is equal to the number of permutations of congruence classes. There are always &lt;math&gt;m&lt;/math&gt; congruence classes &lt;math&gt;\mod m,&lt;/math&gt; so &lt;math&gt;N = 2! \cdot 3! \cdot 5! = 2 \cdot 6 \cdot 120 = 1440 \equiv \framebox[1.3\width]{440} \mod 1000.&lt;/math&gt;<br /> <br /> ===Solution 3 (2-sec solve)===<br /> Note that &lt;math&gt;30=2\cdot 3\cdot 5&lt;/math&gt;. Since &lt;math&gt;\gcd(2, 3, 5)=1&lt;/math&gt;, by CRT, for each value &lt;math&gt;k=0\ldots 29&lt;/math&gt; modulo &lt;math&gt;30&lt;/math&gt; there exists a unique ordered triple of values &lt;math&gt;(a, b, c)&lt;/math&gt; such that &lt;math&gt;k\equiv a\pmod{2}&lt;/math&gt;, &lt;math&gt;k\equiv b\pmod{3}&lt;/math&gt;, and &lt;math&gt;k\equiv c\pmod{5}&lt;/math&gt;. Therefore, we can independantly assign the residues modulo &lt;math&gt;2, 3, 5&lt;/math&gt;, so &lt;math&gt;N=2!\cdot 3!\cdot 5!=1440&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{440}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=13 | num-a=15}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_14&diff=92061 2011 AIME II Problems/Problem 14 2018-02-19T20:13:17Z <p>Theultimate123: </p> <hr /> <div>==Problem 14==<br /> There are &lt;math&gt;N&lt;/math&gt; [[permutation]]s &lt;math&gt;(a_{1}, a_{2}, ... , a_{30})&lt;/math&gt; of &lt;math&gt;1, 2, \ldots, 30&lt;/math&gt; such that for &lt;math&gt;m \in \left\{{2, 3, 5}\right\}&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;a_{n+m} - a_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;1 \leq n &lt; n+m \leq 30&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> __TOC__<br /> ==Solutions==<br /> ===Solution 1===<br /> Be wary of &quot;position&quot; versus &quot;number&quot; in the solution!<br /> <br /> Each POSITION in the 30-position permutation is uniquely defined by an ordered triple &lt;math&gt;(i, j, k)&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt;th position is defined by this ordered triple where &lt;math&gt;i&lt;/math&gt; is &lt;math&gt;n \mod 2&lt;/math&gt;, &lt;math&gt;j&lt;/math&gt; is &lt;math&gt;n \mod 3&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;n \mod 5&lt;/math&gt;. There are 2 choices for &lt;math&gt;i&lt;/math&gt;, 3 for &lt;math&gt;j&lt;/math&gt;, and 5 for &lt;math&gt;k&lt;/math&gt;, yielding &lt;math&gt;2 \cdot 3 \cdot 5=30&lt;/math&gt; possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if i is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If j is the same, then the two numbers must be equivalent &lt;math&gt;\mod 3&lt;/math&gt;, and if &lt;math&gt;k&lt;/math&gt; is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have &lt;math&gt;1 \mod 2&lt;/math&gt;! It's that the POSITION which NUMBER 1 occupies has &lt;math&gt;1 \mod 2&lt;/math&gt;!<br /> <br /> The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.<br /> <br /> The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.<br /> <br /> Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of &lt;math&gt;1\cdot 1 \cdot 3=3&lt;/math&gt; choices for the placement of 3.<br /> <br /> As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of &lt;math&gt;1 \cdot 1 \cdot 2=2&lt;/math&gt; possible placements for 4.<br /> <br /> 5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.<br /> <br /> All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed. <br /> <br /> Thus, &lt;math&gt;N = 30 \cdot 8 \cdot 3 \cdot 2=30 \cdot 48=1440&lt;/math&gt;. Thus, the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\boxed{440}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> We observe that the condition on the permutations means that two numbers with indices congruent &lt;math&gt;\mod m&lt;/math&gt; are themselves congruent &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m \in \{ 2,3,5\}.&lt;/math&gt; Furthermore, suppose that &lt;math&gt;a_n \equiv k \mod m.&lt;/math&gt; Then, there are &lt;math&gt;30/m&lt;/math&gt; indices congruent to &lt;math&gt;n \mod m,&lt;/math&gt; and &lt;math&gt;30/m&lt;/math&gt; numbers congruent to &lt;math&gt;k \mod m,&lt;/math&gt; because 2, 3, and 5 are all factors of 30. Therefore, since every index congruent to &lt;math&gt;n&lt;/math&gt; must contain a number congruent to &lt;math&gt;k,&lt;/math&gt; and no number can appear twice in the permutation, only the indices congruent to &lt;math&gt;n&lt;/math&gt; contain numbers congruent to &lt;math&gt;k.&lt;/math&gt; In other words, &lt;math&gt;a_i \equiv a_j \mod m \iff i \equiv j \mod m.&lt;/math&gt; But it is not necessary that &lt;math&gt;\textcolor{red}{(a_i\equiv i)\cup (a_j\equiv j)\mod m}&lt;/math&gt;. In fact, if that were the case, there would only be one way to assign the indices, since &lt;math&gt;2,3,5&lt;/math&gt; are relatively prime to each other and &lt;math&gt;30=\text{lcm}(2,3,5)&lt;/math&gt;: &lt;math&gt;\{a_1,a_2,\dots a_{30}\}\in\{1,2,\dots 30\}\text{ }respectively&lt;/math&gt;.<br /> <br /> This tells us that in a valid permutation, the congruence classes &lt;math&gt;\mod m&lt;/math&gt; are simply swapped around, and if the set &lt;math&gt;S&lt;/math&gt; is a congruence class &lt;math&gt;\mod m&lt;/math&gt; for &lt;math&gt;m = &lt;/math&gt; 2, 3, or 5, the set &lt;math&gt;\{ a_i \vert i \in S \}&lt;/math&gt; is still a congruence class &lt;math&gt;\mod m.&lt;/math&gt; Clearly, each valid permutation of the numbers 1 through 30 corresponds to exactly one permutation of the congruence classes modulo 2, 3, and 5. Also, if we choose some permutations of the congruence classes modulo 2, 3, and 5, they correspond to exactly one valid permutation of the numbers 1 through 30. This can be shown as follows: First of all, the choice of permutations of the congruence classes gives us every number in the permutation modulo 2, 3, and 5, so by the Chinese Remainder Theorem, it gives us every number &lt;math&gt;\mod 2\cdot 3\cdot 5 = 30.&lt;/math&gt; Because the numbers must be between 1 and 30 inclusive, it thus uniquely determines what number goes in each index. Furthermore, two different indices cannot contain the same number. We will prove this by contradiction, calling the indices &lt;math&gt;a_i&lt;/math&gt; and &lt;math&gt;a_j&lt;/math&gt; for &lt;math&gt;i \neq j.&lt;/math&gt; If &lt;math&gt;a_i=a_j,&lt;/math&gt; then they must have the same residues modulo 2, 3, and 5, and so &lt;math&gt;i \equiv j&lt;/math&gt; modulo 2, 3, and 5. Again using the Chinese Remainder Theorem, we conclude that &lt;math&gt;i \equiv j \mod 30,&lt;/math&gt; so because &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are both between 1 and 30 inclusive, &lt;math&gt;i = j,&lt;/math&gt; giving us a contradiction. Therefore, every choice of permutations of the congruence classes modulo 2, 3, and 5 corresponds to exactly one valid permutation of the numbers 1 through 30.<br /> <br /> In other words, each set of assignment from &lt;math&gt;a_j\rightarrow j\mod (2,3,5)&lt;/math&gt; determines a unique string of &lt;math&gt;30&lt;/math&gt; numbers. For example:<br /> <br /> &lt;math&gt;\left[(0,1)\rightarrow (1,0)\right]\cap\left[(0,1,2)\rightarrow (0,2,1)\right]\cap\left[(0,1,2,3,4)\rightarrow (4,2,3,0,1)\right]&lt;/math&gt;:<br /> &lt;cmath&gt;\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \hline<br /> 2&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;0\\ \hline<br /> 3&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1&amp;0&amp;2&amp;1\\ \hline<br /> 5&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1&amp;4&amp;2&amp;3&amp;0&amp;1\\ \hline\hline<br /> 30&amp;9&amp;2&amp;13&amp;0&amp;11&amp;4&amp;27&amp;8&amp;25&amp;6&amp;29&amp;22&amp;3&amp;20&amp;1&amp;24&amp;17&amp;28&amp;15&amp;26&amp;19&amp;12&amp;23&amp;10&amp;21&amp;14&amp;7&amp;18&amp;5&amp;16\\ \hline<br /> \end{array} &lt;/cmath&gt;<br /> <br /> We have now established a bijection between valid permutations of the numbers 1 through 30 and permutations of the congruence classes modulo 2, 3, and 5, so &lt;math&gt;N&lt;/math&gt; is equal to the number of permutations of congruence classes. There are always &lt;math&gt;m&lt;/math&gt; congruence classes &lt;math&gt;\mod m,&lt;/math&gt; so &lt;math&gt;N = 2! \cdot 3! \cdot 5! = 2 \cdot 6 \cdot 120 = 1440 \equiv \framebox[1.3\width]{440} \mod 1000.&lt;/math&gt;<br /> <br /> ===Solution 3 (2-sec solve)===<br /> Note that &lt;math&gt;30=2\cdot 3\cdot 5&lt;/math&gt;. Since &lt;math&gt;\gcd(2, 3, 5)=1&lt;/math&gt;, by CRT, for each value &lt;math&gt;k=0\ldots 29&lt;/math&gt; modulo &lt;math&gt;30&lt;/math&gt; there exists a unique ordered triple of values &lt;math&gt;(a, b, c)&lt;/math&gt; such that &lt;math&gt;k\equiv a\pmod{2}&lt;/math&gt;, &lt;math&gt;k\equiv b\pmod{3}&lt;/math&gt;, and &lt;math&gt;k\equiv c\pmod{5}&lt;/math&gt;. Therefore, we can independantly assign the residues modulo &lt;math&gt;2, 3, 5&lt;/math&gt;, giving an answer of &lt;math&gt;2!\cdot 3!\cdot 5!=1\boxed{440}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=13 | num-a=15}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=88939 2015 AMC 12B Problems/Problem 19 2017-12-17T00:10:35Z <p>Theultimate123: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\angle C = 90^\circ&lt;/math&gt; and &lt;math&gt;AB = 12&lt;/math&gt;. Squares &lt;math&gt;ABXY&lt;/math&gt; and &lt;math&gt;ACWZ&lt;/math&gt; are constructed outside of the triangle. The points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle. What is the perimeter of the triangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of &lt;math&gt;WZ&lt;/math&gt; and &lt;math&gt;XY&lt;/math&gt; and finding their intersection point. This point happens to be the midpoint of &lt;math&gt;AB&lt;/math&gt;, the hypotenuse. Let this point be &lt;math&gt;M&lt;/math&gt;. To find the radius, determine &lt;math&gt;MY&lt;/math&gt;, where &lt;math&gt;MY^{2} = MA^2 + AY^2&lt;/math&gt;, &lt;math&gt;MA = \frac{12}{2} = 6&lt;/math&gt;, and &lt;math&gt;AY = AB = 12&lt;/math&gt;. Thus, the radius &lt;math&gt;=r =MY = 6\sqrt5&lt;/math&gt;.<br /> <br /> Next we let &lt;math&gt;AC = b&lt;/math&gt; and &lt;math&gt;BC = a&lt;/math&gt;. Consider the right triangle &lt;math&gt;ACB&lt;/math&gt; first. Using the pythagorean theorem, we find that &lt;math&gt;a^2 + b^2 = 12^2 = 144&lt;/math&gt;. Next, we let &lt;math&gt;M'&lt;/math&gt; to be the midpoint of &lt;math&gt;WZ&lt;/math&gt;, and we consider right triangle &lt;math&gt;ZM'M&lt;/math&gt;. By the pythagorean theorem, we have that &lt;math&gt;\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180&lt;/math&gt;. Expanding this equation, we get that<br /> <br /> &lt;cmath&gt;\frac{1}{4}(a^2+b^2) + b^2 + ab = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{144}{4} + b^2 + ab = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2 + ab = 144 = a^2 + b^2&lt;/cmath&gt;<br /> &lt;cmath&gt;ab = a^2&lt;/cmath&gt;<br /> &lt;cmath&gt;b = a&lt;/cmath&gt;<br /> <br /> This means that &lt;math&gt;ABC&lt;/math&gt; is a 45-45-90 triangle, so &lt;math&gt;a = b = \frac{12}{\sqrt2} = 6\sqrt2&lt;/math&gt;. Thus the perimeter is &lt;math&gt;a + b + AB = 12\sqrt2 + 12&lt;/math&gt; which is answer &lt;math&gt;\boxed{\textbf{(C)}\; 12 + 12\sqrt2}&lt;/math&gt;.<br /> image needed<br /> <br /> ==Solution 2==<br /> The center of the circle on which &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie must be equidistant from each of these four points. Draw the perpendicular bisectors of &lt;math&gt;\overline{XY}&lt;/math&gt; and of &lt;math&gt;\overline{WZ}&lt;/math&gt;. Note that the perpendicular bisector of &lt;math&gt;\overline{WZ}&lt;/math&gt; is parallel to &lt;math&gt;\overline{CW}&lt;/math&gt; and passes through the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. Therefore, the triangle that is formed by &lt;math&gt;A&lt;/math&gt;, the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, and the point at which this perpendicular bisector intersects &lt;math&gt;\overline{AB}&lt;/math&gt; must be similar to &lt;math&gt;\triangle ABC&lt;/math&gt;, and the ratio of a side of the smaller triangle to a side of &lt;math&gt;\triangle ABC&lt;/math&gt; is 1:2. Consequently, the perpendicular bisector of &lt;math&gt;\overline{XY}&lt;/math&gt; passes through the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. The perpendicular bisector of &lt;math&gt;\overline{WZ}&lt;/math&gt; must include the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; as well. Since all points on a perpendicular bisector of any two points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; are equidistant from &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, the center of the circle must be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;.