https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Numbertheorist17&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-28T20:31:25Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_2&diff=48108 2008 USAMO Problems/Problem 2 2012-08-26T22:29:46Z <p>Numbertheorist17: /* Solution 9 (analytical) */</p> <hr /> <div>== Problem ==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute, [[scalene]] triangle, and let &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let the [[perpendicular]] [[bisect]]ors of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect ray &lt;math&gt;AM&lt;/math&gt; in points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; respectively, and let lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect in point &lt;math&gt;F&lt;/math&gt;, inside of triangle &lt;math&gt;ABC&lt;/math&gt;. Prove that points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; all lie on one circle.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 (synthetic) ===<br /> &lt;center&gt;&lt;asy&gt;<br /> /* setup and variables */<br /> size(280);<br /> pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br /> pair B=(0,0),C=(5,0),A=(1,4); /* A.x &gt; C.x/2 */<br /> /* construction and drawing */<br /> pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br /> D(MP(&quot;A&quot;,A,(0,1),s)--MP(&quot;B&quot;,B,SW,s)--MP(&quot;C&quot;,C,SE,s)--A--MP(&quot;M&quot;,M,s));<br /> D(B--D(MP(&quot;D&quot;,D,NE,s))--MP(&quot;P&quot;,P,(-1,0),s)--D(MP(&quot;O&quot;,O,(0,1),s)));<br /> D(D(MP(&quot;E&quot;,E,SW,s))--MP(&quot;N&quot;,N,(1,0),s));<br /> D(C--D(MP(&quot;F&quot;,F,NW,s))); <br /> D(B--O--C,linetype(&quot;4 4&quot;)+linewidth(0.7));<br /> D(M--N,linetype(&quot;4 4&quot;)+linewidth(0.7));<br /> D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br /> D(anglemark(B,A,C)); MP(&quot;y&quot;,A,(0,-6));MP(&quot;z&quot;,A,(4,-6));<br /> D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br /> picture p = new picture;<br /> draw(p,circumcircle(B,O,C),linetype(&quot;1 4&quot;)+linewidth(0.7));<br /> clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br /> <br /> /* D(circumcircle(A,P,N),linetype(&quot;4 4&quot;)+linewidth(0.7)); */<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> [[Without loss of generality]] &lt;math&gt;AB &lt; AC&lt;/math&gt;. The intersection of &lt;math&gt;NE&lt;/math&gt; and &lt;math&gt;PD&lt;/math&gt; is &lt;math&gt;O&lt;/math&gt;, the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\angle BAM = y&lt;/math&gt; and &lt;math&gt;\angle CAM = z&lt;/math&gt;. Note &lt;math&gt;D&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, so &lt;math&gt;AD = BD&lt;/math&gt;. So &lt;math&gt;\angle FBC = \angle B - \angle ABD = B - y&lt;/math&gt;. Similarly, &lt;math&gt;\angle FCB = C - z&lt;/math&gt;, so &lt;math&gt;\angle BFC = 180 - (B + C) + (y + z) = 2A&lt;/math&gt;. Notice that &lt;math&gt;\angle BOC&lt;/math&gt; intercepts the minor arc &lt;math&gt;BC&lt;/math&gt; in the [[circumcircle]] of &lt;math&gt;\triangle ABC&lt;/math&gt;, which is double &lt;math&gt;\angle A&lt;/math&gt;. Hence &lt;math&gt;\angle BFC = \angle BOC&lt;/math&gt;, so &lt;math&gt;BFOC&lt;/math&gt; is cyclic. <br /> <br /> <br /> ====Lemma 1====<br /> &lt;math&gt;\triangle FEO&lt;/math&gt; is directly similar to &lt;math&gt;\triangle NEM&lt;/math&gt;<br /> &lt;cmath&gt;<br /> \angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A<br /> &lt;/cmath&gt;<br /> since &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are collinear, &lt;math&gt;BFOC&lt;/math&gt; is cyclic, and &lt;math&gt;OB = OC&lt;/math&gt;. Also<br /> &lt;cmath&gt;<br /> \angle ENM = 90 - \angle MNC = 90 - A<br /> &lt;/cmath&gt;<br /> because &lt;math&gt;NE\perp AC&lt;/math&gt;, and &lt;math&gt;MNP&lt;/math&gt; is the medial triangle of &lt;math&gt;\triangle ABC&lt;/math&gt; so &lt;math&gt;AB \parallel MN&lt;/math&gt;. Hence &lt;math&gt;\angle OFE = \angle ENM&lt;/math&gt;.<br /> <br /> Notice that &lt;math&gt;\angle AEN = 90 - z = \angle CEN&lt;/math&gt; since &lt;math&gt;NE\perp BC&lt;/math&gt;. &lt;math&gt;\angle FED = \angle MEC = 2z&lt;/math&gt;. Then<br /> &lt;cmath&gt;<br /> \angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM<br /> &lt;/cmath&gt;<br /> Hence &lt;math&gt;\angle FEO = \angle NEM&lt;/math&gt;. <br /> <br /> Hence &lt;math&gt;\triangle FEO&lt;/math&gt; is similar to &lt;math&gt;\triangle NEM&lt;/math&gt; by AA similarity. It is easy to see that they are oriented such that they are directly similar.<br /> <br /> ====Lemma 2====<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> /* setup and variables */<br /> size(280);<br /> pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br /> pair B=(0,0),C=(5,0),A=(1,4); /* A.x &gt; C.x/2 */<br /> /* construction and drawing */<br /> pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br /> D(MP(&quot;A&quot;,A,(0,1),s)--MP(&quot;B&quot;,B,SW,s)--MP(&quot;C&quot;,C,SE,s)--A--MP(&quot;M&quot;,M,s));<br /> D(B--D(MP(&quot;D&quot;,D,NE,s))--MP(&quot;P&quot;,P,(-1,0),s)--D(MP(&quot;O&quot;,O,(1,0),s)));<br /> D(D(MP(&quot;E&quot;,E,SW,s))--MP(&quot;N&quot;,N,(1,0),s));<br /> D(C--D(MP(&quot;F&quot;,F,NW,s))); <br /> D(B--O--C,linetype(&quot;4 4&quot;)+linewidth(0.7));<br /> D(F--N); D(O--M);<br /> D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br /> <br /> /* commented in above asy<br /> D(circumcircle(A,P,N),linetype(&quot;4 4&quot;)+linewidth(0.7)); <br /> D(anglemark(B,A,C)); MP(&quot;y&quot;,A,(0,-6));MP(&quot;z&quot;,A,(4,-6));<br /> D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br /> picture p = new picture;<br /> draw(p,circumcircle(B,O,C),linetype(&quot;1 4&quot;)+linewidth(0.7));<br /> clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br /> */<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> By the similarity in Lemma 1, &lt;math&gt;FE: EO = NE: EM\implies FE: EN = OE: EM&lt;/math&gt;. &lt;math&gt;\angle FEN = \angle OEM&lt;/math&gt; so &lt;math&gt;\triangle FEN\sim\triangle OEM&lt;/math&gt; by SAS similarity. Hence<br /> &lt;cmath&gt;<br /> \angle EMO = \angle ENF = \angle ONF<br /> &lt;/cmath&gt;<br /> Using essentially the same angle chasing, we can show that &lt;math&gt;\triangle PDM&lt;/math&gt; is directly similar to &lt;math&gt;\triangle FDO&lt;/math&gt;. It follows that &lt;math&gt;\triangle PDF&lt;/math&gt; is directly similar to &lt;math&gt;\triangle MDO&lt;/math&gt;. So<br /> &lt;cmath&gt;<br /> \angle EMO = \angle DMO = \angle DPF = \angle OPF<br /> &lt;/cmath&gt;<br /> Hence &lt;math&gt;\angle OPF = \angle ONF&lt;/math&gt;, so &lt;math&gt;FONP&lt;/math&gt; is cyclic. In other words, &lt;math&gt;F&lt;/math&gt; lies on the circumcircle of &lt;math&gt;\triangle PON&lt;/math&gt;. Note that &lt;math&gt;\angle ONA = \angle OPA = 90&lt;/math&gt;, so &lt;math&gt;APON&lt;/math&gt; is cyclic. In other words, &lt;math&gt;A&lt;/math&gt; lies on the circumcircle of &lt;math&gt;\triangle PON&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; all lie on the circumcircle of &lt;math&gt;\triangle PON&lt;/math&gt;. Hence &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; lie on a circle, as desired.<br /> <br /> === Solution 2 (synthetic) ===<br /> Without Loss of Generality, assume &lt;math&gt;AB &gt;AC&lt;/math&gt;. It is sufficient to prove that &lt;math&gt;\angle OFA = 90^{\circ}&lt;/math&gt;, as this would immediately prove that &lt;math&gt;A,P,O,F,N&lt;/math&gt; are concyclic.<br /> By applying the Menelaus' Theorem in the Triangle &lt;math&gt;\triangle BFC&lt;/math&gt; for the transversal &lt;math&gt;E,M,D&lt;/math&gt;, we have (in magnitude)<br /> &lt;cmath&gt; \frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD} <br /> &lt;/cmath&gt; <br /> Here, we used that &lt;math&gt;BM=MC&lt;/math&gt;, as &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. Now, since &lt;math&gt;EC =EA&lt;/math&gt; and &lt;math&gt;BD=DA&lt;/math&gt;, we have<br /> &lt;cmath&gt; \frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD <br /> &lt;/cmath&gt;<br /> Now, note that &lt;math&gt;OE&lt;/math&gt; bisects the exterior &lt;math&gt;\angle FED&lt;/math&gt; and &lt;math&gt;OD&lt;/math&gt; bisects exterior &lt;math&gt;\angle FDE&lt;/math&gt;, making &lt;math&gt;O&lt;/math&gt; the &lt;math&gt;F&lt;/math&gt;-excentre of &lt;math&gt;\triangle FED&lt;/math&gt;. This implies that &lt;math&gt;OF&lt;/math&gt; bisects interior &lt;math&gt;\angle EFD&lt;/math&gt;, making &lt;math&gt;OF \perp AF&lt;/math&gt;, as was required.<br /> <br /> === Solution 3 (synthetic) ===<br /> Hint: consider &lt;math&gt;CF&lt;/math&gt; intersection with &lt;math&gt;PM&lt;/math&gt;; show that the resulting intersection lies on the desired circle. {{incomplete|solution}}<br /> <br /> === Solution 4 (synthetic) ===<br /> This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because &lt;math&gt;\angle OPA = \angle ONA = 90&lt;/math&gt;. Let the circumcircles of triangles &lt;math&gt;APN&lt;/math&gt; and &lt;math&gt;BOC&lt;/math&gt; intersect at &lt;math&gt;F'&lt;/math&gt; and &lt;math&gt;O&lt;/math&gt;. <br /> <br /> Lemma. If &lt;math&gt;A,B,C&lt;/math&gt; are points on circle &lt;math&gt;\omega&lt;/math&gt; with center &lt;math&gt;O&lt;/math&gt;, and the tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;B,C&lt;/math&gt; intersect at &lt;math&gt;Q&lt;/math&gt;, then &lt;math&gt;AP&lt;/math&gt; is the symmedian from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.