Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 5.7 AIME, harder Prove it Turkey

Let $H$ be the orthocenter of an acute triangle $ABC$ , and let $A_1, B_1, C_1$ be the feet of the altitudes belonging to the vertices $A, B, C$ , respectively. Let $K$ be a point on the smaller $AB_1$ arc of the circle with diameter $AB$ satisfying the condition $\angle HKB = \angle C_1KB$ . Let $M$ be the point of intersection of the line segment $AA_1$ and the circle with center $C$ and radius $CL$ where $KB \cap CC_1 = \{L\}$ . Let $P$ and $Q$ be the points of intersection of the line $CC_1$ and the circle with center $B$ and radius $BM$ . Show that $A, K, P, Q$ are concyclic.

Solution

Since $A, C_1, L, K$ and $A, C_1, H, B_1$ are concyclic, so is $L, K, B_1, H$ . Using these and the fact that $KL$ bisects $\angle C_1KH$ , we get $\angle C_1AL = \angle LB_1H$ and hence $\angle ALC = \angle LB_1C$ . Therefore the triangles $ALC$ and $LB_1C$ are similar. Using this similarity as well as the facts that $CM = CL$ and $A, B_1, A_1, B$ are concyclic, we conclude that $CM^2 = CB \cdot CA_1$ and $\angle BMC = 90^\circ$ . Now we use $BP = BM$ and the fact that $C, A_1, C_1, A$ are concyclic to deduce $BP^2 = BA \cdot BC_1$ and $\angle BPA = 90^\circ$ . Hence $P$ is on the circle with diameter $AB$ .

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.