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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.0 AIME, harder Prove it Soviet Union

Problem:
Given a rectangle $ABCD$ with $AC$ length $e$ and four circles centers $A$ , $B$ , $C$ , $D$ and radii $a$ , $b$ , $c$ , $d$ respectively, satisfying $a + c = b + d < e$ . Prove you can inscribe a circle inside the quadrilateral whose sides are the two outer common tangents to the circles center $A$ and $C$ , and the two outer common tangents to the circles center $B$ and $D$ .

Solution

Solution:
Let $O$ be the center of the rectangle. Let $r = (a + c)/2 = (b + d)/2$ . The required circle has center $O$ , radius $r$ . Let an outer common tangent touch the circle center $A$ at $W$ , and the circle center $C$ at $X$ . Let $P$ be the midpoint of $WX$ , then $OP$ is parallel to $AW$ and $CX$ and has length $r$ , hence the circle center $O$ touches $AW$ at $P$ . Similarly for the other common tangents.

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