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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...")
 
 
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==Solution by TheUltimate123==
 
==Solution by TheUltimate123==
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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 \(S=\overline{AH}\cap\overline{EF}\).
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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;
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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>
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Disregarding the condition \(\overline{BY}\parallel\overline{CX}\), we contend:
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Disregarding the condition <math>\overline{BY}\parallel\overline{CX}</math>, we contend:
  
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<math>\textbf{Claim}:</math> In general, \(BCXY\) is cyclic.
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<math>\textbf{Claim}:</math> In general, <math>BCXY</math> is cyclic.
  
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Proof. Recall that \(\overline{AA}\parallel\overline{EF}\), so the claim follows from Reim's theorem on \(BCEF\), \(BCXY\). \(\blacksquare\)
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<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>
  
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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>.
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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 14:38, 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
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