<br /> <br /> Now the distance between the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, which is equal to the radius of this circle, is &lt;math&gt;\sqrt{12^2 + 6^2} = \sqrt{180}&lt;/math&gt;. Let &lt;math&gt;a=AC&lt;/math&gt;. Then the distance between the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, also equal to the radius of the circle, is given by &lt;math&gt;\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}&lt;/math&gt; (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br /> <br /> &lt;cmath&gt;\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;144 - a^2 = a\sqrt{144-a^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(144-a^2)^2 = a^2(144-a^2)&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by &lt;math&gt;(144-a^2)&lt;/math&gt;, and arrive at &lt;math&gt;a = 6\sqrt{2}&lt;/math&gt;. The length of other leg of the triangle must be &lt;math&gt;\sqrt{144-72} = 6\sqrt{2}&lt;/math&gt;. Thus, the perimeter of the triangle is &lt;math&gt;12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> In order to solve this problem, we can search for similar triangles. Begin by drawing triangle &lt;math&gt;ABC&lt;/math&gt; and squares &lt;math&gt;ABXY&lt;/math&gt; and &lt;math&gt;ACWZ&lt;/math&gt;. Draw segments &lt;math&gt;\overline{YZ}&lt;/math&gt; and &lt;math&gt;\overline{WX}&lt;/math&gt;. Because we are given points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle, we can conclude that &lt;math&gt;WXYZ&lt;/math&gt; forms a cyclic quadrilateral. Take &lt;math&gt;\overline{AC}&lt;/math&gt; and extend it through a point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;\overline{YZ}&lt;/math&gt;. Now, we must do some angle chasing to prove that &lt;math&gt;\triangle WBX&lt;/math&gt; is similar to &lt;math&gt;\triangle YAZ&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\alpha&lt;/math&gt; denote the measure of &lt;math&gt;\angle ABC&lt;/math&gt;. Following this, &lt;math&gt;\angle BAC&lt;/math&gt; measures &lt;math&gt;90 - \alpha&lt;/math&gt;. By our construction, &lt;math&gt;\overline{CAP}&lt;/math&gt; is a straight line, and we know &lt;math&gt;\angle YAB&lt;/math&gt; is a right angle. Therefore, &lt;math&gt;\angle PAY&lt;/math&gt; measures &lt;math&gt;\alpha&lt;/math&gt;. Also, &lt;math&gt;\angle CAZ&lt;/math&gt; is a right angle and thus, &lt;math&gt;\angle ZAP&lt;/math&gt; is a right angle. Sum &lt;math&gt;\angle ZAP&lt;/math&gt; and &lt;math&gt;\angle PAY&lt;/math&gt; to find &lt;math&gt;\angle ZAY&lt;/math&gt;, which measures &lt;math&gt;90 + \alpha&lt;/math&gt;. We also know that &lt;math&gt;\angle WBY&lt;/math&gt; measures &lt;math&gt;90 + \alpha&lt;/math&gt;. Therefore, &lt;math&gt;\angle ZAY = \angle WBX&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\beta&lt;/math&gt; denote the measure of &lt;math&gt;\angle AZY&lt;/math&gt;. It follows that &lt;math&gt;\angle WZY&lt;/math&gt; measures &lt;math&gt;90 + \beta^\circ&lt;/math&gt;. Because &lt;math&gt;WXYZ&lt;/math&gt; is a cyclic quadrilateral, &lt;math&gt;\angle WZY + \angle YXW = 180^\circ&lt;/math&gt;. Therefore, &lt;math&gt;\angle YXW&lt;/math&gt; must measure &lt;math&gt;90 - \beta&lt;/math&gt;, and &lt;math&gt;\angle BXW&lt;/math&gt; must measure &lt;math&gt;\beta&lt;/math&gt;. Therefore, &lt;math&gt;\angle AZY = \angle BXW&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle ZAY = \angle WBX&lt;/math&gt; and &lt;math&gt;\angle AZY = \angle BXW&lt;/math&gt;, so &lt;math&gt;\triangle AZY \sim \triangle BXW&lt;/math&gt;! Let &lt;math&gt;x = AC = WC&lt;/math&gt;. By Pythagorean theorem, &lt;math&gt;BC = \sqrt{144-x^2}&lt;/math&gt;. Now we have &lt;math&gt;WB = WC + BC = x + \sqrt{144-x^2}&lt;/math&gt;, &lt;math&gt;BX = 12&lt;/math&gt;, &lt;math&gt;YA = 12&lt;/math&gt;, and &lt;math&gt;AZ = x&lt;/math&gt;. <br /> We can set up an equation: <br /> <br /> &lt;cmath&gt;\frac{YA}{AZ} = \frac{WB}{BX}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}&lt;/cmath&gt;<br /> &lt;cmath&gt;144 = x^2 + x\sqrt{144-x^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;12^2 - x^2 = x\sqrt{144-x^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4&lt;/cmath&gt;<br /> &lt;cmath&gt;2x^4 - 3(12^2)x^2 + 12^4 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(2x^2 - 144)(x^2 - 144) = 0&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt;, we find that &lt;math&gt;x = 6\sqrt{2}&lt;/math&gt; or &lt;math&gt;x = 12&lt;/math&gt;, which we omit. The perimeter of the triangle is &lt;math&gt;12 + x + \sqrt{144-x^2}&lt;/math&gt;. Plugging in &lt;math&gt;x = 6\sqrt{2}&lt;/math&gt;, we get &lt;math&gt;\boxed{\textbf{(C)}\; 12+12\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We claim that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle if &lt;math&gt;\triangle ACB&lt;/math&gt; is an isosceles right triangle.<br /> <br /> ''Proof:'' If &lt;math&gt;\triangle ACB&lt;/math&gt; is a right triangle, then &lt;math&gt;\angle WAY=180º&lt;/math&gt;. Therefore, &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear. Since &lt;math&gt;WY&lt;/math&gt; and &lt;math&gt;YX&lt;/math&gt; form a right angle, &lt;math&gt;WX&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle WYX&lt;/math&gt;. Similarly, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear, and &lt;math&gt;ZX&lt;/math&gt; forms a right angle with &lt;math&gt;ZW&lt;/math&gt;. Thus, &lt;math&gt;WX&lt;/math&gt; is also the diameter of the circumcircle of &lt;math&gt;\triangle WZX&lt;/math&gt;. Therefore, since &lt;math&gt;\triangle WYX&lt;/math&gt; and &lt;math&gt;\triangle WZX&lt;/math&gt; share a circumcircle, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle if &lt;math&gt;\triangle ACB&lt;/math&gt; is an isosceles triangle.<br /> <br /> If &lt;math&gt;\triangle ACB&lt;/math&gt; is isosceles, then its legs have length &lt;math&gt;6\sqrt{2}&lt;/math&gt;. The perimeter of &lt;math&gt;\triangle ACB&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C) }12+12\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=88938 2015 AMC 12B Problems/Problem 19 2017-12-17T00:09:30Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\angle C = 90^\circ&lt;/math&gt; and &lt;math&gt;AB = 12&lt;/math&gt;. Squares &lt;math&gt;ABXY&lt;/math&gt; and &lt;math&gt;ACWZ&lt;/math&gt; are constructed outside of the triangle. The points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle. What is the perimeter of the triangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of &lt;math&gt;WZ&lt;/math&gt; and &lt;math&gt;XY&lt;/math&gt; and finding their intersection point. This point happens to be the midpoint of &lt;math&gt;AB&lt;/math&gt;, the hypotenuse. Let this point be &lt;math&gt;M&lt;/math&gt;. To find the radius, determine &lt;math&gt;MY&lt;/math&gt;, where &lt;math&gt;MY^{2} = MA^2 + AY^2&lt;/math&gt;, &lt;math&gt;MA = \frac{12}{2} = 6&lt;/math&gt;, and &lt;math&gt;AY = AB = 12&lt;/math&gt;. Thus, the radius &lt;math&gt;=r =MY = 6\sqrt5&lt;/math&gt;.<br /> <br /> Next we let &lt;math&gt;AC = b&lt;/math&gt; and &lt;math&gt;BC = a&lt;/math&gt;. Consider the right triangle &lt;math&gt;ACB&lt;/math&gt; first. Using the pythagorean theorem, we find that &lt;math&gt;a^2 + b^2 = 12^2 = 144&lt;/math&gt;. Next, we let &lt;math&gt;M'&lt;/math&gt; to be the midpoint of &lt;math&gt;WZ&lt;/math&gt;, and we consider right triangle &lt;math&gt;ZM'M&lt;/math&gt;. By the pythagorean theorem, we have that &lt;math&gt;\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180&lt;/math&gt;. Expanding this equation, we get that<br /> <br /> &lt;cmath&gt;\frac{1}{4}(a^2+b^2) + b^2 + ab = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{144}{4} + b^2 + ab = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2 + ab = 144 = a^2 + b^2&lt;/cmath&gt;<br /> &lt;cmath&gt;ab = a^2&lt;/cmath&gt;<br /> &lt;cmath&gt;b = a&lt;/cmath&gt;<br /> <br /> This means that &lt;math&gt;ABC&lt;/math&gt; is a 45-45-90 triangle, so &lt;math&gt;a = b = \frac{12}{\sqrt2} = 6\sqrt2&lt;/math&gt;. Thus the perimeter is &lt;math&gt;a + b + AB = 12\sqrt2 + 12&lt;/math&gt; which is answer &lt;math&gt;\boxed{\textbf{(C)}\; 12 + 12\sqrt2}&lt;/math&gt;.<br /> image needed<br /> <br /> ==Solution 2==<br /> The center of the circle on which &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie must be equidistant from each of these four points. Draw the perpendicular bisectors of &lt;math&gt;\overline{XY}&lt;/math&gt; and of &lt;math&gt;\overline{WZ}&lt;/math&gt;. Note that the perpendicular bisector of &lt;math&gt;\overline{WZ}&lt;/math&gt; is parallel to &lt;math&gt;\overline{CW}&lt;/math&gt; and passes through the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. Therefore, the triangle that is formed by &lt;math&gt;A&lt;/math&gt;, the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, and the point at which this perpendicular bisector intersects &lt;math&gt;\overline{AB}&lt;/math&gt; must be similar to &lt;math&gt;\triangle ABC&lt;/math&gt;, and the ratio of a side of the smaller triangle to a side of &lt;math&gt;\triangle ABC&lt;/math&gt; is 1:2. Consequently, the perpendicular bisector of &lt;math&gt;\overline{XY}&lt;/math&gt; passes through the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. The perpendicular bisector of &lt;math&gt;\overline{WZ}&lt;/math&gt; must include the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; as well. Since all points on a perpendicular bisector of any two points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; are equidistant from &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, the center of the circle must be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;.<br /> <br /> Now the distance between the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, which is equal to the radius of this circle, is &lt;math&gt;\sqrt{12^2 + 6^2} = \sqrt{180}&lt;/math&gt;. Let &lt;math&gt;a=AC&lt;/math&gt;. Then the distance between the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, also equal to the radius of the circle, is given by &lt;math&gt;\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}&lt;/math&gt; (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br /> <br /> &lt;cmath&gt;\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180&lt;/cmath&gt;<br /> &lt;cmath&gt;144 - a^2 = a\sqrt{144-a^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(144-a^2)^2 = a^2(144-a^2)&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by &lt;math&gt;(144-a^2)&lt;/math&gt;, and arrive at &lt;math&gt;a = 6\sqrt{2}&lt;/math&gt;. The length of other leg of the triangle must be &lt;math&gt;\sqrt{144-72} = 6\sqrt{2}&lt;/math&gt;. Thus, the perimeter of the triangle is &lt;math&gt;12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> In order to solve this problem, we can search for similar triangles. Begin by drawing triangle &lt;math&gt;ABC&lt;/math&gt; and squares &lt;math&gt;ABXY&lt;/math&gt; and &lt;math&gt;ACWZ&lt;/math&gt;. Draw segments &lt;math&gt;\overline{YZ}&lt;/math&gt; and &lt;math&gt;\overline{WX}&lt;/math&gt;. Because we are given points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle, we can conclude that &lt;math&gt;WXYZ&lt;/math&gt; forms a cyclic quadrilateral. Take &lt;math&gt;\overline{AC}&lt;/math&gt; and extend it through a point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;\overline{YZ}&lt;/math&gt;. Now, we must do some angle chasing to prove that &lt;math&gt;\triangle WBX&lt;/math&gt; is similar to &lt;math&gt;\triangle YAZ&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\alpha&lt;/math&gt; denote the measure of &lt;math&gt;\angle ABC&lt;/math&gt;. Following this, &lt;math&gt;\angle BAC&lt;/math&gt; measures &lt;math&gt;90 - \alpha&lt;/math&gt;. By our construction, &lt;math&gt;\overline{CAP}&lt;/math&gt; is a straight line, and we know &lt;math&gt;\angle YAB&lt;/math&gt; is a right angle. Therefore, &lt;math&gt;\angle PAY&lt;/math&gt; measures &lt;math&gt;\alpha&lt;/math&gt;. Also, &lt;math&gt;\angle CAZ&lt;/math&gt; is a right angle and thus, &lt;math&gt;\angle ZAP&lt;/math&gt; is a right angle. Sum &lt;math&gt;\angle ZAP&lt;/math&gt; and &lt;math&gt;\angle PAY&lt;/math&gt; to find &lt;math&gt;\angle ZAY&lt;/math&gt;, which measures &lt;math&gt;90 + \alpha&lt;/math&gt;. We also know that &lt;math&gt;\angle WBY&lt;/math&gt; measures &lt;math&gt;90 + \alpha&lt;/math&gt;. Therefore, &lt;math&gt;\angle ZAY = \angle WBX&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\beta&lt;/math&gt; denote the measure of &lt;math&gt;\angle AZY&lt;/math&gt;. It follows that &lt;math&gt;\angle WZY&lt;/math&gt; measures &lt;math&gt;90 + \beta^\circ&lt;/math&gt;. Because &lt;math&gt;WXYZ&lt;/math&gt; is a cyclic quadrilateral, &lt;math&gt;\angle WZY + \angle YXW = 180^\circ&lt;/math&gt;. Therefore, &lt;math&gt;\angle YXW&lt;/math&gt; must measure &lt;math&gt;90 - \beta&lt;/math&gt;, and &lt;math&gt;\angle BXW&lt;/math&gt; must measure &lt;math&gt;\beta&lt;/math&gt;. Therefore, &lt;math&gt;\angle AZY = \angle BXW&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle ZAY = \angle WBX&lt;/math&gt; and &lt;math&gt;\angle AZY = \angle BXW&lt;/math&gt;, so &lt;math&gt;\triangle AZY \sim \triangle BXW&lt;/math&gt;! Let &lt;math&gt;x = AC = WC&lt;/math&gt;. By Pythagorean theorem, &lt;math&gt;BC = \sqrt{144-x^2}&lt;/math&gt;. Now we have &lt;math&gt;WB = WC + BC = x + \sqrt{144-x^2}&lt;/math&gt;, &lt;math&gt;BX = 12&lt;/math&gt;, &lt;math&gt;YA = 12&lt;/math&gt;, and &lt;math&gt;AZ = x&lt;/math&gt;. <br /> We can set up an equation: <br /> <br /> &lt;cmath&gt;\frac{YA}{AZ} = \frac{WB}{BX}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}&lt;/cmath&gt;<br /> &lt;cmath&gt;144 = x^2 + x\sqrt{144-x^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;12^2 - x^2 = x\sqrt{144-x^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4&lt;/cmath&gt;<br /> &lt;cmath&gt;2x^4 - 3(12^2)x^2 + 12^4 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(2x^2 - 144)(x^2 - 144) = 0&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt;, we find that &lt;math&gt;x = 6\sqrt{2}&lt;/math&gt; or &lt;math&gt;x = 12&lt;/math&gt;, which we omit. The perimeter of the triangle is &lt;math&gt;12 + x + \sqrt{144-x^2}&lt;/math&gt;. Plugging in &lt;math&gt;x = 6\sqrt{2}&lt;/math&gt;, we get &lt;math&gt;\boxed{\textbf{(C)}\; 12+12\sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We claim that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle if &lt;math&gt;\triangle ACB&lt;/math&gt; is an isosceles right triangle.<br /> <br /> Proof: If &lt;math&gt;\triangle ACB&lt;/math&gt; is a right triangle, then &lt;math&gt;\angle WAY=180º&lt;/math&gt;. Therefore, &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear. Since &lt;math&gt;WY&lt;/math&gt; and &lt;math&gt;YX&lt;/math&gt; form a right angle, &lt;math&gt;WX&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle WYX&lt;/math&gt;. Similarly, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear, and &lt;math&gt;ZX&lt;/math&gt; forms a right angle with &lt;math&gt;ZW&lt;/math&gt;. Thus, &lt;math&gt;WX&lt;/math&gt; is also the diameter of the circumcircle of &lt;math&gt;\triangle WZX&lt;/math&gt;. Therefore, since &lt;math&gt;\triangle WYX&lt;/math&gt; and &lt;math&gt;\triangle WZX&lt;/math&gt; share a circumcircle, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;Z&lt;/math&gt;, and &lt;math&gt;W&lt;/math&gt; lie on a circle if &lt;math&gt;\triangle ACB&lt;/math&gt; is an isosceles triangle.<br /> <br /> If &lt;math&gt;\triangle ACB&lt;/math&gt; is isosceles, then its legs have length &lt;math&gt;6\sqrt{2}&lt;/math&gt;. The perimeter of &lt;math&gt;\triangle ACB&lt;/math&gt; is &lt;math&gt;\textbf{(C) }12+12\sqrt{2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems&diff=85234 1954 AHSME Problems 2017-04-15T22:13:30Z <p>Theultimate123: /* Problem 10 */</p> <hr /> <div>== Problem 1==<br /> <br /> The square of &lt;math&gt;5-\sqrt{y^2-25}&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2==<br /> <br /> The equation &lt;math&gt;\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0&lt;/math&gt; can be transformed by eliminating fractions to the equation &lt;math&gt;x^2-5x+4=0&lt;/math&gt;. <br /> The roots of the latter equation are &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Then the roots of the first equation are: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3==<br /> <br /> If &lt;math&gt;x&lt;/math&gt; varies as the cube of &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;y&lt;/math&gt; varies as the fifth root of &lt;math&gt;z&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; varies as the nth power of &lt;math&gt;z&lt;/math&gt;, where n is:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4==<br /> <br /> If the Highest Common Divisor of &lt;math&gt;6432&lt;/math&gt; and &lt;math&gt;132&lt;/math&gt; is diminished by &lt;math&gt;8&lt;/math&gt;, it will equal: <br /> <br /> &lt;math&gt;\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4 &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5==<br /> <br /> A regular hexagon is inscribed in a circle of radius &lt;math&gt;10&lt;/math&gt; inches. Its area is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.} &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6==<br /> <br /> The value of &lt;math&gt;\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16} &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7==<br /> <br /> A housewife saved &lt;math&gt;\textdollar{2.50}&lt;/math&gt; in buying a dress on sale. If she spent &lt;math&gt;\textdollar{25}&lt;/math&gt; for the dress, she saved about: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\% &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8==<br /> <br /> The base of a triangle is twice as long as a side of a square and their areas are the same. <br /> Then the ratio of the altitude of the triangle to the side of the square is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4 &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9==<br /> <br /> A point &lt;math&gt;P&lt;/math&gt; is outside a circle and is &lt;math&gt;13&lt;/math&gt; inches from the center. A secant from &lt;math&gt;P&lt;/math&gt; cuts the circle at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; <br /> so that the external segment of the secant &lt;math&gt;PQ&lt;/math&gt; is &lt;math&gt;9&lt;/math&gt; inches and &lt;math&gt;QR&lt;/math&gt; is &lt;math&gt;7&lt;/math&gt; inches. The radius of the circle is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 3&quot; \qquad \textbf{(B)}\ 4&quot; \qquad \textbf{(C)}\ 5&quot; \qquad \textbf{(D)}\ 6&quot;\qquad\textbf{(E)}\ 7&quot; &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10==<br /> <br /> The sum of the numerical coefficients in the expansion of the binomial &lt;math&gt;(a+b)^6&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11==<br /> <br /> A merchant placed on display some dresses, each with a marked price. He then posted a sign “&lt;math&gt;\frac{1}{3}&lt;/math&gt; off on these dresses.” <br /> The cost of the dresses was &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the price at which he actually sold them. Then the ratio of the cost to the marked price was: <br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12==<br /> <br /> The solution of the equations <br /> <br /> &lt;cmath&gt;\begin{align*}2x-3y &amp;=7 \\ 4x-6y &amp;=20\end{align*}&lt;/cmath&gt; <br /> <br /> is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ x=18, y=12 \qquad <br /> \textbf{(B)}\ x=0, y=0 \qquad <br /> \textbf{(C)}\ \text{There is no solution} \\ <br /> \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad<br /> \textbf{(E)}\ x=8, y=5 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13==<br /> <br /> A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off <br /> by the sides of the quadrilateral, their sum will be: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14==<br /> <br /> When simplified &lt;math&gt;\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}&lt;/math&gt; equals: <br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2} &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15==<br /> <br /> &lt;math&gt;\log 125&lt;/math&gt; equals: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \\<br /> \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5) &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16==<br /> <br /> If &lt;math&gt;f(x) = 5x^2 - 2x - 1&lt;/math&gt;, then &lt;math&gt;f(x + h) - f(x)&lt;/math&gt; equals: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17==<br /> <br /> The graph of the function &lt;math&gt;f(x) = 2x^3 - 7&lt;/math&gt; goes: <br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{up to the right and down to the left} \\ \textbf{(B)}\ \text{down to the right and up to the left}\\ \textbf{(C)}\ \text{up to the right and up to the left}\\ \textbf{(D)}\ \text{down to the right and down to the left}\\ \textbf{(E)}\ \text{none of these ways.} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18==<br /> <br /> Of the following sets, the one that includes all values of &lt;math&gt;x&lt;/math&gt; which will satisfy &lt;math&gt;2x - 3 &gt; 7 - x&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ x &gt; 4 \qquad \textbf{(B)}\ x &lt; \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x &gt;\frac{10}{3}\qquad\textbf{(E)}\ x &lt; 0 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19==<br /> <br /> If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20==<br /> <br /> The equation &lt;math&gt;x^3+6x^2+11x+6=0&lt;/math&gt; has: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21==<br /> <br /> The roots of the equation &lt;math&gt;2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5&lt;/math&gt; can be found by solving: <br /> <br /> &lt;math&gt; \textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0 &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22==<br /> <br /> The expression &lt;math&gt;\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}&lt;/math&gt; cannot be evaluated for &lt;math&gt;x=-1&lt;/math&gt; or &lt;math&gt;x=2&lt;/math&gt;, <br /> since division by zero is not allowed. For other values of &lt;math&gt;x&lt;/math&gt;: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1. &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23==<br /> <br /> If the margin made on an article costing &lt;math&gt;C&lt;/math&gt; dollars and selling for &lt;math&gt;S&lt;/math&gt; dollars is &lt;math&gt;M=\frac{1}{n}C&lt;/math&gt;, then the margin is given by: <br /> <br /> &lt;math&gt; \textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 23|Solution]]<br /> <br /> == Problem 24==<br /> <br /> The values of &lt;math&gt;k&lt;/math&gt; for which the equation &lt;math&gt;2x^2-kx+x+8=0&lt;/math&gt; will have real and equal roots are: <br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> <br /> The two roots of the equation &lt;math&gt;a(b-c)x^2+b(c-a)x+c(a-b)=0&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)} &lt;/math&gt;<br /> <br /> == Problem 26==<br /> <br /> The straight line &lt;math&gt;\overline{AB}&lt;/math&gt; is divided at &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AC=3CB&lt;/math&gt;. Circles are described on &lt;math&gt;\overline{AC}&lt;/math&gt; <br /> and &lt;math&gt;\overline{CB}&lt;/math&gt; as diameters and a common tangent meets &lt;math&gt;AB&lt;/math&gt; produced at &lt;math&gt;D&lt;/math&gt;. Then &lt;math&gt;BD&lt;/math&gt; equals: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27==<br /> <br /> A right circular cone has for its base a circle having the same radius as a given sphere. <br /> The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{2}{1}\qquad\textbf{(E)}\ \sqrt{\frac{5}{4}} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28==<br /> <br /> If &lt;math&gt;\frac{m}{n}=\frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{r}{t}=\frac{9}{14}&lt;/math&gt;, the value of &lt;math&gt;\frac{3mr-nt}{4nt-7mr}&lt;/math&gt; is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29==<br /> <br /> If the ratio of the legs of a right triangle is &lt;math&gt;1: 2&lt;/math&gt;, then the ratio of the corresponding segments of <br /> the hypotenuse made by a perpendicular upon it from the vertex is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5 &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30==<br /> <br /> &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; together can do a job in &lt;math&gt;2&lt;/math&gt; days; &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; can do it in four days; and &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; in &lt;math&gt;2\frac{2}{5}&lt;/math&gt; days. <br /> The number of days required for A to do the job alone is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 30|Solution]]<br /> <br /> == Problem 31==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC&lt;/math&gt;, &lt;math&gt;\angle A=40^\circ&lt;/math&gt;. Point &lt;math&gt;O&lt;/math&gt; is within the triangle with &lt;math&gt;\angle OBC \cong \angle OCA&lt;/math&gt;. <br /> The number of degrees in &lt;math&gt;\angle BOC&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}&lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 31|Solution]]<br /> <br /> == Problem 32==<br /> <br /> The factors of &lt;math&gt;x^4+64&lt;/math&gt; are: <br /> <br /> &lt;math&gt; \textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8) &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 32|Solution]]<br /> <br /> == Problem 33==<br /> <br /> A bank charges &lt;math&gt;\textdollar{6}&lt;/math&gt; for a loan of &lt;math&gt;\textdollar{120}&lt;/math&gt;. The borrower receives &lt;math&gt;\textdollar{114}&lt;/math&gt; and <br /> repays the loan in &lt;math&gt;12&lt;/math&gt; easy installments of &lt;math&gt;\textdollar{10}&lt;/math&gt; a month. The interest rate is approximately: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \% &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 33|Solution]]<br /> <br /> == Problem 34==<br /> <br /> The fraction &lt;math&gt;\frac{1}{3}&lt;/math&gt;: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 34|Solution]]<br /> <br /> == Problem 35==<br /> <br /> In the right triangle shown the sum of the distances &lt;math&gt;BM&lt;/math&gt; and &lt;math&gt;MA&lt;/math&gt; is equal to the sum of the distances &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt;. <br /> If &lt;math&gt;MB = x, CB = h&lt;/math&gt;, and &lt;math&gt;CA = d&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; equals: <br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> draw((0,0)--(8,0)--(0,5)--cycle);<br /> label(&quot;C&quot;,(0,0),SW);<br /> label(&quot;A&quot;,(8,0),SE);<br /> label(&quot;M&quot;,(0,5),N);<br /> dot((0,3.5));<br /> label(&quot;B&quot;,(0,3.5),W);<br /> label(&quot;x$&quot;,(0,4.25),W);<br /> label(&quot;$h$&quot;,(0,1),W);<br /> label(&quot;$d$&quot;,(4,0),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 35|Solution]]<br /> <br /> == Problem 36==<br /> <br /> A boat has a speed of &lt;math&gt;15&lt;/math&gt; mph in still water. In a stream that has a current of &lt;math&gt;5&lt;/math&gt; mph it travels a certain <br /> distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 36|Solution]]<br /> <br /> == Problem 37==<br /> <br /> Given &lt;math&gt;\triangle PQR&lt;/math&gt; with &lt;math&gt;\overline{RS}&lt;/math&gt; bisecting &lt;math&gt;\angle R&lt;/math&gt;, &lt;math&gt;PQ&lt;/math&gt; extended to &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;\angle n&lt;/math&gt; a right angle, then: <br /> <br /> &lt;asy&gt;<br /> path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false)<br /> {<br /> pair M,N;<br /> path mark;<br /> M=t*0.03*unit(A-B)+B;<br /> N=t*0.03*unit(C-B)+B;<br /> if(flip)<br /> mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B));<br /> else<br /> mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B));<br /> return mark;<br /> }<br /> unitsize(1.5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair P=(0,0), R=(3,2), Q=(4,0);<br /> pair S0=bisectorpoint(P,R,Q);<br /> pair Sp=extension(P,Q,S0,R);<br /> pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp);<br /> pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R);<br /> draw(P--R--Q);<br /> draw(R--Sp);<br /> draw(P--D--M);<br /> draw(anglemark2(Sp,P,R,17));<br /> label(&quot;$p$&quot;,P+(0.35,0.1));<br /> draw(anglemark2(R,Q,P,11));<br /> label(&quot;$q$&quot;,Q+(-0.17,0.1));<br /> draw(anglemark2(R,Np,D,8,true));<br /> label(&quot;$n$&quot;,Np+(+0.12,0.07));<br /> draw(anglemark2(R,M,D,13,true));<br /> label(&quot;$m$&quot;,M+(+0.25,0.03));<br /> draw(anglemark2(M,D,P,29));<br /> label(&quot;$d$&quot;,D+(-0.75,0.095));<br /> pen f=fontsize(10pt);<br /> label(&quot;$R$&quot;,R,N,f);<br /> label(&quot;$P$&quot;,P,S,f);<br /> label(&quot;$S$&quot;,Sp,S,f);<br /> label(&quot;$Q$&quot;,Q,S,f);<br /> label(&quot;$D$&quot;,D,S,f);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad <br /> \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad<br /> \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad<br /> \textbf{(E)}\ \text{none of these is correct} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 37|Solution]]<br /> <br /> == Problem 38==<br /> <br /> If &lt;math&gt;\log 2=.3010&lt;/math&gt; and &lt;math&gt;\log 3=.4771&lt;/math&gt;, the value of &lt;math&gt;x&lt;/math&gt; when &lt;math&gt;3^{x+3}=135&lt;/math&gt; is approximately: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 38|Solution]]<br /> <br /> == Problem 39==<br /> <br /> The locus of the midpoint of a line segment that is drawn from a given external point &lt;math&gt;P&lt;/math&gt; to a given circle with center &lt;math&gt;O&lt;/math&gt; and radius &lt;math&gt;r&lt;/math&gt;, is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 39|Solution]]<br /> <br /> == Problem 40==<br /> <br /> If &lt;math&gt;\left (a+\frac{1}{a} \right )^2=3&lt;/math&gt;, then &lt;math&gt;a^3+\frac{1}{a^3}&lt;/math&gt; equals: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 40|Solution]]<br /> <br /> == Problem 41==<br /> <br /> The sum of all the roots of &lt;math&gt;4x^3-8x^2-63x-9=0&lt;/math&gt; is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 41|Solution]]<br /> <br /> == Problem 42==<br /> <br /> Consider the graphs of <br /> &lt;cmath&gt;(1)\qquad y=x^2-\frac{1}{2}x+2&lt;/cmath&gt; <br /> and <br /> &lt;cmath&gt;(2)\qquad y=x^2+\frac{1}{2}x+2&lt;/cmath&gt; <br /> on the same set of axis. <br /> These parabolas are exactly the same shape. Then: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 42|Solution]]<br /> <br /> == Problem 43==<br /> <br /> The hypotenuse of a right triangle is &lt;math&gt;10&lt;/math&gt; inches and the radius of the inscribed circle is &lt;math&gt;1&lt;/math&gt; inch. The perimeter of the triangle in inches is: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 43|Solution]]<br /> <br /> == Problem 44==<br /> <br /> A man born in the first half of the nineteenth century was &lt;math&gt;x&lt;/math&gt; years old in the year &lt;math&gt;x^2&lt;/math&gt;. He was born in: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 44|Solution]]<br /> <br /> == Problem 45==<br /> <br /> In a rhombus, &lt;math&gt;ABCD&lt;/math&gt;, line segments are drawn within the rhombus, parallel to diagonal &lt;math&gt;BD&lt;/math&gt;, <br /> and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment <br /> as a function of its distance from vertex &lt;math&gt;A&lt;/math&gt;. The graph is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 45|Solution]]<br /> <br /> == Problem 46==<br /> <br /> In the diagram, if points &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are points of tangency, then &lt;math&gt;x&lt;/math&gt; equals: <br /> <br /> &lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> dotfactor=3;<br /> pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);<br /> pair O=(0,3/8);<br /> draw((-2/3,9/16)--(2/3,9/16));<br /> draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));<br /> draw(Circle(O,3/16));<br /> draw((-2/3,0)--(2/3,0));<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$\frac{3}{8}$&quot;,O);<br /> draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));<br /> draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));<br /> label(&quot;$\frac{1}{2}$&quot;,(.5,.25));<br /> draw((.5,.33)--(.5,.5),EndArrow(3));<br /> draw((.5,.17)--(.5,0),EndArrow(3));<br /> label(&quot;$x$&quot;,midpoint((.5,.5)--(.5,9/16)));<br /> draw((.5,5/8)--(.5,9/16),EndArrow(3));<br /> label(&quot;$60^{\circ}\$&quot;,(0.01,0.12));<br /> dot(A);<br /> dot(B);<br /> dot(C);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{3}{16}&quot;\qquad\textbf{(B)}\ \frac{1}{8}&quot;\qquad\textbf{(C)}\ \frac{1}{32}&quot;\qquad\textbf{(D)}\ \frac{3}{32}&quot;\qquad\textbf{(E)}\ \frac{1}{16}&quot; &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 46|Solution]]<br /> <br /> == Problem 47==<br /> <br /> At the midpoint of line segment &lt;math&gt;AB&lt;/math&gt; which is &lt;math&gt;p&lt;/math&gt; units long, a perpendicular &lt;math&gt;MR&lt;/math&gt; is erected with length &lt;math&gt;q&lt;/math&gt; units. <br /> An arc is described from &lt;math&gt;R&lt;/math&gt; with a radius equal to &lt;math&gt;\frac{1}{2}AB&lt;/math&gt;, meeting &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;T&lt;/math&gt;. Then &lt;math&gt;AT&lt;/math&gt; and &lt;math&gt;TB&lt;/math&gt; are the roots of: <br /> <br /> &lt;math&gt; \textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x^2+px-q^2=0\\ \textbf{(D)}\ x^2-px-q^2=0\\ \textbf{(E)}\ x^2-px+q=0 &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 47|Solution]]<br /> <br /> == Problem 48==<br /> <br /> A train, an hour after starting, meets with an accident which detains it a half hour, after which it <br /> proceeds at &lt;math&gt;\frac{3}{4}&lt;/math&gt; of its former rate and arrives &lt;math&gt;3\tfrac{1}{2}&lt;/math&gt; hours late. <br /> Had the accident happened &lt;math&gt;90&lt;/math&gt; miles farther along the line, it would have arrived only &lt;math&gt;3&lt;/math&gt; hours late. The length of the trip in miles was: <br /> <br /> &lt;math&gt;\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550 &lt;/math&gt; <br /> <br /> [[1954 AHSME Problems/Problem 48|Solution]]<br /> <br /> == Problem 49==<br /> <br /> The difference of the squares of two odd numbers is always divisible by &lt;math&gt;8&lt;/math&gt;. If &lt;math&gt;a&gt;b&lt;/math&gt;, and &lt;math&gt;2a+1&lt;/math&gt; and &lt;math&gt;2b+1&lt;/math&gt; are the odd numbers, <br /> to prove the given statement we put the difference of the squares in the form: <br /> <br /> &lt;math&gt; \textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b) &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 49|Solution]]<br /> <br /> == Problem 50==<br /> <br /> The times between &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; o'clock, correct to the nearest minute, when the hands of a clock will form an angle of &lt;math&gt;84^{\circ}&lt;/math&gt; are: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{7: 23 and 7: 53}\qquad<br /> \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad<br /> \textbf{(C)}\ \text{7: 22 and 7: 53}\\ <br /> \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad<br /> \textbf{(E)}\ \text{7: 21 and 7: 49} &lt;/math&gt;<br /> <br /> [[1954 AHSME Problems/Problem 50|Solution]]<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME 50p box|year=1954|before=[[1953 AHSME]]|after=[[1955 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_19&diff=84654 2008 AMC 12B Problems/Problem 19 2017-03-11T23:42:45Z <p>Theultimate123: Changed all \text{Re} and \text{Im} to \Re and \Im, respectively. Also added &quot;by the Trivial Inequality.&quot;</p> <hr /> <div>==Problem 19==<br /> A function &lt;math&gt;f&lt;/math&gt; is defined by &lt;math&gt;f(z) = (4 + i) z^2 + \alpha z + \gamma&lt;/math&gt; for all complex numbers &lt;math&gt;z&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; are complex numbers and &lt;math&gt;i^2 = - 1&lt;/math&gt;. Suppose that &lt;math&gt;f(1)&lt;/math&gt; and &lt;math&gt;f(i)&lt;/math&gt; are both real. What is the smallest possible value of &lt;math&gt;| \alpha | + |\gamma |&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad&lt;/math&gt;<br /> <br /> ==Solution==<br /> We need only concern ourselves with the imaginary portions of &lt;math&gt;f(1)&lt;/math&gt; and &lt;math&gt;f(i)&lt;/math&gt; (both of which must be 0). These are:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \Im(f(1)) &amp; = i+i\Im(\alpha)+i\Im(\gamma) \\<br /> \Im(f(i)) &amp; = -i+i\Re(\alpha)+i\Re(\gamma)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;p=\Im(\gamma)&lt;/math&gt; and &lt;math&gt;q=\Re{(\gamma)},&lt;/math&gt; then we know &lt;math&gt;\Im(\alpha)=-p-1&lt;/math&gt; and &lt;math&gt;\Re(\alpha)=1-p.