Competition problem — Geometry — Maths Olympiad Prep

Library / /10 of 19

Geometry Difficulty 5.4 AIME, harder Prove it Soviet Union

Problem:

$A$ and $B$ lie on a circle. $P$ lies on the minor arc $AB$ . $Q$ and $R$ (distinct from $P$ ) also lie on the circle, so that $P$ and $Q$ are equidistant from $A$ , and $P$ and $R$ are equidistant from $B$ . Show that the intersection of $AR$ and $BQ$ is the reflection of $P$ in $AB$ .

Solution

Solution:

Let $AR$ and $BQ$ meet at $X$ . Since arcs $QA$ and $AP$ are equal, we have $\angle ABX = \angle ABP$ . Similarly, $\angle BAX = \angle BAP$ . Side $AB$ is common, so triangles $ABX$ and $ABP$ are congruent. Hence $X$ is the reflection of $P$ .

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