Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Mongolia

Let $ABC$ be an acute triangle with altitudes $AD$ , $BE$ , and $CF$ . Let $\omega$ be the circle with diameter $BC$ , and suppose it intersects the segment $AD$ at point $K$ inside triangle $ABC$ . On ray $KD$ , let $L$ be a point such that $KA = KL$ . Let lines $BL$ and $CL$ intersect the circle $\omega$ again at points $P$ and $Q$ , respectively. Prove that lines $AD$ , $QE$ , and $PF$ are concurrent.

(Bilegdemberel Bat-Amgalan)

Solution

Since $\angle BEC = \angle BFC = 90^\circ$ , points $E$ and $F$ lie on circle $\omega$ .

We observe that:
$\angle LAF = \angle DAB = \angle FCB = \angle FPL,$
so quadrilateral $AFLP$ is cyclic; denote its circumcircle by $\omega_1$ . Similarly, we get that $AQLE$ is cyclic; denote its circumcircle by $\omega_2$ . Now consider the radical axes of these three circles:

- The radical axis of $\omega$ and $\omega_1$ is line $PF$ .
- The radical axis of $\omega$ and $\omega_2$ is line $EQ$ .
- The radical axis of $\omega_1$ and $\omega_2$ is line $AL$ .

By the Radical Axis Theorem, these three radical axes are concurrent. Therefore, the lines $AD$ , $QE$ , and $PF$ meet at a single point.

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