<br /> <br /> It is easy to see &lt;math&gt;Q&lt;/math&gt; (the intersection of ray &lt;math&gt;OM&lt;/math&gt; and the circumcircle of &lt;math&gt;\triangle BOC&lt;/math&gt;) is colinear with &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;F'&lt;/math&gt;, and because line &lt;math&gt;OM&lt;/math&gt; is the diameter of that circle, &lt;math&gt;\angle QBO = \angle QCO = 90&lt;/math&gt;, so &lt;math&gt;Q&lt;/math&gt; is the point &lt;math&gt;Q&lt;/math&gt; in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that &lt;math&gt;F'&lt;/math&gt; satisfies the original construction for &lt;math&gt;F&lt;/math&gt;, showing &lt;math&gt;F=F'&lt;/math&gt;; we are done. {{incomplete|solution}}<br /> <br /> === Solution 5 (trigonometric) ===<br /> By the [[Law of Sines]], &lt;math&gt;\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}&lt;/math&gt;. Since &lt;math&gt;\angle ABF = \angle ABD = \angle BAD = \angle BAM&lt;/math&gt; and similarly &lt;math&gt;\angle ACF = \angle CAM&lt;/math&gt;, we cancel to get &lt;math&gt;\sin\angle AFC = \sin\angle AFB&lt;/math&gt;. Obviously, &lt;math&gt;\angle AFB + \angle AFC &gt; 180^\circ&lt;/math&gt; so &lt;math&gt;\angle AFC = \angle AFB&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF&lt;/math&gt; and &lt;math&gt;\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC&lt;/math&gt;. Subtracting these two equations, &lt;math&gt;\angle FAB - \angle FCA = \angle FCA - \angle FAB&lt;/math&gt; so &lt;math&gt;\angle BAF = \angle ACF&lt;/math&gt;. Therefore, &lt;math&gt;\triangle ABF\sim\triangle CAF&lt;/math&gt; (by AA similarity), so a spiral similarity centered at &lt;math&gt;F&lt;/math&gt; takes &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Therefore, it takes the midpoint of &lt;math&gt;\overline{BA}&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, or &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;N&lt;/math&gt;. So &lt;math&gt;\angle APF = \angle CNF = 180^\circ - \angle ANF&lt;/math&gt; and &lt;math&gt;APFN&lt;/math&gt; is cyclic.<br /> <br /> === Solution 6 (isogonal conjugates) ===<br /> &lt;center&gt;&lt;asy&gt;<br /> /* setup and variables */<br /> size(280);<br /> pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br /> pair B=(0,0),C=(5,0),A=(4,4); /* A.x &gt; C.x/2 */<br /> /* construction and drawing */<br /> pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br /> D(MP(&quot;A&quot;,A,(0,1),s)--MP(&quot;B&quot;,B,SW,s)--MP(&quot;C&quot;,C,SE,s)--A--MP(&quot;M&quot;,M,s));<br /> D(C--D(MP(&quot;E&quot;,E,NW,s))--MP(&quot;N&quot;,N,(1,0),s)--D(MP(&quot;O&quot;,O,SW,s)));<br /> D(D(MP(&quot;D&quot;,D,SE,s))--MP(&quot;P&quot;,P,W,s));<br /> D(B--D(MP(&quot;F&quot;,F,s))); D(O--A--F,linetype(&quot;4 4&quot;)+linewidth(0.7));<br /> D(MP(&quot;O'&quot;,circumcenter(A,P,N),NW,s));<br /> D(circumcircle(A,P,N),linetype(&quot;4 4&quot;)+linewidth(0.7));<br /> D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br /> picture p = new picture;<br /> draw(p,circumcircle(B,O,C),linetype(&quot;1 4&quot;)+linewidth(0.7));<br /> draw(p,circumcircle(A,B,C),linetype(&quot;1 4&quot;)+linewidth(0.7));<br /> clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Construct &lt;math&gt;T&lt;/math&gt; on &lt;math&gt;AM&lt;/math&gt; such that &lt;math&gt;\angle BCT = \angle ACF&lt;/math&gt;. Then &lt;math&gt;\angle BCT = \angle CAM&lt;/math&gt;. Then &lt;math&gt;\triangle AMC\sim\triangle CMT&lt;/math&gt;, so &lt;math&gt;\frac {AM}{CM} = \frac {CM}{TM}&lt;/math&gt;, or &lt;math&gt;\frac {AM}{BM} = \frac {BM}{TM}&lt;/math&gt;. Then &lt;math&gt;\triangle AMB\sim\triangle BMT&lt;/math&gt;, so &lt;math&gt;\angle CBT = \angle BAM = \angle FBA&lt;/math&gt;. Then we have<br /> <br /> &lt;math&gt;\angle CBT = \angle ABF&lt;/math&gt; and &lt;math&gt;\angle BCT = \angle ACF&lt;/math&gt;. So &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are [[isogonal conjugate|isogonally conjugate]]. Thus &lt;math&gt;\angle BAF = \angle CAM&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC&lt;/math&gt;.<br /> <br /> If &lt;math&gt;O&lt;/math&gt; is the [[circumcenter]] of &lt;math&gt;\triangle ABC&lt;/math&gt; then &lt;math&gt;\angle BFC = 2\angle BAC = \angle BOC&lt;/math&gt; so &lt;math&gt;BFOC&lt;/math&gt; is cyclic. Then &lt;math&gt;\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90&lt;/math&gt;. Then &lt;math&gt;\triangle AFO&lt;/math&gt; is a right triangle.<br /> <br /> Now by the [[homothety]] centered at &lt;math&gt;A&lt;/math&gt; with ratio &lt;math&gt;\frac {1}{2}&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; is taken to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is taken to &lt;math&gt;N&lt;/math&gt;. Thus &lt;math&gt;O&lt;/math&gt; is taken to the circumcenter of &lt;math&gt;\triangle APN&lt;/math&gt; and is the midpoint of &lt;math&gt;AO&lt;/math&gt;, which is also the circumcenter of &lt;math&gt;\triangle AFO&lt;/math&gt;, so &lt;math&gt;A,P,N,F,O&lt;/math&gt; all lie on a circle.<br /> <br /> === Solution 7 (symmedians) ===<br /> Median &lt;math&gt;AM&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt; implies &lt;math&gt;\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}&lt;/math&gt;.<br /> Trig ceva for &lt;math&gt;F&lt;/math&gt; shows that &lt;math&gt;AF&lt;/math&gt; is a symmedian.<br /> Then &lt;math&gt;FP&lt;/math&gt; is a median, use the lemma again to show that &lt;math&gt;AFP = C&lt;/math&gt;, and similarly &lt;math&gt;AFN = B&lt;/math&gt;, so you're done. {{incomplete|solution}}<br /> <br /> === Solution 8 (inversion) ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(280);<br /> pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br /> pair B=(0,0),C=(5,0),A=(4,4); /* A.x &gt; C.x/2 */<br /> real r = 1.2; /* inversion radius */<br /> <br /> /* construction and drawing */<br /> pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br /> D(MP(&quot;A&quot;,A,(0,1),s)--MP(&quot;B&quot;,B,SW,s)--MP(&quot;C&quot;,C,SE,s)--A--MP(&quot;M&quot;,M,s));<br /> D(C--D(MP(&quot;E&quot;,E,NW,s))--MP(&quot;N&quot;,N,(1,0),s)--D(MP(&quot;O&quot;,O,SW,s)));<br /> D(D(MP(&quot;D&quot;,D,SE,s))--MP(&quot;P&quot;,P,W,s));<br /> D(B--D(MP(&quot;F&quot;,F,s))); <br /> D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br /> <br /> D(CR(A,r));<br /> pair Pa = A + (P-A)/(r*r); D(MP(&quot;P'&quot;,Pa,NW,s));<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> We consider an [[inversion]] by an arbitrary [[radius]] about &lt;math&gt;A&lt;/math&gt;. We want to show that &lt;math&gt;P', F',&lt;/math&gt; and &lt;math&gt;N'&lt;/math&gt; are [[collinear]]. Notice that &lt;math&gt;D', A,&lt;/math&gt; and &lt;math&gt;P'&lt;/math&gt; lie on a circle with center &lt;math&gt;B'&lt;/math&gt;, and similarly for the other side. We also have that &lt;math&gt;B', D', F', A&lt;/math&gt; form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that &lt;math&gt;A B' F' C'&lt;/math&gt; is a [[parallelogram]], indicating that &lt;math&gt;F'&lt;/math&gt; is the midpoint of &lt;math&gt;P'N'&lt;/math&gt;. {{incomplete|solution}}<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2008|num-b=1|num-a=3}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202907 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_14&diff=47901 2002 AMC 8 Problems/Problem 14 2012-08-14T12:57:01Z <p>Numbertheorist17: /* Solution #1 */</p> <hr /> <div><br /> == Problem 14 ==<br /> <br /> A merchant offers a large group of items at &lt;math&gt;30\%&lt;/math&gt; off. Later, the merchant takes &lt;math&gt;20\%&lt;/math&gt; off these sale prices and claims that the final price of these items is &lt;math&gt;50\%&lt;/math&gt; off the original price. The total discount is<br /> <br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #1==<br /> <br /> Let's assume that each item is &lt;math&gt;100&lt;/math&gt; dollars. First we take off &lt;math&gt;30\%&lt;/math&gt; off of &lt;math&gt;100&lt;/math&gt; dollars. &lt;math&gt;100\cdot0.7=70&lt;/math&gt;<br /> <br /> Next, we take off the extra &lt;math&gt;20\%&lt;/math&gt; as asked by the problem. &lt;math&gt;70\cdot0.80=56&lt;/math&gt;<br /> <br /> So the final price of an item is &amp;#036;56. We have to do &lt;math&gt;100-56&lt;/math&gt; because &lt;math&gt;56&lt;/math&gt; was the final price and we wanted the discount.<br /> <br /> &lt;math&gt;100-56=44&lt;/math&gt; so the final discount was &lt;math&gt;44\%&lt;/math&gt;<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #2==<br /> <br /> Assume the price was &amp;#036;100. We can just do &lt;math&gt;100\cdot0.7\cdot0.8=56&lt;/math&gt; and then do &lt;math&gt;100-56=44&lt;/math&gt; That is the discount percentage wise.