Library / /818 of 1394

, 2015

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:
Let $P$ be a (non-self-intersecting) polygon in the plane. Let $C_{1}, \ldots, C_{n}$ be circles in the plane whose interiors cover the interior of $P$ . For $1 \leq i \leq n$ , let $r_{i}$ be the radius of $C_{i}$ . Prove that there is a single circle of radius $r_{1}+\cdots+r_{n}$ whose interior covers the interior of $P$ .

Solution

Solution:
If $n=1$ , we are done. Suppose $n>1$ . Since $P$ is connected, there must be a point $x$ on the plane which lies in the interiors of two circles, say $C_{i}, C_{j}$ . Let $O_{i}, O_{j}$ , respectively, be the centers of $C_{i}, C_{j}$ . Since $O_{i} O_{j}<r_{i}+r_{j}$ , we can choose $O$ to be a point on segment $O_{i} O_{j}$ such that $O_{i} O \leq r_{j}$ and $O_{j} O \leq r_{i}$ . Replace the two circles $C_{i}$ and $C_{j}$ with the circle $C$ centered at $O$ of radius $r_{i}+r_{j}$ . Note that $C$ covers both $C_{i}$ and $C_{j}$ . Induct to finish.

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