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Japan Mathematical Olympiad Initial Round

Geometry Difficulty 5.9 AIME, harder Prove it Japan

Circle $C_1$ is internally tangent to circle $C_2$ at point $A$ . Let $O$ be the center of circle $C_2$ . A tangent line to circle $C_1$ at point $P$ on $C_1$ goes through point $O$ . Let $Q$ be the point of intersection of half-line $OP$ and circle $C_2$ , and let $R$ be the point of intersection of the line tangent to circle $C_1$ at point $A$ and the line $OP$ . Suppose that the radius of circle $C_2$ is $9$ and that $PQ = QR$ is satisfied. Here we denote the length of the line segment $XY$ also by $XY$ . Determine the value of $OP$ .

Solution

Since lines $RA$ , $RP$ are tangent to circle $C_1$ at $A$ , $P$ , respectively, we have $AR = PR$ . Also, if we let $S$ be the point of intersection, different from $Q$ , of line $OQ$ and circle $C_2$ , then, by the theorem on the power of a point with respect to a circle, we have $AR^2 = SR \cdot QR$ . From $AR = PR = 2QR$ it follows that $SR = \frac{(2QR)^2}{QR} = 4QR$ . Consequently, we obtain $SQ = SR - QR = 3QR$ . Since $SQ$ is the radius of circle $C_2$ , we have $3QR = SQ = 9 \cdot 2 = 18$ , and therefore, $QR = 6$ , from which we conclude that
$OP = OQ - PQ = OQ - QR = 3$
is the desired answer.

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