Competition problem — Geometry — Maths Olympiad Prep

Library / /2 of 24

Geometry Difficulty 5.4 AIME, harder Prove it Belarus

Let $\Omega$ be the circumcircle of a triangle $ABC$ . A circle passing through the vertex $A$ and touching $BC$ at point $X$ meets $\Omega$ at point $Y$ (different from $A$ ). Let point $Z$ (different from $Y$ ) be the intersection point of the ray $YX$ and $\Omega$ .
$\angle CAX = \angle ZAB.$

Solution

Let $\omega$ be the circle passing through $A$ and touching the side $BC$ at $X$ . Since the angle $YAX$ subtends the arc $YX$ , we have $\angle YAX = \frac{1}{2} \angle YX$ .

Further, $\angle YXC = \frac{1}{2} \angle YX$ (as the angle between the tangent $BC$ and the chord $XY$ ). So, $\angle YAX = \angle YXC$ . Let $W$ be the intersection point of the ray $AX$ and $\Omega$ . Then $\angle YAX = \angle YAW$ , so $\angle YAW = \angle YXC$ . We have $\angle BXZ = \angle YXC$ as the vertical angles, so $\angle BXZ = \angle YAW$ . Further, $\angle YAW = \angle YZW$ as inscribed angles of $\Omega$ subtending the same arc. Therefore, $\angle BXZ = \angle YZW$ , so $BC \parallel WZ$ . Hence $CW = ZB$ , and then $\angle CAW = \angle ZAB$ as inscribed angles of $\Omega$ subtending the equal arcs.

Want a route through all this instead of 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.