Difference between revisions of "2021 CIME I Problems/Problem 14"
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Revision as of 21:54, 25 January 2021
Let 
be an acute triangle with orthocenter 
and circumcenter 
. The tangent to the circumcircle of 
at 
intersects lines 
and 
at 
and 
, and 
. Let line 
intersect 
at 
. Suppose that 
, and 
for positive integers 
where 
is not divisible by the square of any prime. Find 
.
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 \(S=\overline{AH}\cap\overline{EF}\).
![[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]](http://latex.artofproblemsolving.com/8/f/6/8f6e1771eea97b73d9356a91f6ec04628cd74c85.png)
Disregarding the condition \(\overline{BY}\parallel\overline{CX}\), we contend:
In general, \(BCXY\) is cyclic.
Proof. Recall that \(\overline{AA}\parallel\overline{EF}\), so the claim follows from Reim's 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![\[\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.\]](http://latex.artofproblemsolving.com/3/8/e/38e4dc72ddd4ca626d4cf8dd09e9ed71459c1c84.png)
But note that \(\triangle AEF\cup H\sim\triangle ABC\cup A'\), so![\[\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,\]](http://latex.artofproblemsolving.com/e/3/1/e313ab85e6f4fe5a2370573c641f480e9a74298b.png)
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 
.
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. 
