PMO 2

Geometry Level 3

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 the 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.

5\sqrt{2}

3\sqrt{3}

4\sqrt{2}

3

6\sqrt{3}

1 solution

Hana Wehbi
Nov 11, 2017

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 $E, F \text { and } G$ 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 $r \implies r=\frac{\sqrt{3}R}{4}$ . With $R=12$ , we get $r=\boxed{3\sqrt{3}}$ .

PMO 2

1 solution

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