<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_14&diff=47900 2002 AMC 8 Problems/Problem 14 2012-08-14T12:56:31Z <p>Numbertheorist17: /* Solution #1 */</p> <hr /> <div><br /> == Problem 14 ==<br /> <br /> A merchant offers a large group of items at &lt;math&gt;30\%&lt;/math&gt; off. Later, the merchant takes &lt;math&gt;20\%&lt;/math&gt; off these sale prices and claims that the final price of these items is &lt;math&gt;50\%&lt;/math&gt; off the original price. The total discount is<br /> <br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #1==<br /> <br /> Let's assume that each item is &lt;math&gt;100&lt;/math&gt; dollars. First we take off &lt;math&gt;30\%&lt;/math&gt; off of &lt;math&gt;100&lt;/math&gt; dollars. &lt;math&gt;&amp;#036;100\cdot0.70=&lt;/math&gt; &amp;#036;70<br /> <br /> Next, we take off the extra &lt;math&gt;20\%&lt;/math&gt; as asked by the problem. &lt;math&gt;70\cdot0.80=56&lt;/math&gt;<br /> <br /> So the final price of an item is &amp;#036;56. We have to do &lt;math&gt;100-56&lt;/math&gt; because &lt;math&gt;56&lt;/math&gt; was the final price and we wanted the discount.<br /> <br /> &lt;math&gt;100-56=44&lt;/math&gt; so the final discount was &lt;math&gt;44\%&lt;/math&gt;<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #2==<br /> <br /> Assume the price was &amp;#036;100. We can just do &lt;math&gt;100\cdot0.7\cdot0.8=56&lt;/math&gt; and then do &lt;math&gt;100-56=44&lt;/math&gt; That is the discount percentage wise.<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_14&diff=47899 2002 AMC 8 Problems/Problem 14 2012-08-14T12:56:05Z <p>Numbertheorist17: /* Solution #1 */</p> <hr /> <div><br /> == Problem 14 ==<br /> <br /> A merchant offers a large group of items at &lt;math&gt;30\%&lt;/math&gt; off. Later, the merchant takes &lt;math&gt;20\%&lt;/math&gt; off these sale prices and claims that the final price of these items is &lt;math&gt;50\%&lt;/math&gt; off the original price. The total discount is<br /> <br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #1==<br /> <br /> Let's assume that each item is &lt;math&gt;100&lt;/math&gt; dollars. First we take off &lt;math&gt;30\%&lt;/math&gt; off of &amp;#036;100. &lt;math&gt;&amp;#036;100\cdot0.70=&lt;/math&gt; &amp;#036;70<br /> <br /> Next, we take off the extra &lt;math&gt;20\%&lt;/math&gt; as asked by the problem. &lt;math&gt;70\cdot0.80=56&lt;/math&gt;<br /> <br /> So the final price of an item is &amp;#036;56. We have to do &lt;math&gt;100-56&lt;/math&gt; because &lt;math&gt;56&lt;/math&gt; was the final price and we wanted the discount.<br /> <br /> &lt;math&gt;100-56=44&lt;/math&gt; so the final discount was &lt;math&gt;44\%&lt;/math&gt;<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #2==<br /> <br /> Assume the price was &amp;#036;100. We can just do &lt;math&gt;100\cdot0.7\cdot0.8=56&lt;/math&gt; and then do &lt;math&gt;100-56=44&lt;/math&gt; That is the discount percentage wise.<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_14&diff=47898 2002 AMC 8 Problems/Problem 14 2012-08-14T12:55:44Z <p>Numbertheorist17: /* Solution #1 */</p> <hr /> <div><br /> == Problem 14 ==<br /> <br /> A merchant offers a large group of items at &lt;math&gt;30\%&lt;/math&gt; off. Later, the merchant takes &lt;math&gt;20\%&lt;/math&gt; off these sale prices and claims that the final price of these items is &lt;math&gt;50\%&lt;/math&gt; off the original price. The total discount is<br /> <br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #1==<br /> <br /> Let's assume that each item is &lt;math&gt;&amp;#036;100&lt;/math&gt;. First we take off &lt;math&gt;30\%&lt;/math&gt; off of &amp;#036;100. &lt;math&gt;&amp;#036;100\cdot0.70=&lt;/math&gt; &amp;#036;70<br /> <br /> Next, we take off the extra &lt;math&gt;20\%&lt;/math&gt; as asked by the problem. &lt;math&gt;70\cdot0.80=56&lt;/math&gt;<br /> <br /> So the final price of an item is &amp;#036;56. We have to do &lt;math&gt;100-56&lt;/math&gt; because &lt;math&gt;56&lt;/math&gt; was the final price and we wanted the discount.<br /> <br /> &lt;math&gt;100-56=44&lt;/math&gt; so the final discount was &lt;math&gt;44\%&lt;/math&gt;<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #2==<br /> <br /> Assume the price was &amp;#036;100. We can just do &lt;math&gt;100\cdot0.7\cdot0.8=56&lt;/math&gt; and then do &lt;math&gt;100-56=44&lt;/math&gt; That is the discount percentage wise.<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_14&diff=47897 2002 AMC 8 Problems/Problem 14 2012-08-14T12:55:23Z <p>Numbertheorist17: /* Solution #1 */</p> <hr /> <div><br /> == Problem 14 ==<br /> <br /> A merchant offers a large group of items at &lt;math&gt;30\%&lt;/math&gt; off. Later, the merchant takes &lt;math&gt;20\%&lt;/math&gt; off these sale prices and claims that the final price of these items is &lt;math&gt;50\%&lt;/math&gt; off the original price. The total discount is<br /> <br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #1==<br /> <br /> Let's assume that each item is &lt;math&gt;&amp;#036/100&lt;/math&gt;. First we take off &lt;math&gt;30\%&lt;/math&gt; off of &amp;#036;100. &lt;math&gt;&amp;#036;100\cdot0.70=&lt;/math&gt; &amp;#036;70<br /> <br /> Next, we take off the extra &lt;math&gt;20\%&lt;/math&gt; as asked by the problem. &lt;math&gt;70\cdot0.80=56&lt;/math&gt;<br /> <br /> So the final price of an item is &amp;#036;56. We have to do &lt;math&gt;100-56&lt;/math&gt; because &lt;math&gt;56&lt;/math&gt; was the final price and we wanted the discount.<br /> <br /> &lt;math&gt;100-56=44&lt;/math&gt; so the final discount was &lt;math&gt;44\%&lt;/math&gt;<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;<br /> <br /> ==Solution #2==<br /> <br /> Assume the price was &amp;#036;100. We can just do &lt;math&gt;100\cdot0.7\cdot0.8=56&lt;/math&gt; and then do &lt;math&gt;100-56=44&lt;/math&gt; That is the discount percentage wise.<br /> <br /> &lt;math&gt; \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\% &lt;/math&gt;</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_4&diff=47849 1983 AIME Problems/Problem 4 2012-08-07T22:23:46Z <p>Numbertheorist17: /* Solution 2: Synthetic Solution */</p> <hr /> <div>== Problem ==<br /> A machine shop cutting tool is in the shape of a notched [[circle]], as shown. The [[radius]] of the circle is &lt;math&gt;\sqrt{50}&lt;/math&gt; cm, the [[length]] of &lt;math&gt;AB&lt;/math&gt; is 6 cm, and that of &lt;math&gt;BC&lt;/math&gt; is 2 cm. The [[angle]] &lt;math&gt;ABC&lt;/math&gt; is a [[right angle]]. Find the [[square]] of the distance (in centimeters) from &lt;math&gt;B&lt;/math&gt; to the center of the circle.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.6)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--O--A--B);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,SE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt; and let the foot of the perpendicular from &lt;math&gt;O&lt;/math&gt; to the [[line]] &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;OE=x&lt;/math&gt; and &lt;math&gt;OD=y&lt;/math&gt;. We're trying to find &lt;math&gt;x^2+y^2&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.6)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> pair D=(A.x,0),F=(0,B.y);<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--O--A--B);<br /> draw(D--O--F--B,dashed);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,F,SW);<br /> &lt;/asy&gt;&lt;/center&gt;&lt;!-- Asymptote replacement for Image:AIME_83_-4_Modified.JPG by bpms --&gt;<br /> <br /> Applying the Pythagorean Theorem, &lt;math&gt;OA^2 = OD^2 + AD^2&lt;/math&gt; and &lt;math&gt;OC^2 = EC^2 + EO^2&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;(\sqrt{50})^2 = y^2 + (6-x)^2&lt;/math&gt;, and &lt;math&gt;(\sqrt{50})^2 = x^2 + (y+2)^2&lt;/math&gt;. We solve this system to get &lt;math&gt;x = 1&lt;/math&gt; and &lt;math&gt;y = 5&lt;/math&gt;, resulting in &lt;math&gt;1^2 + 5^2 = \boxed{026}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Drop perpendiculars from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; (&lt;math&gt;T_1&lt;/math&gt;), &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;OT_1&lt;/math&gt; (&lt;math&gt;T_2&lt;/math&gt;), and &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; (&lt;math&gt;T_3&lt;/math&gt;).<br /> Also, draw the midpoint &lt;math&gt;M&lt;/math&gt; of &lt;math&gt;AC&lt;/math&gt;.<br /> <br /> Then the problem is trivialized. Why?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> pair dl(string name, pair loc, pair offset) {<br /> dot(loc);<br /> label(name,loc,offset);<br /> return loc;<br /> };<br /> pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)};<br /> string n[] = {&quot;O&quot;,&quot;$T_1$&quot;,&quot;B&quot;,&quot;C&quot;,&quot;M&quot;,&quot;A&quot;,&quot;$T_3$&quot;,&quot;M&quot;,&quot;$T_2$&quot;};<br /> for(int i=0;i&lt;a.length;++i) {<br /> dl(n[i],a[i],dir(degrees(a[i],false) ) );<br /> draw(a[(i-1)%a.length]--a[i]);<br /> };<br /> dot(a);<br /> draw(a--a);<br /> draw(a--a);<br /> draw(a--a);<br /> draw(a--a);<br /> draw(a--a);<br /> <br /> draw(a--a--a--cycle,blue+linewidth(0.7));<br /> draw(a--a--a--cycle,blue+linewidth(0.7));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> First notice that by computation, &lt;math&gt;OAC&lt;/math&gt; is a &lt;math&gt;\sqrt {50} - \sqrt {40} - \sqrt {50}&lt;/math&gt; isosceles triangle; thus &lt;math&gt;AC = MO&lt;/math&gt;.