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Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H.$ Let $E$ be the intersection of $BH$ and $AC$ and let $M$ and $N$ be the midpoints of $HB$ and $HO,$ respectively. Let $I$ be the incenter of $AEM$ and $J$ be the interseciton of $ME$ and $AI.$ If $AO=20,$ $AN=17,$ and $\angle{ANM}=90^{\circ},$ then $\frac{AI}{AJ}=\frac{m}{n}$ for relatively prime postive integers $m$ and $N$. Compute $100m+n.$

 

So for this problem I constructed the nine point circle centered at $N$ and I also got $AM=\sqrt{389}.$ But from here I don't really know how to solve it, can anyone give me hints or solutions? Thanks. 

 Jul 3, 2021
edited by hellothere  Jul 3, 2021
 #2
avatar+320 
+3

Hey there, HT!

 

So...

 

Assume O to be the centre of triangle ABC, OT cross BC at M, link XM and YM. Let P be the middle point of BT and Q be the middle point of CT, so we have \(MT=3\sqrt{15}\). Since \( \angle A=\angle CBT=\angle BCT\) we have \( \cos A=\frac{11}{16}\). Notice that \(\angle XTY=180^{\circ}-A\), so \(\cos XYT=-\cos A,\) and this gives us \( 1143-2XY^2=\frac{-11}{8}XT\cdot YT.\) Since TM is perpendicular to BC, BXTM,  and CYTM cocycle (respectively), so \( \theta_1=\angle ABC=\angle MTX\) and \( \theta_2=\angle ACB=\angle YTM\). So \( \angle XPM=2\theta_1\), so \(\frac{\frac{XM}{2}}{XP}=\sin \theta_1\), which yields \(XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.\) So same we have YM=XT. Apply Ptolemy theorem in BXTM we have \( 16TY=11TX+3\sqrt{15}BX\), and use the Pythagoras theorem we have \( BX^2+XT^2=16^2\). Same in YTMC and triangle CYT we have \( 16TX=11TY+3\sqrt{15}CY\) and \( CY^2+YT^2=16^2\). Solve this for XT and TY and submit it into the equation about \( \cos XYT\), we can obtain the result \( XY^2=\boxed{717}\).

 

Hope this helped! :)

( ゚д゚)つ Bye

 Jul 3, 2021
 #3
avatar+68 
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You are writing solution to a different problem

hellothere  Jul 3, 2021
 #4
avatar+320 
+1

Yikes, sorry, HT.

TaliaArticula  Jul 10, 2021

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