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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
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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
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You are writing solution to a different problem

hellothere  Jul 3, 2021
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Yikes, sorry, HT.

TaliaArticula  Jul 10, 2021