Track / Stage 5 / 316 of 400 #916 of 2000

Problem 916

AIME late

Geometry Difficulty 5.8 Prove it

Example 15 In the acute-angled $\triangle ABC$ , the angle bisector of $\angle A$ intersects the circumcircle at another point $A_{1}; B_{1}, C_{1}$ are similarly defined. The line $A_{1} A$ intersects the external angle bisectors of $\angle B$ and $\angle C$ at point $A_{0}; B_{0}, C_{0}$ are similarly defined. Prove that the area of $\triangle A_{0} B_{0} C_{0}$ is twice the area of the hexagon $A C_{1} B A_{1} C B_{1}$ . (Refer to Example 7 in Chapter 11 of the second volume and Figure 2.11.5)

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

Proof: Let $I$ be the incenter of $\triangle ABC$ , then by Theorem $18(1)$ , $I$ is the orthocenter of $\triangle A_{0} B_{0} C_{0}$ . To prove $S_{\triangle A_{0} B_{0} C_{0}}=2 S_{A C_{1} B A_{1} C B_{1}}$ , it suffices to prove $S_{\triangle H_{0} B}=2 S_{\triangle I A_{1} B}$ , and it further suffices to prove that $A_{1}$ is the midpoint of $I A_{0}$ . Since the circle passing through $A, B, C$ is the nine-point circle of $\triangle A_{0} B_{0} C_{0}$ (it passes through the feet of the three altitudes), the intersection point $A_{1}$ of this circle with $I A_{0}$ is the midpoint of $I A_{0}$ , thus the proposition is proved.

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