&lt;/math&gt; Therefore &lt;cmath&gt;|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},&lt;/cmath&gt; which reaches its minimum &lt;math&gt;\sqrt 2&lt;/math&gt; when &lt;math&gt;p=q=0&lt;/math&gt; by the Trivial Inequality. Thus, the answer is &lt;math&gt;\boxed B.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&diff=84434 2011 AIME II Problems/Problem 13 2017-03-06T03:27:56Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of [[square]] &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; be the [[circumcenter]]s of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_{1}PO_{2} = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{a} + \sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Denote the [[midpoint]] of &lt;math&gt;\overline{DC}&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;F&lt;/math&gt;. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and these bisectors go through &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;.<br /> <br /> It is given that &lt;math&gt;\angle O_{1}PO_{2}=120^{\circ}&lt;/math&gt;. Because &lt;math&gt;O_{1}P&lt;/math&gt; and &lt;math&gt;O_{1}B&lt;/math&gt; are [[radius|radii]] of the same circle, the have the same length. This is also true of &lt;math&gt;O_{2}P&lt;/math&gt; and &lt;math&gt;O_{2}D&lt;/math&gt;. Because &lt;math&gt;m\angle CAB=m\angle ACD=45^{\circ}&lt;/math&gt;, &lt;math&gt;m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}&lt;/math&gt;. Thus, &lt;math&gt;O_{1}PB&lt;/math&gt; and &lt;math&gt;O_{2}PD&lt;/math&gt; are isosceles right triangles. Using the given information above and symmetry, &lt;math&gt;m\angle DPB = 120^{\circ}&lt;/math&gt;. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.<br /> <br /> Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; have measures of 30 degrees. Thus, both triangles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, &lt;math&gt;DO_{2}=BO_{1}=4\sqrt{3}&lt;/math&gt;. Because of 45-45-90 right triangles, &lt;math&gt;PB=PD=4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, using [[Law of Cosines]] on &lt;math&gt;\triangle ABP&lt;/math&gt; and letting &lt;math&gt;x = AP&lt;/math&gt;, <br /> <br /> &lt;math&gt;96=144+x^{2}-24x\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;96=144+x^{2}-12x\sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;0=x^{2}-12x\sqrt{2}+48&lt;/math&gt;<br /> <br /> Using quadratic formula,<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{288-192}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{96}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{2 \sqrt{72} \pm 2 \sqrt{24}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \sqrt{72} \pm \sqrt{24}&lt;/math&gt;<br /> <br /> <br /> Because it is given that &lt;math&gt;AP &gt; CP&lt;/math&gt;, &lt;math&gt;AP&gt;6\sqrt{2}&lt;/math&gt;, so the minus version of the above equation is too small.<br /> Thus, &lt;math&gt;AP=\sqrt{72}+ \sqrt{24}&lt;/math&gt; and a + b = 24 + 72 = &lt;math&gt;\framebox[1.5\width]{096.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This takes a slightly different route than Solution 1.<br /> <br /> Solution 1 proves that &lt;math&gt;\angle{DPB}=120^{\circ}&lt;/math&gt; and that &lt;math&gt;\overline{BP} = \overline{DP}&lt;/math&gt;.<br /> Construct diagonal &lt;math&gt;\overline{BD}&lt;/math&gt; and using the two statements above it quickly becomes clear that &lt;math&gt;\angle{BDP} = \angle{DBP} = 30^{\circ}&lt;/math&gt; by isosceles triangle base angles.<br /> Let the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;, and since the diagonals are perpendicular, both triangle &lt;math&gt;DMP&lt;/math&gt; and triangle &lt;math&gt;BMP&lt;/math&gt; are 30-60-90 right triangles.<br /> Since &lt;math&gt;\overline{AB} = 12&lt;/math&gt;, &lt;math&gt;\overline{AC} = \overline{BD} = 12\sqrt{2}&lt;/math&gt; and &lt;math&gt;\overline{BM} = \overline{DM} = 6\sqrt{2}&lt;/math&gt;.<br /> 30-60-90 triangles' sides are in the ratio &lt;math&gt;1 : \sqrt{3} : 2&lt;/math&gt;, so &lt;math&gt;\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}&lt;/math&gt;.<br /> &lt;math&gt;\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}&lt;/math&gt;.<br /> Hence, &lt;math&gt;72 + 24 = \framebox[1.5\width]{096}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Use vectors. In an &lt;math&gt;xy&lt;/math&gt; plane, let &lt;math&gt;(-s,0)&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;(0,s)&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;(s,0)&lt;/math&gt; be &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;(0,-s)&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;(p,0)&lt;/math&gt; be P, where &lt;math&gt;s=|AB|/\sqrt{2}=6\sqrt{2}&lt;/math&gt;. It remains to find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> The line &lt;math&gt;y=-x&lt;/math&gt; is the [[perpendicular bisector]] of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, so &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; lies on the line. Now compute the [[perpendicular bisector]] of &lt;math&gt;AP&lt;/math&gt;. The center has coordinate &lt;math&gt;(\frac{p-s}{2},0)&lt;/math&gt;, and the segment is part of the &lt;math&gt;x&lt;/math&gt;-axis, so the perpendicular bisector has equation &lt;math&gt;x=\frac{p-s}{2}&lt;/math&gt;. Since &lt;math&gt;O_1&lt;/math&gt; is the [[circumcenter]] of triangle &lt;math&gt;ABP&lt;/math&gt;, it lies on the perpendicular bisector of both &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> O_1=(\frac{p-s}{2},-\frac{p-s}{2})<br /> &lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;<br /> O_2=(\frac{p+s}{2},-\frac{p+s}{2})<br /> &lt;/cmath&gt;<br /> The relation &lt;math&gt;\angle O_1PO_2=120^\circ&lt;/math&gt; can now be written using [[Vectors|dot product]] as<br /> &lt;cmath&gt;<br /> \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|<br /> &lt;/cmath&gt;<br /> Computation of both sides yields<br /> &lt;cmath&gt;<br /> \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}<br /> &lt;/cmath&gt;<br /> Solve for &lt;math&gt;p&lt;/math&gt; gives &lt;math&gt;p=s/\sqrt{3}=2\sqrt{6}&lt;/math&gt;, so &lt;math&gt;AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}&lt;/math&gt;. The answer is 72+24&lt;math&gt;\Rightarrow\boxed{096}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Translate &lt;math&gt;\triangle{ABP}&lt;/math&gt; so that the image of &lt;math&gt;AB&lt;/math&gt; coincides &lt;math&gt;DC&lt;/math&gt;. Let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P’&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle{DPC}=\angle{CPB}&lt;/math&gt; by symmetry, and &lt;math&gt;\angle{APB}=\angle{DP’C}&lt;/math&gt; because translation preserves angles. Thus &lt;math&gt;\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ&lt;/math&gt;. Therefore, quadrilateral &lt;math&gt;CPDP’&lt;/math&gt; is cyclic. Thus the image of &lt;math&gt;O_1&lt;/math&gt; coincides with &lt;math&gt;O_2&lt;/math&gt;. <br /> <br /> &lt;math&gt;O_1P&lt;/math&gt; is parallel to &lt;math&gt;O_2P’&lt;/math&gt; so &lt;math&gt;\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ&lt;/math&gt;, so &lt;math&gt;\angle{PDP’}=60^\circ&lt;/math&gt; and &lt;math&gt;\angle{PDC}=15^\circ&lt;/math&gt;, thus &lt;math&gt;\angle{ADP}=75^{\circ}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;\triangle{AMD}&lt;/math&gt; is a 45-45-90 triangle and &lt;math&gt;\triangle{DMP}&lt;/math&gt; is a 30-60-90 triangle. Thus<br /> <br /> &lt;math&gt;AM=6\sqrt{2}&lt;/math&gt; and &lt;math&gt;MP=\frac{6\sqrt{2}}{\sqrt{3}}&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;AP=AM+MP=\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{96}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Reflect &lt;math&gt;O_1&lt;/math&gt; across &lt;math&gt;AP&lt;/math&gt; to &lt;math&gt;O_1'&lt;/math&gt;. By symmetry &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle{DO_1’P}&lt;/math&gt; = &lt;math&gt;2*\angle{DAP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_1’PD}=45^\circ&lt;/math&gt;<br /> <br /> similarly &lt;math&gt;\angle{DO_2P}&lt;/math&gt; = &lt;math&gt;2*\angle{DCP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_2PD}=45^\circ&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\angle{O_1’PO_2}=90^\circ&lt;/math&gt;, so that &lt;math&gt;\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ&lt;/math&gt;<br /> <br /> By symmetry, &lt;math&gt;\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ&lt;/math&gt;<br /> <br /> Therefore, since &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;\angle{ADP}&lt;/math&gt; = &lt;math&gt;0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ&lt;/math&gt;<br /> <br /> Using sine rule in &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{96}.&lt;/math&gt;<br /> <br /> By Kris17<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&diff=84433 2011 AIME II Problems/Problem 13 2017-03-06T03:27:45Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of [[square]] &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; be the [[circumcenter]]s of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_{1}PO_{2} = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{a} + \sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> [geogebra]7b0d7e3170597705121a87857a112a90dff8cac9[/geogebra]<br /> <br /> Denote the [[midpoint]] of &lt;math&gt;\overline{DC}&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;F&lt;/math&gt;. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and these bisectors go through &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;.<br /> <br /> It is given that &lt;math&gt;\angle O_{1}PO_{2}=120^{\circ}&lt;/math&gt;. Because &lt;math&gt;O_{1}P&lt;/math&gt; and &lt;math&gt;O_{1}B&lt;/math&gt; are [[radius|radii]] of the same circle, the have the same length. This is also true of &lt;math&gt;O_{2}P&lt;/math&gt; and &lt;math&gt;O_{2}D&lt;/math&gt;. Because &lt;math&gt;m\angle CAB=m\angle ACD=45^{\circ}&lt;/math&gt;, &lt;math&gt;m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}&lt;/math&gt;. Thus, &lt;math&gt;O_{1}PB&lt;/math&gt; and &lt;math&gt;O_{2}PD&lt;/math&gt; are isosceles right triangles. Using the given information above and symmetry, &lt;math&gt;m\angle DPB = 120^{\circ}&lt;/math&gt;. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.<br /> <br /> Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; have measures of 30 degrees. Thus, both triangles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, &lt;math&gt;DO_{2}=BO_{1}=4\sqrt{3}&lt;/math&gt;. Because of 45-45-90 right triangles, &lt;math&gt;PB=PD=4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, using [[Law of Cosines]] on &lt;math&gt;\triangle ABP&lt;/math&gt; and letting &lt;math&gt;x = AP&lt;/math&gt;, <br /> <br /> &lt;math&gt;96=144+x^{2}-24x\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;96=144+x^{2}-12x\sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;0=x^{2}-12x\sqrt{2}+48&lt;/math&gt;<br /> <br /> Using quadratic formula,<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{288-192}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{12 \sqrt{2} \pm \sqrt{96}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{2 \sqrt{72} \pm 2 \sqrt{24}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \sqrt{72} \pm \sqrt{24}&lt;/math&gt;<br /> <br /> <br /> Because it is given that &lt;math&gt;AP &gt; CP&lt;/math&gt;, &lt;math&gt;AP&gt;6\sqrt{2}&lt;/math&gt;, so the minus version of the above equation is too small.<br /> Thus, &lt;math&gt;AP=\sqrt{72}+ \sqrt{24}&lt;/math&gt; and a + b = 24 + 72 = &lt;math&gt;\framebox[1.5\width]{096.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This takes a slightly different route than Solution 1.<br /> <br /> Solution 1 proves that &lt;math&gt;\angle{DPB}=120^{\circ}&lt;/math&gt; and that &lt;math&gt;\overline{BP} = \overline{DP}&lt;/math&gt;.<br /> Construct diagonal &lt;math&gt;\overline{BD}&lt;/math&gt; and using the two statements above it quickly becomes clear that &lt;math&gt;\angle{BDP} = \angle{DBP} = 30^{\circ}&lt;/math&gt; by isosceles triangle base angles.<br /> Let the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;, and since the diagonals are perpendicular, both triangle &lt;math&gt;DMP&lt;/math&gt; and triangle &lt;math&gt;BMP&lt;/math&gt; are 30-60-90 right triangles.<br /> Since &lt;math&gt;\overline{AB} = 12&lt;/math&gt;, &lt;math&gt;\overline{AC} = \overline{BD} = 12\sqrt{2}&lt;/math&gt; and &lt;math&gt;\overline{BM} = \overline{DM} = 6\sqrt{2}&lt;/math&gt;.