Track / Stage 6 / 77 of 400 #1077 of 2000

Problem 1077

National olympiad, first round

Geometry Difficulty 6.1 Prove it

[ Four points lying on one circle]

The inscribed circle touches the sides $A B$ and $A C$ of triangle $A B C$ at points $M$ and $N$ . Let $P$ be the point of intersection of the line $M N$ and the bisector of angle $B$ (or its extension). Prove that:

a) $\angle B P C=90^{\circ}$ ;

b) $S_{\mathrm{ABP}}: S_{\mathrm{ABC}}=1: 2$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

a) It is sufficient to prove that if $P_{1}$ is a point on the bisector of angle $B$ (or its extension), from which the segment $B C$ is seen at an angle of $90^{\circ}$ , then $P_{1}$ lies on the line $M N$ . Points $P_{1}$ and $N$ lie on the circle with diameter $C O$ , where $O-$ is the intersection point of the bisectors, so $\angle\left(P_{1} N, N C\right)=\angle\left(P_{1} O, O C\right)=\left(180^{\circ}-\angle A\right) / 2=\angle(M N, N C)$ .

b) Since $\angle B P C=90^{\circ}$ , then $B P=B C \cos (B / 2)$ , so $S_{\mathrm{ABP}}: S_{\mathrm{ABC}}=(B P \sin (B / 2)):(B C \sin B)=1: 2$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.