Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National Olympiad Prove it JBMO

Problem:

$\omega_{1}$ and $\omega_{2}$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\omega_{3}$ at the points $K$ and $L$ , respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\omega_{1}$ and $\omega_{2}$ intersects $\omega_{3}$ . Show that if $\angle K A B = \angle L A B$ then the line segment $A B$ is a diameter of $\omega_{3}$ .

Solution

Solution:

Let $C$ be the intersection point of the tangent lines to the circles $\omega_{1}$ at $K$ and $\omega_{2}$ at $L$ . Point $C$ lies on the radical axis of circles $\omega_{1}$ and $\omega_{3}$ , and also on the radical axis of the circles $\omega_{2}$ and $\omega_{3}$ . Therefore $C$ lies on the radical axis of the circles $\omega_{1}$ and $\omega_{2}$ too. Therefore the points $A, B, C$ are collinear.

Since $\angle K A B = \angle L A B$ , the chords $K B$ and $B L$ have the same length. As we also have $C K = C L$ , the triangles $K B C$ and $L B C$ are congruent. In particular, $\angle K B A = \angle L B A$ . Therefore, $\angle B K A = 180^{\circ} - (\angle A B K + \angle B A K) = 180^{\circ} - (\angle L B K + \angle L A K) / 2 = 180^{\circ} - 90^{\circ} = 90^{\circ}$ , and $A B$ is a diameter.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.