<br /> 30-60-90 triangles' sides are in the ratio &lt;math&gt;1 : \sqrt{3} : 2&lt;/math&gt;, so &lt;math&gt;\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}&lt;/math&gt;.<br /> &lt;math&gt;\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}&lt;/math&gt;.<br /> Hence, &lt;math&gt;72 + 24 = \framebox[1.5\width]{096}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Use vectors. In an &lt;math&gt;xy&lt;/math&gt; plane, let &lt;math&gt;(-s,0)&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;(0,s)&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;(s,0)&lt;/math&gt; be &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;(0,-s)&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;(p,0)&lt;/math&gt; be P, where &lt;math&gt;s=|AB|/\sqrt{2}=6\sqrt{2}&lt;/math&gt;. It remains to find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> The line &lt;math&gt;y=-x&lt;/math&gt; is the [[perpendicular bisector]] of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, so &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; lies on the line. Now compute the [[perpendicular bisector]] of &lt;math&gt;AP&lt;/math&gt;. The center has coordinate &lt;math&gt;(\frac{p-s}{2},0)&lt;/math&gt;, and the segment is part of the &lt;math&gt;x&lt;/math&gt;-axis, so the perpendicular bisector has equation &lt;math&gt;x=\frac{p-s}{2}&lt;/math&gt;. Since &lt;math&gt;O_1&lt;/math&gt; is the [[circumcenter]] of triangle &lt;math&gt;ABP&lt;/math&gt;, it lies on the perpendicular bisector of both &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> O_1=(\frac{p-s}{2},-\frac{p-s}{2})<br /> &lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;<br /> O_2=(\frac{p+s}{2},-\frac{p+s}{2})<br /> &lt;/cmath&gt;<br /> The relation &lt;math&gt;\angle O_1PO_2=120^\circ&lt;/math&gt; can now be written using [[Vectors|dot product]] as<br /> &lt;cmath&gt;<br /> \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|<br /> &lt;/cmath&gt;<br /> Computation of both sides yields<br /> &lt;cmath&gt;<br /> \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}<br /> &lt;/cmath&gt;<br /> Solve for &lt;math&gt;p&lt;/math&gt; gives &lt;math&gt;p=s/\sqrt{3}=2\sqrt{6}&lt;/math&gt;, so &lt;math&gt;AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}&lt;/math&gt;. The answer is 72+24&lt;math&gt;\Rightarrow\boxed{096}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Translate &lt;math&gt;\triangle{ABP}&lt;/math&gt; so that the image of &lt;math&gt;AB&lt;/math&gt; coincides &lt;math&gt;DC&lt;/math&gt;. Let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P’&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle{DPC}=\angle{CPB}&lt;/math&gt; by symmetry, and &lt;math&gt;\angle{APB}=\angle{DP’C}&lt;/math&gt; because translation preserves angles. Thus &lt;math&gt;\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ&lt;/math&gt;. Therefore, quadrilateral &lt;math&gt;CPDP’&lt;/math&gt; is cyclic. Thus the image of &lt;math&gt;O_1&lt;/math&gt; coincides with &lt;math&gt;O_2&lt;/math&gt;. <br /> <br /> &lt;math&gt;O_1P&lt;/math&gt; is parallel to &lt;math&gt;O_2P’&lt;/math&gt; so &lt;math&gt;\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ&lt;/math&gt;, so &lt;math&gt;\angle{PDP’}=60^\circ&lt;/math&gt; and &lt;math&gt;\angle{PDC}=15^\circ&lt;/math&gt;, thus &lt;math&gt;\angle{ADP}=75^{\circ}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;\triangle{AMD}&lt;/math&gt; is a 45-45-90 triangle and &lt;math&gt;\triangle{DMP}&lt;/math&gt; is a 30-60-90 triangle. Thus<br /> <br /> &lt;math&gt;AM=6\sqrt{2}&lt;/math&gt; and &lt;math&gt;MP=\frac{6\sqrt{2}}{\sqrt{3}}&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;AP=AM+MP=\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{96}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Reflect &lt;math&gt;O_1&lt;/math&gt; across &lt;math&gt;AP&lt;/math&gt; to &lt;math&gt;O_1'&lt;/math&gt;. By symmetry &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle{DO_1’P}&lt;/math&gt; = &lt;math&gt;2*\angle{DAP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_1’PD}=45^\circ&lt;/math&gt;<br /> <br /> similarly &lt;math&gt;\angle{DO_2P}&lt;/math&gt; = &lt;math&gt;2*\angle{DCP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_2PD}=45^\circ&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\angle{O_1’PO_2}=90^\circ&lt;/math&gt;, so that &lt;math&gt;\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ&lt;/math&gt;<br /> <br /> By symmetry, &lt;math&gt;\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ&lt;/math&gt;<br /> <br /> Therefore, since &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;\angle{ADP}&lt;/math&gt; = &lt;math&gt;0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ&lt;/math&gt;<br /> <br /> Using sine rule in &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{96}.&lt;/math&gt;<br /> <br /> By Kris17<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_10&diff=84376 2017 AMC 12B Problems/Problem 10 2017-03-03T02:38:39Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> At Typico High School, &lt;math&gt;60\%&lt;/math&gt; of the students like dancing, and the rest dislike it. Of those who like dancing, &lt;math&gt;80\%&lt;/math&gt; say that they like it, and the rest say that they dislike it. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt; say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ \frac{100}{3}\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> WLOG, let there be &lt;math&gt;100&lt;/math&gt; students. &lt;math&gt;60&lt;/math&gt; of them like dancing, and &lt;math&gt;40&lt;/math&gt; do not. Of those who like dancing, &lt;math&gt;20\%&lt;/math&gt;, or &lt;math&gt;12&lt;/math&gt; of them say they dislike dancing. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt;, or &lt;math&gt;36&lt;/math&gt; of them say they dislike it. Thus, &lt;math&gt;\frac{12}{12+36} = \frac{12}{48} = \frac{1}{4} = 25\% \boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_15&diff=84375 2011 AIME I Problems/Problem 15 2017-03-03T02:33:50Z <p>Theultimate123: </p> <hr /> <div>== Problem ==<br /> For some integer &lt;math&gt;m&lt;/math&gt;, the polynomial &lt;math&gt;x^3 - 2011x + m&lt;/math&gt; has the three integer roots &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. Find &lt;math&gt;|a| + |b| + |c|&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> With Vieta's formulas, we know that &lt;math&gt;a+b+c = 0&lt;/math&gt;, and &lt;math&gt;ab+bc+ac = -2011&lt;/math&gt;. <br /> <br /> <br /> &lt;br /&gt;<br /> &lt;math&gt;a,b,c\neq 0&lt;/math&gt; since any one being zero will make the other &lt;math&gt;2 \pm \sqrt{2011}&lt;/math&gt;. <br /> <br /> &lt;math&gt;a = -(b+c)&lt;/math&gt;. WLOG, let &lt;math&gt;|a| \ge |b| \ge |c|&lt;/math&gt;. <br /> <br /> Then if &lt;math&gt;a &gt; 0&lt;/math&gt;, then &lt;math&gt;b,c &lt; 0&lt;/math&gt; and if &lt;math&gt;a &lt; 0&lt;/math&gt;, &lt;math&gt;b,c &gt; 0&lt;/math&gt;.<br /> <br /> &lt;br /&gt;<br /> &lt;math&gt;ab+bc+ac = -2011 = a(b+c)+bc = -a^2+bc&lt;/math&gt;<br /> <br /> &lt;br /&gt;<br /> &lt;math&gt;a^2 = 2011 + bc&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; have the same sign. So &lt;math&gt;|a| \ge 45&lt;/math&gt;. (&lt;math&gt;44^2&lt;2011&lt;/math&gt; and &lt;math&gt;45^2 = 2025&lt;/math&gt;)<br /> <br /> Also, &lt;math&gt;bc&lt;/math&gt; maximize when &lt;math&gt;b = c&lt;/math&gt; if we fixed &lt;math&gt;a&lt;/math&gt;. Hence, &lt;math&gt;2011 = a^2 - bc &gt; \frac{3}{4}a^2&lt;/math&gt;.<br /> <br /> So &lt;math&gt;a ^2 &lt; \frac{(4)2011}{3} = 2681+\frac{1}{3}&lt;/math&gt;. <br /> <br /> &lt;math&gt;52^2 = 2704&lt;/math&gt; so &lt;math&gt;|a| \le 51&lt;/math&gt;.<br /> <br /> <br /> &lt;br /&gt;<br /> Now we have limited &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;45\le |a| \le 51&lt;/math&gt;.<br /> <br /> Let's us analyze &lt;math&gt;a^2 = 2011 + bc&lt;/math&gt;.<br /> <br /> &lt;br /&gt;<br /> <br /> Here is a table:<br /> <br /> &lt;table border = 1 cellspacing = 0 cellpadding = 5 style = &quot;text-align:center;&quot;&gt;<br /> <br /> &lt;tr&gt;&lt;th&gt;&lt;math&gt;|a|&lt;/math&gt;&lt;/th&gt;&lt;th&gt;&lt;math&gt;a^2 = 2011 + bc&lt;/math&gt;&lt;/th&gt;&lt;/tr&gt;<br /> <br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;45&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;14&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;46&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;14 + 91 =105&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;47&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;105 + 93 = 198&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;48&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;198 + 95 = 293&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;tr&gt;&lt;td&gt;&lt;math&gt;49&lt;/math&gt;&lt;/td&gt;&lt;td&gt;&lt;math&gt;293 + 97 = 390&lt;/math&gt;&lt;/td&gt;&lt;/tr&gt;<br /> &lt;/table&gt;<br /> &lt;br /&gt;<br /> <br /> We can tell we don't need to bother with &lt;math&gt;45&lt;/math&gt;, <br /> <br /> &lt;math&gt;105 = (3)(5)(7)&lt;/math&gt;, So &lt;math&gt;46&lt;/math&gt; won't work. &lt;math&gt;198/47 &gt; 4&lt;/math&gt;, <br /> <br /> &lt;math&gt;198&lt;/math&gt; is not divisible by &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;198/6 = 33&lt;/math&gt;, which is too small to get &lt;math&gt;47&lt;/math&gt;.<br /> <br /> &lt;math&gt;293/48 &gt; 6&lt;/math&gt;, &lt;math&gt;293&lt;/math&gt; is not divisible by &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, we can clearly tell that &lt;math&gt;10&lt;/math&gt; is too much.<br /> <br /> <br /> Hence, &lt;math&gt;|a| = 49&lt;/math&gt;, &lt;math&gt;a^2 -2011 = 390&lt;/math&gt;. &lt;math&gt;b = 39&lt;/math&gt;, &lt;math&gt;c = 10&lt;/math&gt;.<br /> <br /> Answer: &lt;math&gt;\boxed{098}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> <br /> Starting off like the previous solution, we know that &lt;math&gt;a + b + c = 0&lt;/math&gt;, and &lt;math&gt;ab + bc + ac = -2011&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;c = -b-a&lt;/math&gt;.<br /> <br /> Substituting, &lt;math&gt;ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011&lt;/math&gt;.<br /> <br /> Factoring the perfect square, we get: &lt;math&gt;ab-(b+a)^2=-2011&lt;/math&gt; or &lt;math&gt;(b+a)^2-ab=2011&lt;/math&gt;.<br /> <br /> Therefore, a sum (&lt;math&gt;a+b&lt;/math&gt;) squared minus a product (&lt;math&gt;ab&lt;/math&gt;) gives &lt;math&gt;2011&lt;/math&gt;..<br /> <br /> &lt;br/&gt;<br /> <br /> We can guess and check different &lt;math&gt;a+b&lt;/math&gt;’s starting with &lt;math&gt;45&lt;/math&gt; since &lt;math&gt;44^2 &lt; 2011&lt;/math&gt;.<br /> <br /> &lt;math&gt;45^2 = 2025&lt;/math&gt; therefore &lt;math&gt;ab = 2025-2011 = 14&lt;/math&gt;. <br /> <br /> Since no factors of &lt;math&gt;14&lt;/math&gt; can sum to &lt;math&gt;45&lt;/math&gt; (&lt;math&gt;1+14&lt;/math&gt; being the largest sum), a + b cannot equal &lt;math&gt;45&lt;/math&gt;.<br /> <br /> &lt;math&gt;46^2 = 2116&lt;/math&gt; making &lt;math&gt;ab = 105 = 3 * 5 * 7&lt;/math&gt;.<br /> <br /> &lt;math&gt;5 * 7 + 3 &lt; 46&lt;/math&gt; and &lt;math&gt;3 * 5 * 7 &gt; 46&lt;/math&gt; so &lt;math&gt;46&lt;/math&gt; cannot work either.<br /> <br /> &lt;br/&gt;<br /> <br /> We can continue to do this until we reach &lt;math&gt;49&lt;/math&gt;.<br /> <br /> &lt;math&gt;49^2 = 2401&lt;/math&gt; making &lt;math&gt;ab = 390 = 2 * 3 * 5* 13&lt;/math&gt;.<br /> <br /> &lt;math&gt;3 * 13 + 2* 5 = 49&lt;/math&gt;, so one root is &lt;math&gt;10&lt;/math&gt; and another is &lt;math&gt;39&lt;/math&gt;. The roots sum to zero, so the last root must be &lt;math&gt;-49&lt;/math&gt;.<br /> <br /> &lt;br/&gt;<br /> <br /> &lt;math&gt;|-49|+10+39 = \boxed{098}&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=84374 2011 AIME I Problems/Problem 1 2017-03-03T02:33:25Z <p>Theultimate123: </p> <hr /> <div>== Problem 1 ==<br /> Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is &lt;math&gt;k\%&lt;/math&gt; acid. From jar C, &lt;math&gt;\frac{m}{n}&lt;/math&gt; liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;k + m + n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> There are &lt;math&gt;\frac{45}{100}(4)=\frac{9}{5}&lt;/math&gt; L of acid in Jar A. There are &lt;math&gt;\frac{48}{100}(5)=\frac{12}{5}&lt;/math&gt; L of acid in Jar B. And there are &lt;math&gt;\frac{k}{100}&lt;/math&gt; L of acid in Jar C. After transfering the solutions from jar C, there will be<br /> &lt;br&gt; &lt;math&gt;4+\frac{m}{n}&lt;/math&gt; L of solution in Jar A and &lt;math&gt;\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}&lt;/math&gt; L of acid in Jar A.