Library / /1257 of 1394

, 2016

Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:
Let $ABC$ be a triangle with incenter $I$ , incircle $\gamma$ and circumcircle $\Gamma$ . Let $M, N, P$ be the midpoints of sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ and let $E, F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$ , respectively. Let $U$ , $V$ be the intersections of line $EF$ with line $MN$ and line $MP$ , respectively, and let $X$ be the midpoint of arc $\widehat{BAC}$ of $\Gamma$ . Given that $AB=5$ , $AC=8$ , and $\angle A=60^\circ$ , compute the area of triangle $XUV$ .

Solution

Solution:
Let segments $AI$ and $EF$ meet at $K$ . Extending $AK$ to meet the circumcircle again at $Y$ , we see that $X$ and $Y$ are diametrically opposite, and it follows that $AX$ and $EF$ are parallel. Therefore the height from $X$ to $\overline{UV}$ is merely $AK$ . Observe that $AE=AF$ , so $\triangle AEF$ is equilateral; since $MN, MP$ are parallel to $AF, AE$ respectively, it follows that $\triangle MVU, \triangle UEN, \triangle FPV$ are equilateral as well. Then $MV=MP-PV=\frac{1}{2}AC-FP=\frac{1}{2}AC-AF+AP=\frac{1}{2}AC-AF+\frac{1}{2}AB=\frac{1}{2}BC$ , since $E, F$ are the tangency points of the incircle. Since $\triangle MVU$ is equilateral, we have $UV=MU=MV=\frac{1}{2}BC$ .
Now we can compute $BC=7$ , whence $UV=\frac{7}{2}$ and
$AK=\frac{AB+AC-BC}{2} \cdot \cos 30^\circ=\frac{3 \sqrt{3}}{2}$
Hence, the answer is $\frac{21 \sqrt{3}}{8}$ .

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