HMMT February

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:

Let $ABC$ be an equilateral triangle with side length $8$ . Let $X$ be on side $AB$ so that $AX = 5$ and $Y$ be on side $AC$ so that $AY = 3$ . Let $Z$ be on side $BC$ so that $AZ$ , $BY$ , $CX$ are concurrent. Let $ZX$ , $ZY$ intersect the circumcircle of $AXY$ again at $P$ , $Q$ respectively. Let $XQ$ and $YP$ intersect at $K$ . Compute $KX \cdot KQ$ .

Solution

Solution:

Let $BY$ and $CX$ meet at $O$ . $O$ is on the circumcircle of $AXY$ , since $\triangle AXC \cong \triangle CYB$ .

We claim that $KA$ and $KO$ are tangent to the circumcircle of $AXY$ . Let $XY$ and $BC$ meet at $L$ . Then, $LBZC$ is harmonic. A perspectivity at $X$ gives $AYOP$ is harmonic. Similarly, a perspectivity at $Y$ gives $AXOQ$ is harmonic. Thus, $K$ is the pole of chord $AO$ .

Now we compute. Denote $r$ as the radius and $\theta$ as $\angle AXO$ . Then,

$\begin{gathered} r = \frac{XY}{\sqrt{3}} = \frac{\sqrt{5^2 + 3^2 - 3 \cdot 5}}{\sqrt{3}} = \sqrt{\frac{19}{3}} \\ \sin \theta = \sin 60^\circ \cdot \frac{AC}{XC} = \frac{\sqrt{3}}{2} \cdot \frac{8}{\sqrt{5^2 + 8^2 - 5 \cdot 8}} = \frac{4}{7} \sqrt{3} \\ KX \cdot KQ = KA^2 = (r \cdot \tan \theta)^2 = \left(\sqrt{\frac{19}{3}} \cdot 4 \sqrt{3}\right)^2 = 304. \end{gathered}$

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.