HMMT February

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:
Let $ABC$ be a triangle with $AB = 13$ , $AC = 14$ , and $BC = 15$ . Let $G$ be the point on $AC$ such that the reflection of $BG$ over the angle bisector of $\angle B$ passes through the midpoint of $AC$ . Let $Y$ be the midpoint of $GC$ and $X$ be a point on segment $AG$ such that $\frac{AX}{XG} = 3$ . Construct $F$ and $H$ on $AB$ and $BC$ , respectively, such that $FX \parallel BG \parallel HY$ . If $AH$ and $CF$ concur at $Z$ and $W$ is on $AC$ such that $WZ \parallel BG$ , find $WZ$ .

Solution

Solution:
Observe that $BG$ is the $B$ -symmedian, and thus $\frac{AG}{GC} = \frac{c^2}{a^2}$ . Stewart's theorem gives us
$BG = \sqrt{\frac{2a^2c^2b}{b(a^2 + c^2)} - \frac{a^2b^2c^2}{a^2 + c^2}} = \frac{ac}{a^2 + c^2} \sqrt{2(a^2 + c^2) - b^2} = \frac{390\sqrt{37}}{197}$
Then by similar triangles,
$ZW = HY \frac{ZA}{HA} = BG \frac{YC}{GC} \frac{ZA}{HA} = BG \frac{1}{2} \frac{6}{7} = \frac{1170\sqrt{37}}{1379}$
where $\frac{ZA}{HA}$ is found with mass points or Ceva.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.