Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it Austria

Let ABCABC be a triangle and II its incenter. The circumcircle of ACIACI intersects the line BCBC a second time in the point XX and the circumcircle of BCIBCI intersects the line ACAC a second time in the point YY.
Prove that the segments AYAY and BXBX are of equal length.

Solution

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

Figure 1
Figure 3: Problem 10

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

The inscribed angle theorem gives
∠AXB=∠AXC=∠AIC=−∠CIA=180∘−∠CIA=∠IAC+∠ACI=12(α+γ). \angle AXB = \angle AXC = \angle AIC = -\angle CIA = 180^\circ - \angle CIA = \angle IAC + \angle ACI = \frac{1}{2}(\alpha + \gamma).
This immediately implies
∠BAX=−∠AXB−∠XBA=−12(α+γ)−β=12(α+γ). \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.

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