Track / Stage 5 / 368 of 400 #968 of 2000

Problem 968

AIME late

Geometry Difficulty 5.9 Prove it

Example 13 Let $I$ be the incenter of $\triangle ABC$ , and the incircle of $\triangle ABC$ touches the sides $BC$ , $CA$ , and $AB$ at points $K$ , $L$ , and $M$ , respectively. The line through $B$ parallel to $MK$ intersects lines $LM$ and $LK$ at points $R$ and $S$ , respectively. Prove that $\angle RIS$ is acute.
(IMO - 39 problem)

Sister problem Let $I$ be the excenter of $\triangle ABC$ , and the excircle of $\triangle ABC$ touches side $AC$ at point $L$ , and the extensions of sides $BA$ and $BC$ at points $M$ and $K$ , respectively. The line through $B$ parallel to $MK$ intersects lines $LM$ and $LK$ at points $R$ and $S$ , respectively. Prove that $\angle RIS$ is acute.

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

Prove as shown in Figure $16-15$ , connect $B I, M I, K I$ .
In $\triangle B K S$ and $\triangle L M K$ , $\angle L K M = \angle K S B$ , $\angle B K S = \angle L M K$ .
Therefore, $\triangle B K S \sim \triangle L M K$ , hence $\frac{K S}{B K} = \frac{L K}{L M}$ .
Similarly, we can prove $\frac{B R}{B M} = \frac{L M}{L K}$ .
Thus, $B S \cdot B R = B K \cdot B M$ .
Clearly, $B I \perp M K$ , $B K = B M$ , so $B S \cdot B R = B M^2 < B I^2$ , which means $B S \cdot B R < B I^2$ . This indicates that there exists a point $P$ on $B I$ such that $B R \cdot B S = B P^2$ .

In $\triangle R P S$ , applying the inverse of the projection theorem, we know $\angle R P S = 90^{\circ}$ , thus point $I$ lies outside the circle with $R S$ as its diameter, hence $\angle R I S < 90^{\circ}$ .

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