<br /> Then, notice that &lt;math&gt;\angle MOT_2 = \angle T_3MO = \angle BAC&lt;/math&gt;. Thus the two blue triangles are congruent.<br /> <br /> So, &lt;math&gt;MT_2 = 2,OT_2 = 6&lt;/math&gt;. As &lt;math&gt;T_3B = 3, MT_3 = 1&lt;/math&gt;, we subtract and get &lt;math&gt;OT_1 = 5,T_1B = 1&lt;/math&gt;. Then the Pythagorean Theorem shows &lt;math&gt;OB^2 = \boxed{026}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=Cyclic_sum&diff=47795 Cyclic sum 2012-08-04T21:39:10Z <p>Numbertheorist17: /* Rigorous definition */</p> <hr /> <div>A '''cyclic''' sum is a [[summation]] that cycles through all the values of a function and takes their sum, so to speak.<br /> <br /> ==Rigorous definition==<br /> Consider a function &lt;math&gt;f(a_1,a_2,a_3,\ldots a_n)&lt;/math&gt;. The cyclic sum &lt;math&gt;\sum_{cyc} f(a_1,a_2,a_3,\ldots a_n)&lt;/math&gt; is equal to <br /> <br /> &lt;cmath&gt;f(a_1,a_2,a_3,\ldots a_n)+f(a_2,a_3,a_4,\ldots a_n,a_1)+f(a_3,a_4,\ldots a_n,a_1,a_2)\ldots+f(a_n,a_1,a_2,\ldots a_{n-1})&lt;/cmath&gt;<br /> <br /> Note that not all permutations of the variables are used; they are just cycled through.<br /> <br /> ==Notation==<br /> A cyclic sum is often specified by having the variables to cycle through underneath the sigma, as follows: &lt;math&gt;\sum_{a,b,c}\frac{ab}{cd}&lt;/math&gt;. Note that a cyclic sum need not cycle through all of the variables. <br /> <br /> A cyclic sum is also sometimes specified by &lt;math&gt;\sum_{cyc}&lt;/math&gt;. This notation implies that all variables are cycled through.<br /> <br /> ==See also==<br /> *[[Summation]]<br /> *[[Symmetric sum]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Definition]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=47791 2012 AIME I Problems/Problem 14 2012-08-03T23:35:44Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solutions ==<br /> <br /> ==Solution 1==<br /> <br /> By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Assume &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, so at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and thus &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=47790 2012 AIME I Problems/Problem 14 2012-08-03T23:35:29Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> ==Solution 1==<br /> <br /> By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Assume &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, so at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and thus &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1989_USAMO_Problems/Problem_5&diff=47778 1989 USAMO Problems/Problem 5 2012-08-01T22:45:47Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;u&lt;/math&gt; and &lt;math&gt;v&lt;/math&gt; be real numbers such that<br /> &lt;cmath&gt; (u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8. &lt;/cmath&gt;<br /> Determine, with proof, which of the two numbers, &lt;math&gt;u&lt;/math&gt; or &lt;math&gt;v&lt;/math&gt;, is larger.<br /> <br /> ==Solution==<br /> <br /> The answer is &lt;math&gt;v&lt;/math&gt;.<br /> <br /> We define real functions &lt;math&gt;U&lt;/math&gt; and &lt;math&gt;V&lt;/math&gt; as follows:<br /> &lt;cmath&gt;\begin{align*}<br /> U(x) &amp;= (x+x^2 + \dotsb + x^8) + 10x^9 = \frac{x^{10}-x}{x-1} + 9x^9 \\<br /> V(x) &amp;= (x+x^2 + \dotsb + x^{10}) + 10x^{11} = \frac{x^{12}-x}{x-1} + 9x^{11} .<br /> \end{align*} &lt;/cmath&gt;<br /> We wish to show that if &lt;math&gt;U(u)=V(v)=8&lt;/math&gt;, then &lt;math&gt;u &lt;v&lt;/math&gt;.<br /> <br /> We first note that when &lt;math&gt;x \le 0&lt;/math&gt;, &lt;math&gt;x^{12}-x \ge 0&lt;/math&gt;, &lt;math&gt;x-1 &lt; 0&lt;/math&gt;, and &lt;math&gt;9x^9 \le 0&lt;/math&gt;, so<br /> &lt;cmath&gt; U(x) = \frac{x^{10}-x}{x-1} + 9x^9 \le 0 &lt; 8 .&lt;/cmath&gt;<br /> Similarly, &lt;math&gt;V(x) \le 0 &lt; 8&lt;/math&gt;.<br /> <br /> We also note that if &lt;math&gt;x \ge 9/10 &lt;/math&gt;, then<br /> &lt;cmath&gt; \begin{align*}<br /> U(x) &amp;= \frac{x-x^{10}}{1-x} + 9x^9 \ge \frac{9/10 - 9^9/10^9}{1/10} + 9 \cdot \frac{9^{9}}{10^9} \\<br /> &amp;= 9 - 10 \cdot \frac{9^9}{10^9} + 9 \cdot \frac{9^9}{10^9} = 9 - \frac{9^9}{10^9} &gt; 8.<br /> \end{align*} &lt;/cmath&gt;<br /> Similarly &lt;math&gt;V(x) &gt; 8&lt;/math&gt;. It then follows that &lt;math&gt;u, v \in (0,9/10)&lt;/math&gt;.<br /> <br /> Now, for all &lt;math&gt;x \in (0,9/10)&lt;/math&gt;,<br /> &lt;cmath&gt; \begin{align*}<br /> V(x) &amp;= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\<br /> &amp;= U(x) + x^9 (10x -9) (x+1) &lt; U(x) .<br /> \end{align*} &lt;/cmath&gt;<br /> Since &lt;math&gt;V&lt;/math&gt; and &lt;math&gt;U&lt;/math&gt; are both strictly increasing functions over the nonnegative reals, it then follows that<br /> &lt;cmath&gt; V(u) &lt; U(u) = 8 = V(v), &lt;/cmath&gt;<br /> so &lt;math&gt;u&lt;v&lt;/math&gt;, as desired. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> == Resources ==<br /> <br /> {{USAMO box|year=1989|num-b=4|after=Final Question}}<br /> <br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356639#356639 Discussion on AoPS/MathLinks]<br /> <br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47691 1987 IMO Problems/Problem 1 2012-07-21T17:22:51Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ==Solution 2==<br /> <br /> The probability of any number &lt;math&gt;i&lt;/math&gt; where &lt;math&gt;1\le i\le n&lt;/math&gt; being a fixed point is &lt;math&gt;\frac{1}{n}&lt;/math&gt;. Thus, the expected value of the number of fixed points is &lt;math&gt;n\times \frac{1}{n}=1&lt;/math&gt;. <br /> <br /> The expected value is also &lt;math&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}&lt;/math&gt;.<br /> <br /> Thus, &lt;cmath&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1&lt;/cmath&gt; or &lt;cmath&gt;\sum_{k=0}^{n} k \cdot p_n (k) = n!.&lt;/cmath&gt;<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47690 1987 IMO Problems/Problem 1 2012-07-21T17:22:38Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ==Solution 2==<br /> <br /> The probability of any number &lt;math&gt;i&lt;/math&gt; where &lt;math&gt;1\le i\le n&lt;/math&gt; being a fixed point is &lt;math&gt;\frac{1}{n}&lt;/math&gt;. Thus, the expected value of the number of fixed points is &lt;math&gt;n\times \frac{1}{n}=1&lt;/math&gt;. <br /> <br /> The expected value is also &lt;math&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}&lt;/math&gt;.<br /> <br /> Thus, &lt;cmath&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1&lt;/cmath&gt; or &lt;cmath&gt;\sum_{k=0}^{n} k \cdot p_n (k) = n!&lt;/cmath&gt;.<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47689 1987 IMO Problems/Problem 1 2012-07-21T17:22:25Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ==Solution 2==<br /> <br /> The probability of any number &lt;math&gt;i&lt;/math&gt; where &lt;math&gt;1\le i\le n&lt;/math&gt; being a fixed point is &lt;math&gt;\frac{1}{n}&lt;/math&gt;. Thus, the expected value of the number of fixed points is &lt;math&gt;n\times \frac{1}{n}=1&lt;/math&gt;. <br /> <br /> The expected value is also &lt;math&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}&lt;/math&gt;.<br /> <br /> Thus, &lt;cmath&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1&lt;/cmath&gt; or &lt;cmath&gt;\sum_{k=0}^{n} k \cdot p_n (k) = n!&lt;/cmath&gt;.<br /> {{alternate solutions}}<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47688 1987 IMO Problems/Problem 1 2012-07-21T17:22:13Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ==Solution 2==<br /> <br /> The probability of any number &lt;math&gt;i&lt;/math&gt; where &lt;math&gt;1\le i\le n&lt;/math&gt; being a fixed point is &lt;math&gt;\frac{1}{n}&lt;/math&gt;. Thus, the expected value of the number of fixed points is &lt;math&gt;n\times \frac{1}{n}=1&lt;/math&gt;. <br /> <br /> The expected value is also &lt;math&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}&lt;/math&gt;.<br /> <br /> Thus, &lt;cmath&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1&lt;/cmath&gt;, or &lt;cmath&gt;\sum_{k=0}^{n} k \cdot p_n (k) = n!&lt;/cmath&gt;.<br /> {{alternate solutions}}<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47687 1987 IMO Problems/Problem 1 2012-07-21T17:21:29Z <p>Numbertheorist17: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ==Solution 2==<br /> <br /> The probability of any number &lt;math&gt;i&lt;/math&gt; where &lt;math&gt;1\le i\le n&lt;/math&gt; being a fixed point is &lt;math&gt;\frac{1}{n}&lt;/math&gt;. Thus, the expected value of the number of fixed points is &lt;math&gt;n\times \frac{1}{n}=1&lt;/math&gt;. <br /> <br /> The expected value is also &lt;math&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}&lt;/math&gt;.