Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

Let $A B C$ be a triangle with $A B=13, A C=14$ , and $B C=15$ . Let $G$ be the point on $A C$ such that the reflection of $B G$ over the angle bisector of $\angle B$ passes through the midpoint of $A C$ . Let $Y$ be the midpoint of $G C$ and $X$ be a point on segment $A G$ such that $\frac{A X}{X G}=3$ . Construct $F$ and $H$ on $A B$ and $B C$ , respectively, such that $F X\|B G\| H Y$ . If $A H$ and $C F$ concur at $Z$ and $W$ is on $A C$ such that $W Z \| B G$ , find $W Z$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Observe that $B G$ is the $B$ -symmedian, and thus $\frac{A G}{G C}=\frac{c^{2}}{a^{2}}$ . Stewart's theorem gives us $B G=\sqrt{\frac{2 a^{2} c^{2} b}{b\left(a^{2}+c^{2}\right)}-\frac{a^{2} b^{2} c^{2}}{a^{2}+c^{2}}}=\frac{a c}{a^{2}+c^{2}} \sqrt{2\left(a^{2}+c^{2}\right)-b^{2}}=\frac{390 \sqrt{37}}{197}$ Then by similar triangles, $Z W=H Y \frac{Z A}{H A}=B G \frac{Y C}{G C} \frac{Z A}{H A}=B G \frac{1}{2} \frac{6}{7}=\frac{1170 \sqrt{37}}{1379}$ where $\frac{Z A}{H A}$ is found with mass points or Ceva.

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