&lt;br&gt;<br /> &lt;br&gt; &lt;math&gt;6-\frac{m}{n}&lt;/math&gt; L of solution in Jar B and &lt;math&gt;\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}&lt;/math&gt; of acid in Jar B.<br /> &lt;br&gt;Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.&lt;br&gt;<br /> &lt;cmath&gt;\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}&lt;/cmath&gt;<br /> Add the equations to get<br /> &lt;cmath&gt;\frac{42}{5}+\frac{k}{50}=10&lt;/cmath&gt; Solving gives &lt;math&gt;k=80&lt;/math&gt;.<br /> &lt;br&gt;If we substitute back in the original equation we get &lt;math&gt;\frac{m}{n}=\frac{2}{3}&lt;/math&gt; so &lt;math&gt;3m=2n&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;. Thus &lt;math&gt;k+m+n=80+2+3=\boxed{085}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> One might cleverly change the content of both Jars. <br /> <br /> Since the end result of both Jars are &lt;math&gt;50\%&lt;/math&gt; acid, we can turn Jar A into a 1 gallon liquid with &lt;math&gt;50\%-4(5\%) = 30\%&lt;/math&gt; acid <br /> <br /> and Jar B into 1 gallon liquid with &lt;math&gt;50\%-5(2\%) =40\%&lt;/math&gt; acid.<br /> <br /> Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so &lt;math&gt;\dfrac{2}{3}&lt;/math&gt; of Jar C will be pour into Jar A.<br /> <br /> Thus, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;.<br /> <br /> &lt;math&gt;\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; yields &lt;math&gt;k=80&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;80+2+3 = \boxed{085}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> One may first combine all three jars in to a single container. That container will have &lt;math&gt;10&lt;/math&gt; liters of liquid, and it should be &lt;math&gt;50\%&lt;/math&gt; acidic. Thus there must be &lt;math&gt;5&lt;/math&gt; liters of acid. <br /> <br /> Jug A contained &lt;math&gt;45\% \cdot 4L&lt;/math&gt;, or &lt;math&gt;1.8L&lt;/math&gt; of acid, and jug B &lt;math&gt;48\% \cdot 5L&lt;/math&gt; or &lt;math&gt;2.4L&lt;/math&gt;. Solving for the amount of acid in jug C, &lt;math&gt;k = (5 - 2.4 - 1.8) = .8&lt;/math&gt;, or &lt;math&gt;80\%&lt;/math&gt;.<br /> <br /> Once one knows that the jug C is &lt;math&gt;80\%&lt;/math&gt; acid, use solution 1 to figure out m and n.<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=84373 2011 AIME I Problems/Problem 1 2017-03-03T02:32:40Z <p>Theultimate123: </p> <hr /> <div>== Problem 1 ==<br /> Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is &lt;math&gt;k\%&lt;/math&gt; acid. From jar C, &lt;math&gt;\frac{m}{n}&lt;/math&gt; liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;k + m + n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> There are &lt;math&gt;\frac{45}{100}(4)=\frac{9}{5}&lt;/math&gt; L of acid in Jar A. There are &lt;math&gt;\frac{48}{100}(5)=\frac{12}{5}&lt;/math&gt; L of acid in Jar B. And there are &lt;math&gt;\frac{k}{100}&lt;/math&gt; L of acid in Jar C. After transfering the solutions from jar C, there will be<br /> &lt;br&gt; &lt;math&gt;4+\frac{m}{n}&lt;/math&gt; L of solution in Jar A and &lt;math&gt;\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}&lt;/math&gt; L of acid in Jar A.&lt;br&gt;<br /> &lt;br&gt; &lt;math&gt;6-\frac{m}{n}&lt;/math&gt; L of solution in Jar B and &lt;math&gt;\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}&lt;/math&gt; of acid in Jar B.<br /> &lt;br&gt;Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.&lt;br&gt;<br /> &lt;cmath&gt;\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}&lt;/cmath&gt;<br /> Add the equations to get<br /> &lt;cmath&gt;\frac{42}{5}+\frac{k}{50}=10&lt;/cmath&gt; Solving gives &lt;math&gt;k=80&lt;/math&gt;.<br /> &lt;br&gt;If we substitute back in the original equation we get &lt;math&gt;\frac{m}{n}=\frac{2}{3}&lt;/math&gt; so &lt;math&gt;3m=2n&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;. Thus &lt;math&gt;k+m+n=80+2+3=\boxed{085}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> One might cleverly change the content of both Jars. <br /> <br /> Since the end result of both Jars are &lt;math&gt;50\%&lt;/math&gt; acid, we can turn Jar A into a 1 gallon liquid with &lt;math&gt;50\%-4(5\%) = 30\%&lt;/math&gt; acid <br /> <br /> and Jar B into 1 gallon liquid with &lt;math&gt;50\%-5(2\%) =40\%&lt;/math&gt; acid.<br /> <br /> Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so &lt;math&gt;\dfrac{2}{3}&lt;/math&gt; of Jar C will be pour into Jar A.<br /> <br /> Thus, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;.<br /> <br /> &lt;math&gt;\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; yields &lt;math&gt;k=80&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;80+2+3 = \boxed{085}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> One may first combine all three jars in to a single container. That container will have &lt;math&gt;10&lt;/math&gt; liters of liquid, and it should be &lt;math&gt;50\%&lt;/math&gt; acidic. Thus there must be &lt;math&gt;5&lt;/math&gt; liters of acid. <br /> <br /> Jug A contained &lt;math&gt;45\% \cdot 4L&lt;/math&gt;, or &lt;math&gt;1.8L&lt;/math&gt; of acid, and jug B &lt;math&gt;48\% \cdot 5L&lt;/math&gt; or &lt;math&gt;2.4L&lt;/math&gt;. Solving for the amount of acid in jug C, &lt;math&gt;k = (5 - 2.4 - 1.8) = .8&lt;/math&gt;, or &lt;math&gt;80\%&lt;/math&gt;.<br /> <br /> Once one knows that the jug C is &lt;math&gt;80\%&lt;/math&gt; acid, use solution 1 to figure out m and n.<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=||before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=84372 2011 AIME I Problems/Problem 1 2017-03-03T02:32:08Z <p>Theultimate123: </p> <hr /> <div>== Problem 1 ==<br /> Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is &lt;math&gt;k\%&lt;/math&gt; acid. From jar C, &lt;math&gt;\frac{m}{n}&lt;/math&gt; liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;k + m + n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> There are &lt;math&gt;\frac{45}{100}(4)=\frac{9}{5}&lt;/math&gt; L of acid in Jar A. There are &lt;math&gt;\frac{48}{100}(5)=\frac{12}{5}&lt;/math&gt; L of acid in Jar B. And there are &lt;math&gt;\frac{k}{100}&lt;/math&gt; L of acid in Jar C. After transfering the solutions from jar C, there will be<br /> &lt;br&gt; &lt;math&gt;4+\frac{m}{n}&lt;/math&gt; L of solution in Jar A and &lt;math&gt;\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}&lt;/math&gt; L of acid in Jar A.&lt;br&gt;<br /> &lt;br&gt; &lt;math&gt;6-\frac{m}{n}&lt;/math&gt; L of solution in Jar B and &lt;math&gt;\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}&lt;/math&gt; of acid in Jar B.<br /> &lt;br&gt;Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.&lt;br&gt;<br /> &lt;cmath&gt;\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}&lt;/cmath&gt;<br /> Add the equations to get<br /> &lt;cmath&gt;\frac{42}{5}+\frac{k}{50}=10&lt;/cmath&gt; Solving gives &lt;math&gt;k=80&lt;/math&gt;.<br /> &lt;br&gt;If we substitute back in the original equation we get &lt;math&gt;\frac{m}{n}=\frac{2}{3}&lt;/math&gt; so &lt;math&gt;3m=2n&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;. Thus &lt;math&gt;k+m+n=80+2+3=\boxed{085}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> One might cleverly change the content of both Jars. <br /> <br /> Since the end result of both Jars are &lt;math&gt;50\%&lt;/math&gt; acid, we can turn Jar A into a 1 gallon liquid with &lt;math&gt;50\%-4(5\%) = 30\%&lt;/math&gt; acid <br /> <br /> and Jar B into 1 gallon liquid with &lt;math&gt;50\%-5(2\%) =40\%&lt;/math&gt; acid.<br /> <br /> Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so &lt;math&gt;\dfrac{2}{3}&lt;/math&gt; of Jar C will be pour into Jar A.<br /> <br /> Thus, &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=3&lt;/math&gt;.<br /> <br /> &lt;math&gt;\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; yields &lt;math&gt;k=80&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;80+2+3 = \boxed{085}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> One may first combine all three jars in to a single container. That container will have &lt;math&gt;10&lt;/math&gt; liters of liquid, and it should be &lt;math&gt;50\%&lt;/math&gt; acidic. Thus there must be &lt;math&gt;5&lt;/math&gt; liters of acid. <br /> <br /> Jug A contained &lt;math&gt;45\% \cdot 4L&lt;/math&gt;, or &lt;math&gt;1.8L&lt;/math&gt; of acid, and jug B &lt;math&gt;48\% \cdot 5L&lt;/math&gt; or &lt;math&gt;2.4L&lt;/math&gt;. Solving for the amount of acid in jug C, &lt;math&gt;k = (5 - 2.4 - 1.8) = .8&lt;/math&gt;, or &lt;math&gt;80\%&lt;/math&gt;.<br /> <br /> Once one knows that the jug C is &lt;math&gt;80\%&lt;/math&gt; acid, use solution 1 to figure out m and n.<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=First Problem|before=-|num-a=2}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=84371 2011 AIME I Problems/Problem 11 2017-03-03T02:26:41Z <p>Theultimate123: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;R&lt;/math&gt; be the set of all possible remainders when a number of the form &lt;math&gt;2^n&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; a nonnegative integer, is divided by &lt;math&gt;1000&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the sum of the elements in &lt;math&gt;R&lt;/math&gt;. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Note that &lt;math&gt;x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}&lt;/math&gt; and &lt;math&gt;x \equiv y \pmod{8}&lt;/math&gt;. So we must find the first two integers &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; such that &lt;math&gt;2^i \equiv 2^j \pmod{125}&lt;/math&gt; and &lt;math&gt;2^i \equiv 2^j \pmod{8}&lt;/math&gt; and &lt;math&gt;i \neq j&lt;/math&gt;. Note that &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; will be greater than 2 since remainders of &lt;math&gt;1, 2, 4&lt;/math&gt; will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that &lt;math&gt;2^{100}\equiv 1\pmod{125}&lt;/math&gt; (see [[Euler's Totient Theorem|Euler's theorem]]) and &lt;math&gt;2^0,2^1,2^2,\ldots,2^{99}&lt;/math&gt; are all distinct modulo 125 (proof below). Thus, &lt;math&gt;i = 103&lt;/math&gt; and &lt;math&gt;j =3&lt;/math&gt; are the first two integers such that &lt;math&gt;2^i \equiv 2^j \pmod{1000}&lt;/math&gt;. All that is left is to find &lt;math&gt;S&lt;/math&gt; in mod &lt;math&gt;1000&lt;/math&gt;. After some computation:<br /> &lt;cmath&gt;<br /> S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.<br /> &lt;/cmath&gt;<br /> To show that &lt;math&gt;2^0, 2^1,\ldots, 2^{99}&lt;/math&gt; are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of &lt;math&gt;2^{20}\equiv 1\pmod{125}&lt;/math&gt; or &lt;math&gt;2^{50}\equiv 1\pmod{125}&lt;/math&gt;. However, writing &lt;math&gt;2^{10}\equiv 25 - 1\pmod{125}&lt;/math&gt;, we can easily verify that &lt;math&gt;2^{20}\equiv -49\pmod{125}&lt;/math&gt; and &lt;math&gt;2^{50}\equiv -1\pmod{125}&lt;/math&gt;, giving us the needed contradiction.<br /> <br /> == Solution 2 ==<br /> Notice that our sum of remainders looks like &lt;cmath&gt;S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots.&lt;/cmath&gt; We want to find the remainder of &lt;math&gt;S&lt;/math&gt; upon division by &lt;math&gt;1000.&lt;/math&gt; Since &lt;math&gt;1000&lt;/math&gt; decomposes into primes as &lt;math&gt;1000 = 2^3 \cdot 5^3&lt;/math&gt;, we can check the remainders of &lt;math&gt;S&lt;/math&gt; modulo &lt;math&gt;2^3&lt;/math&gt; and modulo &lt;math&gt;5^3&lt;/math&gt; separately. <br /> <br /> Checking &lt;math&gt;S&lt;/math&gt; modulo &lt;math&gt;2^3&lt;/math&gt; is easy, so lets start by computing the remainder of &lt;math&gt;S&lt;/math&gt; upon division by &lt;math&gt;5^3.&lt;/math&gt; To do this, let's figure out when our sequence finally repeats.<br /> Notice that since the remainder when dividing any term of &lt;math&gt;S&lt;/math&gt; (after the third term) by &lt;math&gt;1000&lt;/math&gt; will be a multiple of &lt;math&gt;2^3&lt;/math&gt;, when this summation finally repeats, the first term to repeat will be ''not'' be &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2^3 \nmid 1.&lt;/math&gt; Instead, the first term to repeat will be &lt;math&gt;2^3&lt;/math&gt;, and then the sequence will continue once again &lt;math&gt;2^4, 2^5, \cdots.