Library / Mongolian National Mathematical Olympiad /2 of 31

Mongolian National Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Mongolia

Let $M$ be the centroid and $O$ be the circumcenter of a triangle $ABC$ . Take a point $K$ on the line $OM$ in such a way that $\angle KAB = \angle ABC$ and the points $K$ , $C$ are on the same side of the line $AB$ . Prove that $\angle KCB = \angle ABC$ .

Solution

Let $P$ , $Q$ be the midpoints of the segments $AB$ , $BC$ respectively. Let $F := CK \cap AB$ and let $E := AK \cap BC$ . Since $\angle BAK = \angle ABC$ , the triangle $AEB$ is an isosceles triangle. Hence $EP$ is an altitude of the triangle $\triangle AEB$ . Therefore $O \in EP$ .

By assumption, we have $AQ \cap PC = M$ and $CF \cap AE = K$ . Let $O' := FQ \cap PE$ . The Pappus' theorem for AFPQEC yields that the points $M$ , $K$ and $O'$ are collinear.

On the other hand, we have $MK \cap EP = O$ . Therefore $O = O'$ . Thus $F \in OQ$ and $FB = FC$ . Hence $\angle FBC = \angle BCF$ .

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.