Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it Philippines

Problem:
Two circles of radius 1212 have their centers on each other. As shown in the figure, AA is the center of the left circle, and ABAB is a diameter of the right circle. A smaller circle is constructed tangent to ABAB 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.

Figure 1

Solution

Solution:
Figure 2
Let RR be the common radius of the larger circles, and rr that of the small circle. Let CC and DD be the centers of the right large circle and the small circle, respectively. Let EE, FF and GG be the points of tangency of the small circle with ABAB, the left large circle, and the right large circle, respectively. Since the centers of tangent circles are collinear with the point of tangency, then AA-FF-DD and CC-DD-GG are collinear.

From △AED\triangle AED, AE2=(R+r)2−r2=R2+2RrAE^2 = (R + r)^2 - r^2 = R^2 + 2Rr. Therefore, CE=AE−R=R2+2Rr−RCE = AE - R = \sqrt{R^2 + 2Rr} - R.

From △CED\triangle CED, CE2=(R−r)2−r2=R2−2RrCE^2 = (R - r)^2 - r^2 = R^2 - 2Rr.

Therefore, R2+2Rr−R=R2−2Rr\sqrt{R^2 + 2Rr} - R = \sqrt{R^2 - 2Rr}. Solving this for rr yields r=34Rr = \frac{\sqrt{3}}{4} R. With R=12R = 12, we get r=33r = 3\sqrt{3}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.