Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 CIME I Problems/Problem 14"

(Created page with "Let <math>ABC</math> be an acute triangle with orthocenter <math>H</math> and circumcenter <math>O</math>. The tangent to the circumcircle of <math>\triangle ABC</math> at <ma...")
 
 
Line 2: Line 2:
  
 
==Solution by TheUltimate123==
 
==Solution by TheUltimate123==
−
Let \(H\) be the orthocenter of \(\triangle ABC\), and let \(E\), \(F\) be the feet of the altitudes from \(B\), \(C\). Also let \(A'\) be the antipode of \(A\) on the circumcircle and let \(S=\overline{AH}\cap\overline{EF}\).
+
Let <math>H</math> be the orthocenter of <math>\triangle ABC</math>, and let <math>E</math>, <math>F</math> be the feet of the altitudes from <math>B, C</math>. Also let <math>A'</math> be the antipode of <math>A</math> on the circumcircle and let <math>S=\overline{AH}\cap\overline{EF}</math>, as shown below:
 
<asy>
 
<asy>
 
size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
 
size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
Line 11: Line 11:
  
 
dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S);</asy>
 
dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S);</asy>
−
Disregarding the condition \(\overline{BY}\parallel\overline{CX}\), we contend:
+
Disregarding the condition <math>\overline{BY}\parallel\overline{CX}</math>, we contend:
  
−
<math>\textbf{Claim}:</math> In general, \(BCXY\) is cyclic.
+
<math>\textbf{Claim}:</math> In general, <math>BCXY</math> is cyclic.
  
−
Proof. Recall that \(\overline{AA}\parallel\overline{EF}\), so the claim follows from Reim's theorem on \(BCEF\), \(BCXY\). \(\blacksquare\)
+
<math>\textbf{Proof}.</math> Recall that <math>\overline{AA}\parallel\overline{EF}</math>, so the claim follows from Reims' theorem on <math>BCEF, BCXY. \blacksquare</math>
  
−
With \(\overline{BY}\parallel\overline{CX}\), it follows that \(BCXY\) is an isosceles trapezoid. In particular, \(HB=HY\) and \(HC=HX\). Since \(\overline{SF}\parallel\overline{AY}\), we have<cmath>\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.</cmath>But note that \(\triangle AEF\cup H\sim\triangle ABC\cup A'\), so<cmath>\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,</cmath>i.e.\ \(AD=2R(1-\cos A)\). We are given \(R=25\), and by the law of sines, \(\sin A=\frac{49}{50}\), so \(\cos A=\frac{3\sqrt{11}}{50}\), and \(AD=50-3\sqrt{11}\), so <math>50+3+11=\boxed{064}</math>.
+
With <math>\overline{BY}\parallel\overline{CX}</math>, it follows that <math>BCXY</math> is an isosceles trapezoid. In particular, <math>HB=HY</math> and <math>HC=HX</math>. Since <math>\overline{SF}\parallel\overline{AY}</math>, we have <cmath>\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.</cmath> But note that <math>\triangle AEF\cup H\sim\triangle ABC\cup A'</math>, so <cmath>\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,</cmath> i.e.\ <math>AD=2R(1-\cos A)</math>. We are given <math>R=25</math>, and by the law of sines, <math>\sin A=\frac{49}{50}</math>, so <math>\cos A=\frac{3\sqrt{11}}{50}</math>, and <math>AD=50-3\sqrt{11}</math>, so <math>50+3+11=\boxed{064}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 13:38, 26 January 2021

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$, and $BY\parallel CX$. Let line $AO$ intersect $\overline{BC}$ at $D$. Suppose that $AO=25, BC=49$, and $AD=a-b\sqrt{c}$ for positive integers $a, b, c,$ where $c$ is not divisible by the square of any prime. Find $a+b+c$.

Solution by TheUltimate123

Let $H$ be the orthocenter of $\triangle ABC$, and let $E$, $F$ be the feet of the altitudes from $B, C$. Also let $A'$ be the antipode of $A$ on the circumcircle and let $S=\overline{AH}\cap\overline{EF}$, as shown below: [asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,X,Y,EE,F,Ap,SS,D; O=(0,0); A=dir(147.55); B=dir(195); C=dir(345); H=A+B+C; X=extension(B,H,A,A+rotate(90)*A); Y=extension(C,H,A,X); EE=foot(B,C,A); F=foot(C,A,B); Ap=-A; SS=extension(A,H,EE,F); D=extension(A,O,B,C); draw(A--Ap,qua+Dotted); draw(B--Ap--C,qua); draw(B--X,tri); draw(C--Y,tri); draw(EE--F,sec+Dotted); draw(X--Y,sec); draw(B--Y,sec); draw(C--X,sec); filldraw(circumcircle(B,C,X),sfil,sec); filldraw(A--B--C--cycle,fil,pri); filldraw(circle(O,1),fil,pri); dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S);[/asy] Disregarding the condition $\overline{BY}\parallel\overline{CX}$, we contend:

$\textbf{Claim}:$ In general, $BCXY$ is cyclic.

$\textbf{Proof}.$ Recall that $\overline{AA}\parallel\overline{EF}$, so the claim follows from Reims' theorem on $BCEF, BCXY. \blacksquare$

With $\overline{BY}\parallel\overline{CX}$, it follows that $BCXY$ is an isosceles trapezoid. In particular, $HB=HY$ and $HC=HX$. Since $\overline{SF}\parallel\overline{AY}$, we have \[\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.\] But note that $\triangle AEF\cup H\sim\triangle ABC\cup A'$, so \[\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,\] i.e.\ $AD=2R(1-\cos A)$. We are given $R=25$, and by the law of sines, $\sin A=\frac{49}{50}$, so $\cos A=\frac{3\sqrt{11}}{50}$, and $AD=50-3\sqrt{11}$, so $50+3+11=\boxed{064}$.

See also

2021 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_CIME_I_Problems/Problem_14&oldid=143327"