<br /> <br /> Thus, &lt;cmath&gt;\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1&lt;/cmath&gt;, or \[\sum_{k=0}^{n} k \cdot p_n (k) = n!]\.<br /> {{alternate solutions}}<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_1&diff=47686 1987 IMO Problems/Problem 1 2012-07-21T17:16:13Z <p>Numbertheorist17: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p_n (k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt;\{ 1, \ldots , n \} , \; n \ge 1 &lt;/math&gt;, which have exactly &lt;math&gt;k &lt;/math&gt; fixed points. Prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> (Remark: A permutation &lt;math&gt;f &lt;/math&gt; of a set &lt;math&gt;S &lt;/math&gt; is a one-to-one mapping of &lt;math&gt;S &lt;/math&gt; onto itself. An element &lt;math&gt;i &lt;/math&gt; in &lt;math&gt;S &lt;/math&gt; is called a fixed point of the permutation &lt;math&gt;f &lt;/math&gt; if &lt;math&gt;f(i) = i &lt;/math&gt;.)<br /> <br /> == Solution ==<br /> <br /> The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element &lt;math&gt;i &lt;/math&gt; of the set, there are &lt;math&gt;(n-1)! &lt;/math&gt; permutations which have &lt;math&gt;i &lt;/math&gt; as a fixed point. Therefore<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \sum_{k=0}^{n} k \cdot p_n (k) = n!<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> <br /> <br /> <br /> {{alternate solutions}}<br /> <br /> {{IMO box|before=First question|num-a=2|year=1987}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=Expected_value&diff=47685 Expected value 2012-07-21T17:12:53Z <p>Numbertheorist17: /* Olympiad */</p> <hr /> <div>Given an event with a variety of different possible outcomes, the '''expected value''' is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is ''not'' the same as the &quot;most likely outcome.&quot;<br /> <br /> <br /> <br /> ==Formal Definition==<br /> More formally, we can define expected value as follows: if we have an event &lt;math&gt;Z&lt;/math&gt; whose outcomes have a [[discrete]] [[probability]] distribution, the expected value &lt;math&gt;E(Z) = \sum_z P(z) \cdot z&lt;/math&gt; where the sum is over all outcomes &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;P(z)&lt;/math&gt; is the probability of that particular outcome. If the event &lt;math&gt;Z&lt;/math&gt; has a [[continuous]] probability distribution, then &lt;math&gt;E(Z) = \int_z P(z)\cdot z\ dz&lt;/math&gt;.<br /> <br /> == Uses ==<br /> *Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.<br /> *Expected value can be used to calculate winning strategies in games of chance, such as board games.<br /> <br /> <br /> ==Example==<br /> As an example, flipping a fair coin has two possible outcomes, heads (denoted here by &lt;math&gt;H&lt;/math&gt;) or tails (&lt;math&gt;T&lt;/math&gt;). If we flip a fair coin repeatedly, we expect that we will get about the same number of heads as tails, or half as many as the total number of flips. Thus, the average outcome is &lt;math&gt;\frac 12 H + \frac 12 T&lt;/math&gt;. Note that not only is this not the most likely outcome, it is not even a possible outcome for a single flip.<br /> ==Problems==<br /> ===Introductory===<br /> *Find the expected value of a dice roll.<br /> **Find the expected value of a weighted dice roll, where each dot has equal probability of being on top.<br /> ===Intermediate===<br /> *In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt;<br /> &lt;div align=&quot;right&quot;&gt;([[Mock_AIME_2_2006-2007/Problem_13|Source]])&lt;/div&gt;<br /> *An equilateral triangle is tiled with &lt;math&gt;n^2&lt;/math&gt; smaller congruent equilateral triangles such that there are &lt;math&gt;n&lt;/math&gt; smaller triangles along each of the sides of the original triangle. The case &lt;math&gt;n = 11&lt;/math&gt; is shown [http://usamts.org/Tests/USAMTSProblems_17_2.pdf here, #3]. For each of the small equilateral triangles, we randomly choose a vertex &lt;math&gt;V&lt;/math&gt; of the triangle and draw an arc with that vertex as center connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full circles formed, in terms of &lt;math&gt;n&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 2, problem 3)&lt;/div&gt;<br /> *We play a game. The pot starts at &lt;dollar/&gt;0. On every turn, you flip a fair coin. If you flip heads, I add &lt;dollar/&gt;100 to the pot. If you flip tails, I take all of the money out of the pot, and you are assessed a “strike.” You can stop the game before any flip and collect the contents of the pot, but if you get 3 strikes, the game is over and you win nothing. Find, with proof, the expected value of your winnings if you follow an optimal strategy.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 4, problem 3)&lt;/div&gt;<br /> ===Olympiad===<br /> *Let &lt;math&gt; p_{n}(k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt; \{1,2,3,\ldots,n\} &lt;/math&gt; which have exactly &lt;math&gt;k&lt;/math&gt; fixed points. Prove that &lt;math&gt; \sum_{k=0}^{n}k p_{n}(k)=n! &lt;/math&gt;. <br /> &lt;div align=&quot;right&quot;&gt;(IMO 1987, #1)&lt;/div&gt;<br /> *Prove that there exists a 4-coloring of the set &lt;math&gt;M=\{1,2,3,\cdots,1987\}&lt;/math&gt; such that any 10-term arithmetic progression in the set &lt;math&gt;M&lt;/math&gt; is not monochromatic. <br /> &lt;div align=&quot;right&quot;&gt;(IMO Shortlist, 1987)&lt;/div&gt;<br /> <br /> ==See Also==<br /> *[[Probability]]<br /> *[[Combinatorics]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Combinatorics]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=Expected_value&diff=47684 Expected value 2012-07-21T17:12:37Z <p>Numbertheorist17: /* Olympiad */</p> <hr /> <div>Given an event with a variety of different possible outcomes, the '''expected value''' is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is ''not'' the same as the &quot;most likely outcome.&quot;<br /> <br /> <br /> <br /> ==Formal Definition==<br /> More formally, we can define expected value as follows: if we have an event &lt;math&gt;Z&lt;/math&gt; whose outcomes have a [[discrete]] [[probability]] distribution, the expected value &lt;math&gt;E(Z) = \sum_z P(z) \cdot z&lt;/math&gt; where the sum is over all outcomes &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;P(z)&lt;/math&gt; is the probability of that particular outcome. If the event &lt;math&gt;Z&lt;/math&gt; has a [[continuous]] probability distribution, then &lt;math&gt;E(Z) = \int_z P(z)\cdot z\ dz&lt;/math&gt;.<br /> <br /> == Uses ==<br /> *Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.<br /> *Expected value can be used to calculate winning strategies in games of chance, such as board games.<br /> <br /> <br /> ==Example==<br /> As an example, flipping a fair coin has two possible outcomes, heads (denoted here by &lt;math&gt;H&lt;/math&gt;) or tails (&lt;math&gt;T&lt;/math&gt;). If we flip a fair coin repeatedly, we expect that we will get about the same number of heads as tails, or half as many as the total number of flips. Thus, the average outcome is &lt;math&gt;\frac 12 H + \frac 12 T&lt;/math&gt;. Note that not only is this not the most likely outcome, it is not even a possible outcome for a single flip.<br /> ==Problems==<br /> ===Introductory===<br /> *Find the expected value of a dice roll.<br /> **Find the expected value of a weighted dice roll, where each dot has equal probability of being on top.<br /> ===Intermediate===<br /> *In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt;<br /> &lt;div align=&quot;right&quot;&gt;([[Mock_AIME_2_2006-2007/Problem_13|Source]])&lt;/div&gt;<br /> *An equilateral triangle is tiled with &lt;math&gt;n^2&lt;/math&gt; smaller congruent equilateral triangles such that there are &lt;math&gt;n&lt;/math&gt; smaller triangles along each of the sides of the original triangle. The case &lt;math&gt;n = 11&lt;/math&gt; is shown [http://usamts.org/Tests/USAMTSProblems_17_2.pdf here, #3]. For each of the small equilateral triangles, we randomly choose a vertex &lt;math&gt;V&lt;/math&gt; of the triangle and draw an arc with that vertex as center connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full circles formed, in terms of &lt;math&gt;n&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 2, problem 3)&lt;/div&gt;<br /> *We play a game. The pot starts at &lt;dollar/&gt;0. On every turn, you flip a fair coin. If you flip heads, I add &lt;dollar/&gt;100 to the pot. If you flip tails, I take all of the money out of the pot, and you are assessed a “strike.” You can stop the game before any flip and collect the contents of the pot, but if you get 3 strikes, the game is over and you win nothing. Find, with proof, the expected value of your winnings if you follow an optimal strategy.