Competition problem — Geometry — Olympiad Maths Prep

Library / /11 of 45

Geometry Difficulty 5.4 AIME, harder Prove it Ukraine

Point $I$ is the incenter of triangle $ABC$ , where $AB < AC$ . On the external angle bisector of angle $ABC$ , a point $X$ is chosen such that $IC = IX$ . Let the tangent to the circumcircle of triangle $BXC$ at point $X$ intersect line $AB$ at point $Y$ . Prove that $AC = AY$ .

Solution

From the fact that $XY$ is tangent to the circumcircle of $\triangle BXC$ , we have $\angle BXY = \angle BCX$ . Combining this with the fact that $\angle YBX = \angle XBC$ , we have the similarity $\triangle BXY \sim \triangle BCX$ . Therefore, we have $\frac{BX}{BC} = \frac{BY}{BX}$ , so $BY = \frac{BX^2}{BC}$ . Notice that $\angle IBX = 90^\circ$ , so (Fig. 9)

$BY = \frac{BX^2}{BC} = \frac{IX^2 - BI^2}{BC} = \frac{CI^2 - BI^2}{BC}.$

Let $K$ be the point of tangency of the incircle of $\triangle ABC$ with side $BC$ . Then
$BY = \frac{CI^2 - BI^2}{BC} = \frac{(IK^2 + CK^2) - (IK^2 + BK^2)}{BC} = \frac{CK^2 - BK^2}{BC} = CK - BK.$

$AY = AB + BY = AB + CK - BK = c + (p - c) - (p - b) = b = AC$ ,

as desired.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.