Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Ukraine

Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$ . Tangents to this circle at points $A$ and $B$ meet at point $P$ , and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$ . Prove, that $PA = PK$ .

(Danylo Khilko)

Solution

First note, that as $\angle ABC > 90^\circ$ , point $K$ lies on $AC$ (fig. 1). Also, it's clear that $PA = PB$ . We will show that $K$ lies on the circle $\omega$ with a center $P$ and radius $PA$ . It's enough to prove, that $\angle APB = 2(180^\circ - \angle AKB)$ . Indeed, take on the larger arc of circle $\omega$ any point $X$ . Then $\angle AXB = \frac{1}{2}\angle APB$ . If this condition holds, then $\angle AKB + \angle AXB = 180^\circ$ , so points $A, K, B$ and $X$ lie on a circle $\omega$ . As $P$ is the center of this circle, $PK = PA$ .

So, let's prove that $\angle APB = 2(180^\circ - \angle AKB)$ . As $PA = PB$ , and also using the theorem about the angle between the chord and a tangent, we get the following: $\angle APB = 180^\circ - 2\angle ABP = 180^\circ - 2\angle ACB = 180^\circ - 2(90^\circ - \angle CKB) = 2\angle CKB = 2(180^\circ - \angle AKB)$ .

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