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Philippine Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Philippines

Problem:
Two circles of radius $12$ have their centers on each other. As shown in the figure, $A$ is the center of the left circle, and $AB$ is a diameter of the right circle. A smaller circle is constructed tangent to $AB$ and the two given circles, internally to the right circle and externally to the left circle, as shown. Find the radius of the smaller circle.

Solution

Solution:

Let $R$ be the common radius of the larger circles, and $r$ that of the small circle. Let $C$ and $D$ be the centers of the right large circle and the small circle, respectively. Let $E$ , $F$ and $G$ be the points of tangency of the small circle with $AB$ , the left large circle, and the right large circle, respectively. Since the centers of tangent circles are collinear with the point of tangency, then $A$ - $F$ - $D$ and $C$ - $D$ - $G$ are collinear.

From $\triangle AED$ , $AE^2 = (R + r)^2 - r^2 = R^2 + 2Rr$ . Therefore, $CE = AE - R = \sqrt{R^2 + 2Rr} - R$ .

From $\triangle CED$ , $CE^2 = (R - r)^2 - r^2 = R^2 - 2Rr$ .

Therefore, $\sqrt{R^2 + 2Rr} - R = \sqrt{R^2 - 2Rr}$ . Solving this for $r$ yields $r = \frac{\sqrt{3}}{4} R$ . With $R = 12$ , we get $r = 3\sqrt{3}$ .

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