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 4, problem 3)&lt;/div&gt;<br /> ===Olympiad===<br /> *Let &lt;math&gt; p_{n}(k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt; \{1,2,3,\ldots,n\} &lt;/math&gt; which have exactly &lt;math&gt;k&lt;/math&gt; fixed points. Prove that &lt;math&gt; \sum_{k=0}^{n}k p_{n}(k)=n! &lt;/math&gt;. <br /> (IMO 1987, #1)<br /> &lt;div align=&quot;right&quot;&gt;(IMO 1987, #1)&lt;/div&gt;<br /> *Prove that there exists a 4-coloring of the set &lt;math&gt;M=\{1,2,3,\cdots,1987\}&lt;/math&gt; such that any 10-term arithmetic progression in the set &lt;math&gt;M&lt;/math&gt; is not monochromatic. (IMO Shortlist, 1987)<br /> &lt;div align=&quot;right&quot;&gt;(IMO Shortlist, 1987)&lt;/div&gt;<br /> <br /> ==See Also==<br /> *[[Probability]]<br /> *[[Combinatorics]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Combinatorics]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=Expected_value&diff=47683 Expected value 2012-07-21T17:11:58Z <p>Numbertheorist17: /* Olympiad */</p> <hr /> <div>Given an event with a variety of different possible outcomes, the '''expected value''' is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is ''not'' the same as the &quot;most likely outcome.&quot;<br /> <br /> <br /> <br /> ==Formal Definition==<br /> More formally, we can define expected value as follows: if we have an event &lt;math&gt;Z&lt;/math&gt; whose outcomes have a [[discrete]] [[probability]] distribution, the expected value &lt;math&gt;E(Z) = \sum_z P(z) \cdot z&lt;/math&gt; where the sum is over all outcomes &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;P(z)&lt;/math&gt; is the probability of that particular outcome. If the event &lt;math&gt;Z&lt;/math&gt; has a [[continuous]] probability distribution, then &lt;math&gt;E(Z) = \int_z P(z)\cdot z\ dz&lt;/math&gt;.<br /> <br /> == Uses ==<br /> *Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.<br /> *Expected value can be used to calculate winning strategies in games of chance, such as board games.<br /> <br /> <br /> ==Example==<br /> As an example, flipping a fair coin has two possible outcomes, heads (denoted here by &lt;math&gt;H&lt;/math&gt;) or tails (&lt;math&gt;T&lt;/math&gt;). If we flip a fair coin repeatedly, we expect that we will get about the same number of heads as tails, or half as many as the total number of flips. Thus, the average outcome is &lt;math&gt;\frac 12 H + \frac 12 T&lt;/math&gt;. Note that not only is this not the most likely outcome, it is not even a possible outcome for a single flip.<br /> ==Problems==<br /> ===Introductory===<br /> *Find the expected value of a dice roll.<br /> **Find the expected value of a weighted dice roll, where each dot has equal probability of being on top.<br /> ===Intermediate===<br /> *In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt;<br /> &lt;div align=&quot;right&quot;&gt;([[Mock_AIME_2_2006-2007/Problem_13|Source]])&lt;/div&gt;<br /> *An equilateral triangle is tiled with &lt;math&gt;n^2&lt;/math&gt; smaller congruent equilateral triangles such that there are &lt;math&gt;n&lt;/math&gt; smaller triangles along each of the sides of the original triangle. The case &lt;math&gt;n = 11&lt;/math&gt; is shown [http://usamts.org/Tests/USAMTSProblems_17_2.pdf here, #3]. For each of the small equilateral triangles, we randomly choose a vertex &lt;math&gt;V&lt;/math&gt; of the triangle and draw an arc with that vertex as center connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full circles formed, in terms of &lt;math&gt;n&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 2, problem 3)&lt;/div&gt;<br /> *We play a game. The pot starts at &lt;dollar/&gt;0. On every turn, you flip a fair coin. If you flip heads, I add &lt;dollar/&gt;100 to the pot. If you flip tails, I take all of the money out of the pot, and you are assessed a “strike.” You can stop the game before any flip and collect the contents of the pot, but if you get 3 strikes, the game is over and you win nothing. Find, with proof, the expected value of your winnings if you follow an optimal strategy.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 4, problem 3)&lt;/div&gt;<br /> ===Olympiad===<br /> *Let &lt;math&gt; p_{n}(k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt; \{1,2,3,\ldots,n\} &lt;/math&gt; which have exactly &lt;math&gt;k&lt;/math&gt; fixed points. Prove that &lt;math&gt; \sum_{k=0}^{n}k p_{n}(k)=n! &lt;/math&gt;. <br /> (IMO 1987, #1)<br /> *Prove that there exists a 4-coloring of the set &lt;math&gt;M=\{1,2,3,\cdots,1987\}&lt;/math&gt; such that any 10-term arithmetic progression in the set &lt;math&gt;M&lt;/math&gt; is not monochromatic. (IMO Shortlist, 1987)<br /> &lt;div align=&quot;right&quot;&gt;(IMO Shortlist, 1987)&lt;/div&gt;<br /> <br /> ==See Also==<br /> *[[Probability]]<br /> *[[Combinatorics]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Combinatorics]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=Expected_value&diff=47682 Expected value 2012-07-21T17:11:33Z <p>Numbertheorist17: /* Olympiad */</p> <hr /> <div>Given an event with a variety of different possible outcomes, the '''expected value''' is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is ''not'' the same as the &quot;most likely outcome.&quot;<br /> <br /> <br /> <br /> ==Formal Definition==<br /> More formally, we can define expected value as follows: if we have an event &lt;math&gt;Z&lt;/math&gt; whose outcomes have a [[discrete]] [[probability]] distribution, the expected value &lt;math&gt;E(Z) = \sum_z P(z) \cdot z&lt;/math&gt; where the sum is over all outcomes &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;P(z)&lt;/math&gt; is the probability of that particular outcome. If the event &lt;math&gt;Z&lt;/math&gt; has a [[continuous]] probability distribution, then &lt;math&gt;E(Z) = \int_z P(z)\cdot z\ dz&lt;/math&gt;.<br /> <br /> == Uses ==<br /> *Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.<br /> *Expected value can be used to calculate winning strategies in games of chance, such as board games.<br /> <br /> <br /> ==Example==<br /> As an example, flipping a fair coin has two possible outcomes, heads (denoted here by &lt;math&gt;H&lt;/math&gt;) or tails (&lt;math&gt;T&lt;/math&gt;). If we flip a fair coin repeatedly, we expect that we will get about the same number of heads as tails, or half as many as the total number of flips. Thus, the average outcome is &lt;math&gt;\frac 12 H + \frac 12 T&lt;/math&gt;. Note that not only is this not the most likely outcome, it is not even a possible outcome for a single flip.<br /> ==Problems==<br /> ===Introductory===<br /> *Find the expected value of a dice roll.<br /> **Find the expected value of a weighted dice roll, where each dot has equal probability of being on top.<br /> ===Intermediate===<br /> *In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt;<br /> &lt;div align=&quot;right&quot;&gt;([[Mock_AIME_2_2006-2007/Problem_13|Source]])&lt;/div&gt;<br /> *An equilateral triangle is tiled with &lt;math&gt;n^2&lt;/math&gt; smaller congruent equilateral triangles such that there are &lt;math&gt;n&lt;/math&gt; smaller triangles along each of the sides of the original triangle. The case &lt;math&gt;n = 11&lt;/math&gt; is shown [http://usamts.org/Tests/USAMTSProblems_17_2.pdf here, #3]. For each of the small equilateral triangles, we randomly choose a vertex &lt;math&gt;V&lt;/math&gt; of the triangle and draw an arc with that vertex as center connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full circles formed, in terms of &lt;math&gt;n&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 2, problem 3)&lt;/div&gt;<br /> *We play a game. The pot starts at &lt;dollar/&gt;0. On every turn, you flip a fair coin. If you flip heads, I add &lt;dollar/&gt;100 to the pot. If you flip tails, I take all of the money out of the pot, and you are assessed a “strike.” You can stop the game before any flip and collect the contents of the pot, but if you get 3 strikes, the game is over and you win nothing. Find, with proof, the expected value of your winnings if you follow an optimal strategy.