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Problem 1242

AIME late

Geometry Difficulty 5.3 Prove it Berkeley Math Circle Monthly Contest 1 · United States

Circles $j$ and $k$ , centered at $O$ and $P$ respectively, do not intersect. The two tangent rays from $O$ to $k$ meet $j$ at $A$ and $B$ , respectively, and the two tangent rays from $P$ to $j$ meet $k$ at $C$ and $D$ , respectively. Prove that $A, B, C$ , and $D$ are the vertices of a rectangle.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Without loss of generality, we may assume that $A, B, C$ , and $D$ have the relative positions shown. We label the points of tangency $W, X, Y$ , and $Z$ . We also note that the entire construction is symmetric about the line of centers $OP$ , which therefore perpendicularly bisects segments $AB$ , $CD$ , $WX$ , and $YZ$ at their respective midpoints $K, L, M$ , and $N$ . Let $r_1$ and $r_2$ be the respective radii of $j$ and $k$ . We will first prove that $AK = CL$ . Since $\triangle OAK \sim \triangle OYN \sim \triangle OPY$ , we have
$\frac{AK}{OA} = \frac{PY}{OP}, \quad \text{so } AK = \frac{OA \cdot PY}{OP} = \frac{r_1 r_2}{OP}$
Symmetrically, $CL = r_1 r_2 / OP$ so $AK = CL$ . Now quadrilateral $AKLC$ has right angles at $K$ and $L$ and equal, parallel sides $AK = CL$ , so it is a rectangle. Symmetrically, $BKLD$ is a rectangle so $ABDC$ is a rectangle.

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