AUT ABooklet

Geometry Difficulty 5.0 AIME, harder Prove it Austria

Let $ABC$ be a triangle and $I$ its incenter. The circumcircle of $ACI$ intersects the line $BC$ a second time in the point $X$ and the circumcircle of $BCI$ intersects the line $AC$ a second time in the point $Y$ .
Prove that the segments $AY$ and $BX$ are of equal length.

Solution

We shall show that $AB = BX$ holds. Since $AB = AY$ then follows by the same argument, this completes the proof (see Figure 3).

Figure 3: Problem 10

In this solution, we use oriented angles between lines (modulo $180^\circ$ ) with the notation $\angle PQR$ . As usual the angles of the triangle $ABC$ are denoted by $\alpha = \angle BAC$ , $\beta = \angle CBA$ and $\gamma = \angle ACB$ .

The inscribed angle theorem gives
$\angle AXB = \angle AXC = \angle AIC = -\angle CIA = 180^\circ - \angle CIA = \angle IAC + \angle ACI = \frac{1}{2}(\alpha + \gamma).$
This immediately implies
$\angle BAX = -\angle AXB - \angle XBA = -\frac{1}{2}(\alpha + \gamma) - \beta = \frac{1}{2}(\alpha + \gamma).$
Therefore, the triangle ABX is indeed isosceles, and we are done.

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.