<br /> &lt;div align=&quot;right&quot;&gt;(USAMTS year 17, round 4, problem 3)&lt;/div&gt;<br /> ===Olympiad===<br /> *Let &lt;math&gt; p_{n}(k) &lt;/math&gt; be the number of permutations of the set &lt;math&gt; \{1,2,3,\ldots,n\} &lt;/math&gt; which have exactly &lt;math&gt;k&lt;/math&gt; fixed points. Prove that &lt;math&gt; \sum_{k=0}^{n}k p_{n}(k)=n! &lt;/math&gt;. (IMO 1987, #1)<br /> *Prove that there exists a 4-coloring of the set &lt;math&gt;M=\{1,2,3,\cdots,1987\}&lt;/math&gt; such that any 10-term arithmetic progression in the set &lt;math&gt;M&lt;/math&gt; is not monochromatic. (IMO Shortlist, 1987)<br /> &lt;div align=&quot;right&quot;&gt;(IMO Shortlist, 1987)&lt;/div&gt;<br /> <br /> ==See Also==<br /> *[[Probability]]<br /> *[[Combinatorics]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Combinatorics]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=47621 2001 IMO Problems/Problem 1 2012-07-11T23:40:00Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>== Problem ==<br /> Consider an acute triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude of triangle &lt;math&gt;\triangle ABC&lt;/math&gt; issuing from the vertex &lt;math&gt;A&lt;/math&gt;, and let &lt;math&gt;O&lt;/math&gt; be the [[circumcenter]] of triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Assume that &lt;math&gt;\angle C \geq \angle B+30^{\circ}&lt;/math&gt;. Prove that &lt;math&gt;\angle A+\angle COP &lt; 90^{\circ}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Take &lt;math&gt;D&lt;/math&gt; on the circumcircle with &lt;math&gt;AD \parallel BC&lt;/math&gt;. Notice that &lt;math&gt;\angle CBD = \angle BCA&lt;/math&gt;, so &lt;math&gt;\angle ABD \ge 30^\circ&lt;/math&gt;. Hence &lt;math&gt;\angle AOD \ge 60^\circ&lt;/math&gt;. Let &lt;math&gt;Z&lt;/math&gt; be the midpoint of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; the midpoint of &lt;math&gt;BC&lt;/math&gt;. Then &lt;math&gt;AZ \ge R/2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle. But &lt;math&gt;AZ = YX&lt;/math&gt; (since &lt;math&gt;AZYX&lt;/math&gt; is a rectangle).<br /> <br /> Now &lt;math&gt;O&lt;/math&gt; cannot coincide with &lt;math&gt;Y&lt;/math&gt; (otherwise &lt;math&gt;\angle A&lt;/math&gt; would be &lt;math&gt;90^\circ&lt;/math&gt; and the triangle would not be acute-angled). So &lt;math&gt;OX &gt; YX \ge R/2&lt;/math&gt;. But &lt;math&gt;XC = YC - YX &lt; R - YX \le R/2&lt;/math&gt;. So &lt;math&gt;OX &gt; XC&lt;/math&gt;.<br /> <br /> Hence &lt;math&gt;\angle COX &lt; \angle OCX&lt;/math&gt;. Let &lt;math&gt;CE&lt;/math&gt; be a diameter of the circle, so that &lt;math&gt;\angle OCX = \angle ECB&lt;/math&gt;. But &lt;math&gt;\angle ECB = \angle EAB&lt;/math&gt; and &lt;math&gt;\angle EAB + \angle BAC = \angle EAC = 90^\circ&lt;/math&gt;, since &lt;math&gt;EC&lt;/math&gt; is a diameter. Hence &lt;math&gt;\angle COX + \angle BAC &lt; 90^\circ&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{IMO box|year=2001|before=First question|num-a=2}}<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_1&diff=47620 2001 IMO Problems/Problem 1 2012-07-11T23:39:05Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>== Problem ==<br /> Consider an acute triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude of triangle &lt;math&gt;\triangle ABC&lt;/math&gt; issuing from the vertex &lt;math&gt;A&lt;/math&gt;, and let &lt;math&gt;O&lt;/math&gt; be the [[circumcenter]] of triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Assume that &lt;math&gt;\angle C \geq \angle B+30^{\circ}&lt;/math&gt;. Prove that &lt;math&gt;\angle A+\angle COP &lt; 90^{\circ}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Take &lt;math&gt;D&lt;/math&gt; on the circumcircle with &lt;math&gt;AD \parallel to BC&lt;/math&gt;. Notice that &lt;math&gt;\angle CBD = \angle BCA&lt;/math&gt;, so &lt;math&gt;\angle ABD \ge 30^\circ&lt;/math&gt;. Hence &lt;math&gt;\angle AOD \ge 60^\circ&lt;/math&gt;. Let &lt;math&gt;Z&lt;/math&gt; be the midpoint of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; the midpoint of &lt;math&gt;BC&lt;/math&gt;. Then &lt;math&gt;AZ \ge R/2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle. But &lt;math&gt;AZ = YX&lt;/math&gt; (since &lt;math&gt;AZYX&lt;/math&gt; is a rectangle).<br /> <br /> Now &lt;math&gt;O&lt;/math&gt; cannot coincide with &lt;math&gt;Y&lt;/math&gt; (otherwise &lt;math&gt;\angle A&lt;/math&gt; would be &lt;math&gt;90^\circ&lt;/math&gt; and the triangle would not be acute-angled). So &lt;math&gt;OX &gt; YX \ge R/2&lt;/math&gt;. But &lt;math&gt;XC = YC - YX &lt; R - YX \le R/2&lt;/math&gt;. So &lt;math&gt;OX &gt; XC&lt;/math&gt;.<br /> <br /> Hence &lt;math&gt;\angle COX &lt; \angle OCX&lt;/math&gt;. Let &lt;math&gt;CE&lt;/math&gt; be a diameter of the circle, so that &lt;math&gt;\angle OCX = \angle ECB&lt;/math&gt;. But &lt;math&gt;\angle ECB = \angle EAB&lt;/math&gt; and &lt;math&gt;\angle EAB + \angle BAC = \angle EAC = 90^\circ&lt;/math&gt;, since &lt;math&gt;EC&lt;/math&gt; is a diameter. Hence &lt;math&gt;\angle COX + \angle BAC &lt; 90^\circ&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{IMO box|year=2001|before=First question|num-a=2}}<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=1994_IMO_Problems/Problem_3&diff=47503 1994 IMO Problems/Problem 3 2012-07-01T19:36:37Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> For any positive integer k, let f(k) be the number of elements in the set<br /> {k + 1; k + 2;....; 2k} whose base 2 representation has precisely three 1s.<br /> * (a) Prove that, for each positive integer m, there exists at least one positive integer k such that f(k) = m.<br /> * (b) Determine all positive integers m for which there exists exactly one k with f(k) = m.<br /> <br /> <br /> == Solution ==<br /> <br /> <br /> === Solution 1 ===<br /> <br /> '''a) Surjectivity of f'''<br /> <br /> For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.<br /> <br /> It's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number) <br /> * &lt;math&gt;f(2^n) = \tbinom n2 = \sum_{i=1}^{n-1} i&lt;/math&gt; whence <br /> * &lt;math&gt;f(2^{n+1}) = f(2^n)+n &lt;/math&gt;<br /> <br /> Now consider &lt;math&gt;k=2^n + 2&lt;/math&gt; with &lt;math&gt;n \ge 2&lt;/math&gt; and the corrisponding set &lt;math&gt;S(2^n+2)=\{2^n+3,...,2^{n+1},2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\}&lt;/math&gt;<br /> whose subset &lt;math&gt;\{2^n+3,...,2^{n+1}\}&lt;/math&gt; contains &lt;math&gt;f(2^n)&lt;/math&gt; T-numbers since &lt;math&gt;S(2^n)=\{2^n+1,...,2^{n+1}\}=\{2^n+1,2^n+2\} \cup \{2^n+3,...,2^{n+1}\}&lt;/math&gt; contains &lt;math&gt;f(2^n)&lt;/math&gt; T-numbers by definition but &lt;math&gt;\{2^n+1,2^n+2\}&lt;/math&gt; has none. So &lt;math&gt;S(2^n+2)=\{2^n+3,...,2^{n+1}\} \cup \{2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\}&lt;/math&gt; but the last set has the only T-number &lt;math&gt;2^{n+1}+3&lt;/math&gt;. We conclude that:<br /> *&lt;math&gt;f(2^n + 2)=f(2^n)+1&lt;/math&gt;<br /> <br /> Now consider &lt;math&gt;k=2^n +2^r + 2^s &lt;/math&gt; with &lt;math&gt;n&gt;r&gt;s\ge 0&lt;/math&gt; and &lt;math&gt;n\ge 2&lt;/math&gt;. <br /> We explicitly calculate f(k) for such numbers.<br /> So we have to calculate how many T-numbers are in the set &lt;math&gt;S(k)=\{2^n +2^r + 2^s+1;...;2^{n+1} +2^{r+1} + 2^{s+1}\}&lt;/math&gt;. <br /> <br /> The T-numbers less than &lt;math&gt;2^{n+1}&lt;/math&gt; in &lt;math&gt;S(k)&lt;/math&gt; are &lt;math&gt; T_1 = \{2^n +2^j + 2^i : (j=r \wedge r&gt;i&gt;s) \vee <br /> (n&gt;j&gt;r \wedge j&gt;i&gt;0) \}&lt;/math&gt; <br /> whence &lt;math&gt; \# (T_1)= r-s-1 + \sum_{h=r+1}^{n-1} h &lt;/math&gt;.<br /> <br /> The T-numbers greater than &lt;math&gt;2^{n+1}&lt;/math&gt; in &lt;math&gt;S(k)&lt;/math&gt; are &lt;math&gt; T_2 = \{2^{n+1} +2^j + 2^i : (j=r+1 \wedge s+1 \ge i\ge 0) \vee <br /> (r\ge j \ge 1 \wedge j&gt;i \ge 0) \}&lt;/math&gt; <br /> whence &lt;math&gt; \# (T_2)= s+2 + \sum_{h=1}^{r} h &lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;S(k)&lt;/math&gt; contains &lt;math&gt; \# (T_1) + \# (T_2)=r-s-1 + s+2 + \sum_{h=1}^{r}h +\sum_{h=r+1}^{n-1} h = \sum_{h=1}^{n-1} h + r + 1 &lt;/math&gt; T-numbers and we have: <br /> *&lt;math&gt;f(2^n +2^r + 2^s)=f(2^n) + r +1 &lt;/math&gt; where &lt;math&gt;r=1,2,...,n-1&lt;/math&gt; ans &lt;math&gt;r &gt; s \ge 0&lt;/math&gt; .<br /> Summarizing <br /> * &lt;math&gt;f(2^{n+1}) = f(2^n)+n &lt;/math&gt;<br /> * &lt;math&gt;f(2^n + 2)=f(2^n)+1&lt;/math&gt;<br /> * &lt;math&gt;f(2^n +2^r + 2^s)=f(2^n) + r +1 &lt;/math&gt; where &lt;math&gt;r=1,2,...,n-1&lt;/math&gt; and &lt;math&gt;r &gt; s \ge 0&lt;/math&gt; .<br /> That's to say that f takes all the values from &lt;math&gt;f(2^n)&lt;/math&gt; to &lt;math&gt;f(2^{n+1})&lt;/math&gt; for every &lt;math&gt; n \ge 2 &lt;/math&gt; and<br /> then takes any positive integer value since &lt;math&gt;f(4)=1&lt;/math&gt; and &lt;math&gt;f(2^n)&lt;/math&gt; becomes arbitrarily large.<br /> <br /> '''b) one-from-one values'''<br /> <br /> By the fact that &lt;math&gt;f(2^n +2^r + 2^s)=f(2^n) + r +1 &lt;/math&gt; where &lt;math&gt;r=1,2,...,n-1&lt;/math&gt; and &lt;math&gt;r &gt; s \ge 0&lt;/math&gt; <br /> it follows that each of the values &lt;math&gt;f(2^n) + r +1&lt;/math&gt; where &lt;math&gt;r=2,...,n-1&lt;/math&gt; come from at least two different k because there are at least two different choices for &lt;math&gt;s&lt;/math&gt;. These values are &lt;math&gt;f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})&lt;/math&gt;.<br /> <br /> Thus the only possible one-from-one values are &lt;math&gt;f(2^n)+1&lt;/math&gt; that come from &lt;math&gt;k=2^n + 2&lt;/math&gt; and &lt;math&gt;f(2^n)+2&lt;/math&gt; that come from &lt;math&gt;k=2^n + 3&lt;/math&gt;.<br /> <br /> If &lt;math&gt;k=2^n + 3&lt;/math&gt; we have &lt;math&gt;f(k)=f(k+1)&lt;/math&gt; because k+1 is not T and &lt;math&gt;2k+1=2^{n+1} + 7&lt;/math&gt; and &lt;math&gt;2k+2=2^{n+1} + 8&lt;/math&gt; are not T so f maps &lt;math&gt;2^n + 3&lt;/math&gt; and &lt;math&gt;2^n + 4&lt;/math&gt; to &lt;math&gt;f(2^n)+2&lt;/math&gt;.