&lt;/math&gt; <br /> <br /> Now, to compute &lt;math&gt;S&lt;/math&gt; modulo &lt;math&gt;125&lt;/math&gt;, we want to find the least positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;2^d \equiv 1 \pmod {125}&lt;/math&gt; since then &lt;math&gt;d&lt;/math&gt; will just be the number of terms of &lt;math&gt;S&lt;/math&gt; (after the third term!) before the sequence repeats. In other words, our sequence will be of the form &lt;math&gt;S = 1 + 2 + 4 + \left(2^3 + 2^4 + \cdots + 2^{2 + d}\right)&lt;/math&gt; and then we will have &lt;math&gt;2^{d + 3} \equiv 2^3 \pmod {125}&lt;/math&gt;, and the sequence will repeat from there. Here, &lt;math&gt;d&lt;/math&gt; simply represents the order of &lt;math&gt;2&lt;/math&gt; modulo &lt;math&gt;125&lt;/math&gt;, denoted by &lt;math&gt;d = \text{ord}_{125}(2).&lt;/math&gt; To begin with, we'll use a well-known property of the order to get a bound on &lt;math&gt;d.&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\gcd(2, 125) = 1&lt;/math&gt; and &lt;math&gt;\phi(125) = 100&lt;/math&gt;, we know by Euler's Theorem that &lt;math&gt;2^{100} \equiv 1 \pmod {125}.&lt;/math&gt; However, we do not know that &lt;math&gt;100&lt;/math&gt; is the ''least'' &lt;math&gt;d&lt;/math&gt; satisfying &lt;math&gt;2^d \equiv 1 \pmod {125}.&lt;/math&gt; Nonetheless, it is a well known property of the order that &lt;math&gt;\text{ord}_{125}(2) = d | \phi(125) = 100.&lt;/math&gt; Therefore, we can conclude that &lt;math&gt;d&lt;/math&gt; must be a positive divisor of &lt;math&gt;100.&lt;/math&gt;<br /> <br /> Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say &lt;math&gt;\mod 5.&lt;/math&gt; Clearly, we must have that &lt;math&gt;2^d \equiv 1 \pmod 5.&lt;/math&gt; Since &lt;math&gt;2^4 \equiv 1 \pmod 5&lt;/math&gt; and powers of two will then cycle every four terms, we know that &lt;math&gt;2^d \equiv 1 \pmod 5 \iff 4 | d.&lt;/math&gt; Combining this relation with &lt;math&gt;d | 100&lt;/math&gt;, it follows that &lt;math&gt;d \in \{4, 20, 100\}.&lt;/math&gt; <br /> <br /> Now, it is trivial to verify that &lt;math&gt;d \ne 4.&lt;/math&gt; In addition, we know that &lt;cmath&gt;2^{20} = \left(2^{10}\right)^2 = \left(1024\right)^2 \equiv 24^2 = 576 \not\equiv 1 \pmod {125}.&lt;/cmath&gt; Therefore, we conclude that &lt;math&gt;d \ne 20.&lt;/math&gt; Hence, we must have &lt;math&gt;d = 100.&lt;/math&gt; (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly).<br /> <br /> Now, since we have found &lt;math&gt;d = 100&lt;/math&gt;, we know that &lt;cmath&gt;S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots + 2^{102}.&lt;/cmath&gt; There are two good ways to finish from here: <br /> <br /> The first way is to use a trick involving powers of &lt;math&gt;2.&lt;/math&gt; Notice that &lt;cmath&gt;S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.&lt;/cmath&gt; Certainly &lt;cmath&gt;S = 2^{103} - 1 \equiv -1 \equiv 7 \pmod {8}.&lt;/cmath&gt; In addition, since we already computed &lt;math&gt;\text{ord}_{125}(2) = d = 100&lt;/math&gt;, we know that &lt;cmath&gt;S = 2^{103} - 1 = 2^{100} \cdot 2^3 - 1 \equiv 2^3 - 1 \equiv 7 \pmod {125}.&lt;/cmath&gt; Therefore, since &lt;math&gt;S \equiv 7 \pmod{8}&lt;/math&gt; and &lt;math&gt;S \equiv 7 \pmod{125}&lt;/math&gt;, we conclude that &lt;math&gt;S \equiv \boxed{007} \pmod {1000}.&lt;/math&gt;<br /> <br /> The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of &lt;math&gt;S&lt;/math&gt; into groups: &lt;cmath&gt;R = 1 + 2 + 4; \quad T = 2^3 + 2^4 + \cdots + 2^{102}.&lt;/cmath&gt; It is easy to see that &lt;math&gt;R&lt;/math&gt; is congruent to &lt;math&gt;7&lt;/math&gt; modulo &lt;math&gt;1000.&lt;/math&gt; <br /> <br /> Now, for &lt;math&gt;T&lt;/math&gt;, notice that there are &lt;math&gt;100&lt;/math&gt; terms in the summation, each with a different remainder upon division by &lt;math&gt;125.&lt;/math&gt; Since each of these remainders is certainly relatively prime to &lt;math&gt;125&lt;/math&gt;, these &lt;math&gt;100&lt;/math&gt; remainders correspond to the &lt;math&gt;100&lt;/math&gt; positive integers less than &lt;math&gt;125&lt;/math&gt; that are relatively prime to &lt;math&gt;125.&lt;/math&gt; Therefore, <br /> &lt;cmath&gt;\begin{align*}T &amp;\equiv 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + \cdots + 124 \pmod{125} \\<br /> &amp;= \left(1 + 2 + 3 + \cdots + 125\right) - \left(5 + 10 + 15 + \cdots 125\right) \\<br /> &amp;= \frac{125 \cdot 126}{2} - 5\left(1 + 2 + 3 + \cdots 25\right) \\<br /> &amp;= 125 \cdot 63 - 5\left(\frac{25 \cdot 26}{2}\right) \\<br /> &amp;= 125\left(63 - 13\right) \\<br /> &amp;\equiv 0 \pmod{125}.\end{align*}&lt;/cmath&gt;<br /> Then, since &lt;math&gt;T&lt;/math&gt; is divisible by &lt;math&gt;125&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;, it follows that &lt;math&gt;T&lt;/math&gt; is divisible by &lt;math&gt;1000.&lt;/math&gt; Therefore, &lt;cmath&gt;S = R + T \equiv R \equiv \boxed{007} \pmod{1000}.&lt;/cmath&gt;<br /> <br /> == Solution 3 (Fake-ish) ==<br /> After listing out a few remainders of numbers of the form &lt;math&gt;2^i&lt;/math&gt;, namely &lt;math&gt;1, 2, 4, 8, 16, 64, 128, 512, 024, 048...&lt;/math&gt;, we can guess that, except for the first &lt;math&gt;3&lt;/math&gt;, the remainder will hit every multiple of &lt;math&gt;8&lt;/math&gt; between &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;992&lt;/math&gt;, inclusive. This means that our sum is of the form &lt;math&gt;1 + 2 + 4 + 8(1+2+...+124)&lt;/math&gt;. This can be written as <br /> &lt;math&gt;1 + 2 + 4 + 8(\frac{124\cdot125}{2}) = 7+4(124\cdot125)&lt;/math&gt;. Since the, last term, &lt;math&gt;4(124\cdot125)&lt;/math&gt;, is divisible by &lt;math&gt;1000&lt;/math&gt; it's remainder is &lt;math&gt;0&lt;/math&gt;. Thus the total remainder is just &lt;math&gt;1+2+4 = \boxed{007}.&lt;/math&gt;<br /> <br /> == Solution 4 ==<br /> We know &lt;math&gt;1, 2,&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; are in &lt;math&gt;R&lt;/math&gt;. Any other element in &lt;math&gt;R&lt;/math&gt; must be a multiple of &lt;math&gt;8&lt;/math&gt;. All multiples of &lt;math&gt;8&lt;/math&gt; under &lt;math&gt;1000&lt;/math&gt; sum up to a multiple of &lt;math&gt;1000&lt;/math&gt;. So we can ignore them. We need to remove all multiples of &lt;math&gt;5&lt;/math&gt;, or &lt;math&gt;40&lt;/math&gt; in what we counted because all elements of &lt;math&gt;R&lt;/math&gt; can only be divisible by &lt;math&gt;2&lt;/math&gt;. But, their sum is also a multiple of &lt;math&gt;1000&lt;/math&gt;. Likewise, the sum of all multiples of &lt;math&gt;8k&lt;/math&gt; for some &lt;math&gt;k&lt;/math&gt; will be a multiple of &lt;math&gt;1000&lt;/math&gt;. Thus, our answer is &lt;math&gt;1+2+4=\boxed{007}&lt;/math&gt;.<br /> <br /> Solution by TheUltimate123<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=84131 2014 AIME II Problems/Problem 10 2017-02-21T03:33:16Z <p>Theultimate123: /* Solution 3 (easy) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)==<br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Our equation can be simplified like the following.<br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> We recognize this as the Law of Cosines with angle &lt;math&gt;120&lt;/math&gt; degrees.<br /> Our polygon is an equilateral triangle, say &lt;math&gt;ABC&lt;/math&gt;, with center &lt;math&gt;O&lt;/math&gt; at the origin and &lt;math&gt;AB=AC=BC=2014&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;3*[ABO]=3*(1007*1007\sqrt{3})=3*1007^2*\sqrt{3}=3042147\sqrt{3}&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{147}&lt;/math&gt;.<br /> <br /> Solution by TheUltimate123 (Eric Shen)<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_10&diff=84130 2014 AIME II Problems/Problem 10 2017-02-21T03:32:43Z <p>Theultimate123: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> ==Solution 1 (long but non-bashy)==<br /> <br /> Note that the given equality reduces to<br /> <br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{w^3 - z^3}{w-z} = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;w^3 = z^3, w \neq z&lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;w = r_w e^{i \theta_w}&lt;/math&gt; and likewise for &lt;math&gt;z&lt;/math&gt;. Consider circle &lt;math&gt;O&lt;/math&gt; with the origin as the center and radius 2014 on the complex plane. It is clear that &lt;math&gt;z&lt;/math&gt; must be one of the points on this circle, as &lt;math&gt;|z| = 2014&lt;/math&gt;. <br /> <br /> By DeMoivre's Theorem, the complex modulus of &lt;math&gt;w&lt;/math&gt; is cubed when &lt;math&gt;w&lt;/math&gt; is cubed. Thus &lt;math&gt;w&lt;/math&gt; must lie on &lt;math&gt;O&lt;/math&gt;, since its the cube of its modulus, and thus its modulus, must be equal to &lt;math&gt;z&lt;/math&gt;'s modulus.<br /> <br /> Again, by DeMoivre's Theorem, &lt;math&gt;\theta_w&lt;/math&gt; is tripled when &lt;math&gt;w&lt;/math&gt; is cubed and likewise for &lt;math&gt;z&lt;/math&gt;. For &lt;math&gt;w&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt;, and the origin to lie on the same line, &lt;math&gt;3 \theta_w&lt;/math&gt; must be some multiple of 360 degrees apart from &lt;math&gt;3 \theta_z&lt;/math&gt; , so &lt;math&gt;\theta_w&lt;/math&gt; must differ from &lt;math&gt;\theta_z&lt;/math&gt; by some multiple of 120 degrees.<br /> <br /> Now, without loss of generality, assume that &lt;math&gt;z&lt;/math&gt; is on the real axis. (The circle can be rotated to put &lt;math&gt;z&lt;/math&gt; in any other location.) Then there are precisely two possible distinct locations for &lt;math&gt;w&lt;/math&gt;; one is obtained by going 120 degrees clockwise from &lt;math&gt;z&lt;/math&gt; about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.<br /> <br /> Let the two possible locations for &lt;math&gt;w&lt;/math&gt; be &lt;math&gt;W_1&lt;/math&gt; and &lt;math&gt;W_2&lt;/math&gt; and the location of &lt;math&gt;z&lt;/math&gt; be point &lt;math&gt;Z&lt;/math&gt;. Note that by symmetry, &lt;math&gt;W_1W_2Z&lt;/math&gt; is equilateral, say, with side length &lt;math&gt;x&lt;/math&gt;. We know that the circumradius of this equilateral triangle is &lt;math&gt;2014&lt;/math&gt;, so using the formula &lt;math&gt;\frac{abc}{4R} = [ABC]&lt;/math&gt; and that the area of an equilateral triangle with side length &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, so we have<br /> <br /> &lt;cmath&gt;\frac{x^3}{4R} = \frac{x^2\sqrt{3}}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;x = R \sqrt{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x^2\sqrt{3}}{4} = \frac{3R^2 \sqrt{3}}{4}&lt;/cmath&gt;<br /> <br /> Since we're concerned with the non-radical part of this expression and &lt;math&gt;R = 2014&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\frac{3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}&lt;/cmath&gt;<br /> <br /> and we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2 (short but a little bashy)==<br /> Assume &lt;math&gt;z = 2014&lt;/math&gt;. Then<br /> &lt;cmath&gt;\frac{1}{2014 + w} = \frac{1}{2014} + \frac{1}{w}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = w(2014 + w) + 2014(2014 + w)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2014w = 2014w + w^2 + 2014^2 + 2014w&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = w^2 + 2014w + 2014^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;P&lt;/math&gt; is an isosceles triangle with area &lt;math&gt;\frac{1}{2}(2014 - (-1007))(2\cdot 1007\sqrt{3}) = 3021\cdot 1007\sqrt{3}&lt;/math&gt; and &lt;math&gt;n \equiv 7\cdot 21\equiv \boxed{147} \pmod{1000}.&lt;/math&gt;<br /> <br /> ==Solution 3 (easy)==<br /> Our equation can be simplified like the following.<br /> &lt;cmath&gt;\frac{1}{w+z} = \frac{w+z}{wz}&lt;/cmath&gt;<br /> &lt;cmath&gt;wz = {(w+z)}^2&lt;/cmath&gt;<br /> &lt;cmath&gt;w^2 + wz + z^2 = 0&lt;/cmath&gt;<br /> We recognize this as the Law of Cosines with angle &lt;math&gt;120&lt;/math&gt; degrees.<br /> Our polygon is an equilateral triangle, say &lt;math&gt;ABC&lt;/math&gt;, with center &lt;math&gt;O&lt;/math&gt; at the origin and &lt;math&gt;AB=AC=BC=2014&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;3*[ABO]=3*(1007*1007\sqrt{3})=3*1007^2*\sqrt{3}=3042147\sqrt{3}&lt;/math&gt;. Thus, the answer is \boxed{147}.<br /> <br /> Solution by TheUltimate123 (Eric Shen)<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Theultimate123