<br /> <br /> If &lt;math&gt;k=2^n + 2&lt;/math&gt; then &lt;math&gt;f(k+1)=f(2^n)+2&gt;f(k)=f(2^n)+1&gt;f(k-1)=f(2^n+1)=f(2^n)&lt;/math&gt;. <br /> The function f is non-decreasing.<br /> It is sufficient to prove that &lt;math&gt;f(k+1) \ge f(k)&lt;/math&gt; but this follows by the fact that if k+1 is T then 2k+2 is T too. <br /> By the monotonicity of f &lt;math&gt;f( 2^n + 2)=f(2^n)+1=\tbinom n2 +1= n(n-1)/2 + 1&lt;/math&gt; with &lt;math&gt;n \ge 2&lt;/math&gt; are the only one-from-one values of f.<br /> <br /> === Solution 2 ===<br /> <br /> '''a) Surjectivity of f'''<br /> <br /> For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation and define the sets &lt;math&gt;S(k)=\{k+1;k+2;...;2k\}&lt;/math&gt;.<br /> So &lt;math&gt;f(k)&lt;/math&gt; is the number of T-numbers in &lt;math&gt;S(k)&lt;/math&gt;.<br /> <br /> A positive even integer 2n is T iff n is T.(Fundamental theorem of T numbers)(FTT)<br /> <br /> The function f is non-decreasing.<br /> It is sufficient to prove that &lt;math&gt;f(k+1) \ge f(k)&lt;/math&gt;. This follows by (FTT) if k+1 is T then 2k+2 is T too; so passing from &lt;math&gt;S(k)&lt;/math&gt; to &lt;math&gt;S(k+1)&lt;/math&gt; we can loose a T-number &lt;math&gt;k+1&lt;/math&gt; but we regain it with &lt;math&gt;2k+2&lt;/math&gt; in &lt;math&gt;S(k+1)&lt;/math&gt; so &lt;math&gt;f(k+1)&lt;/math&gt; cannot be less than &lt;math&gt;f(k)&lt;/math&gt;. We also have &lt;math&gt;f(k+1) \le f(k) + 1&lt;/math&gt;. This would be false if &lt;math&gt;k+1&lt;/math&gt; were not T and both &lt;math&gt;2k+1&lt;/math&gt; and &lt;math&gt;2k+2&lt;/math&gt; were T. But if &lt;math&gt;2k+2&lt;/math&gt; were T then &lt;math&gt;k+1&lt;/math&gt; would be T too by the (FTT).<br /> Then we have &lt;math&gt; f(k) \le f(k+1) \le f(k) + 1&lt;/math&gt; that is &lt;math&gt;f(k+1)=f(k)&lt;/math&gt; or &lt;math&gt;f(k+1)=f(k)+1&lt;/math&gt;.<br /> But &lt;math&gt;f(4)=1&lt;/math&gt; and &lt;math&gt;f(2^n)= \tbinom n2&lt;/math&gt; so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values.<br /> <br /> '''b) one-from-one values of f'''<br /> <br /> The one-from-one values &lt;math&gt;f(k)&lt;/math&gt; are such that &lt;math&gt;f(k)=f(k-1)+1&lt;/math&gt; and &lt;math&gt;f(k+1)=f(k)+1&lt;/math&gt; since f is monotone non-decreasing and has step 1.<br /> <br /> By the condition &lt;math&gt;f(k+1)=f(k)+1&lt;/math&gt; since &lt;math&gt;k+1&lt;/math&gt; is T iff &lt;math&gt;2k+2&lt;/math&gt; is T we have that &lt;math&gt;2k+1&lt;/math&gt; must be T.<br /> <br /> By the condition &lt;math&gt;f(k)=f(k-1)+1&lt;/math&gt; since &lt;math&gt;k&lt;/math&gt; is T iff &lt;math&gt;2k&lt;/math&gt; is T we have that &lt;math&gt;2k-1&lt;/math&gt; must be T.<br /> <br /> Then &lt;math&gt;2k-1&lt;/math&gt; and &lt;math&gt;2k+1&lt;/math&gt; must have the form &lt;math&gt;2^{n+1}+2^{r+1}+1&lt;/math&gt; with &lt;math&gt;n&gt;r\ge 0&lt;/math&gt; since they are odd and &lt;math&gt;n \ge 2&lt;/math&gt; since there is only 1 T-number less than 8 and they must have the same number of bits since their diferrence is 2 and both are T (the only two binary numbers wich differs by 2 and have different number of bits are &lt;math&gt;2^n+1&lt;/math&gt; and &lt;math&gt;2^n-1&lt;/math&gt; or &lt;math&gt;2^n&lt;/math&gt; and &lt;math&gt;2^n-2&lt;/math&gt; which are evidently not T). Let &lt;math&gt;2k+1=2^{n+1}+2^{j+1}+1&lt;/math&gt; and &lt;math&gt;2k-1=2^{n+1}+2^{i+1}+1&lt;/math&gt; with &lt;math&gt;j&gt;i \ge 0&lt;/math&gt;. Then &lt;math&gt;2k+ 1-(2k-1)=2^{j+1}-2^{i+1}=2&lt;/math&gt; that is &lt;math&gt;j=1&lt;/math&gt; and &lt;math&gt;i=0&lt;/math&gt;.<br /> We conclude that &lt;math&gt;f(k)&lt;/math&gt; is one-from-one for &lt;math&gt;k=2^n+2^j=2^n+2&lt;/math&gt; with &lt;math&gt;n \ge 2&lt;/math&gt;.<br /> Since &lt;math&gt;f(2^n+1)=f(2^n)=\tbinom n2&lt;/math&gt; we have that &lt;math&gt;f(2^n+2)=\tbinom n2 +1= n(n-1)/2 + 1&lt;/math&gt; with &lt;math&gt;n \ge 2&lt;/math&gt; are the only one-from-one values of f.</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45843 2012 AIME I Problems/Problem 14 2012-04-01T13:57:12Z <p>Numbertheorist17: /* Solution 1 */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and thus &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45842 2012 AIME I Problems/Problem 14 2012-04-01T13:57:01Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> ===Solution 2===<br /> <br /> Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and thus &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45841 2012 AIME I Problems/Problem 14 2012-04-01T13:56:37Z <p>Numbertheorist17: /* Solution 1 */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> ===Solution 2===<br /> By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45748 2012 AIME I Problems/Problem 14 2012-03-25T16:58:13Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45747 2012 AIME I Problems/Problem 14 2012-03-25T16:57:55Z <p>Numbertheorist17: /* Solution 2 */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> This is a more geometric solution. By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45746 2012 AIME I Problems/Problem 14 2012-03-25T16:57:45Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> This is a more geometric solution. By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45745 2012 AIME I Problems/Problem 14 2012-03-25T16:57:28Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and then &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> This is a more geometric solution. By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_14&diff=45703 2012 AIME I Problems/Problem 14 2012-03-23T11:58:26Z <p>Numbertheorist17: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Complex numbers &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are zeros of a polynomial &lt;math&gt;P(z) = z^3 + qz + r,&lt;/math&gt; and &lt;math&gt;|a|^2 + |b|^2 + |c|^2 = 250.&lt;/math&gt; The points corresponding to &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the complex plane are the vertices of a right triangle with hypotenuse &lt;math&gt;h.&lt;/math&gt; Find &lt;math&gt;h^2.&lt;/math&gt;<br /> <br /> == Solution ==<br /> Since &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are real, at least one of &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume &lt;math&gt;a&lt;/math&gt; is real and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;x + yi&lt;/math&gt; and &lt;math&gt;x - yi&lt;/math&gt; respectively. By symmetry, the triangle described by &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be isosceles and is thus an isosceles right triangle with hypotenuse &lt;math&gt;\overline{ab}.&lt;/math&gt; Now since &lt;math&gt;P(z)&lt;/math&gt; has no &lt;math&gt;z^2&lt;/math&gt; term, we must have &lt;math&gt;a+b+c = a + (x + yi) + (x - yi) = 0&lt;/math&gt; and thus &lt;math&gt;a = -2x.&lt;/math&gt; Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, &lt;math&gt;a-x=y&lt;/math&gt; and thus &lt;math&gt;y=-3x.&lt;/math&gt; We can then solve for &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |a|^2 + |b|^2 + |c|^2 &amp;= 250\\<br /> |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &amp;= 250\\<br /> 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &amp;= 250\\<br /> x^2 &amp;= \frac{250}{24}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;h&lt;/math&gt; is the distance between &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; so &lt;math&gt;h = 2y = -6x&lt;/math&gt; and then &lt;math&gt;h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}&lt;/math&gt;<br /> <br /> <br /> == Solution 2 ==<br /> This is a more geometric solution. By Vieta's formula, the sum of the roots is equal to 0, or &lt;math&gt;a+b+c=0&lt;/math&gt;. Therefore, &lt;math&gt;\frac{(a+b+c)}{3}=0&lt;/math&gt;. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be &lt;math&gt;x&lt;/math&gt; and the other leg be &lt;math&gt;y&lt;/math&gt;. Without the loss of generality, let &lt;math&gt;\overline{ac}&lt;/math&gt; be the hypotenuse. The magnitudes of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are just &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the medians because the origin, or the centroid in this case, cuts the median in a ratio of &lt;math&gt;2:1&lt;/math&gt;. So, &lt;math&gt;|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}&lt;/math&gt; because &lt;math&gt;|a|&lt;/math&gt; is two thirds of the median from &lt;math&gt;a&lt;/math&gt;. Similarly, &lt;math&gt;|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. The median from &lt;math&gt;b&lt;/math&gt; is just half the hypotenuse because the hypotenuse of any right triangle is just half the hypotenuse. So, &lt;math&gt;|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}&lt;/math&gt;. Hence, &lt;math&gt;|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250&lt;/math&gt;. Therefore, &lt;math&gt;h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=13|num-a=15}}</div> Numbertheorist17