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Geometry Difficulty 6.2 National Olympiad Prove it United States

Problem:
A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle.
Semicircle $S_{1}$ is inscribed in a semicircle $S_{2}$ , which is inscribed in another semicircle $S_{3}$ . The radii of $S_{1}$ and $S_{3}$ are $1$ and $10$ , respectively, and the diameters of $S_{1}$ and $S_{3}$ are parallel. The endpoints of the diameter of $S_{3}$ are $A$ and $B$ , and $S_{2}$ 's arc is tangent to $AB$ at $C$ . Compute $AC \cdot CB$ .

Solution

Solution:

Let $P$ , $Q$ , and $R$ be the midpoints of the diameters (i.e., the center of the circular arcs) of $S_{3}$ , $S_{2}$ , and $S_{1}$ , respectively. Observe that if one fixes $S_{3}$ , the location of $S_{2}$ is uniquely determined by the angle between the diameters of $S_{2}$ and $S_{3}$ . The same holds for $S_{2}$ and $S_{1}$ . Thus, the figures $S_{3} \cup S_{2}$ and $S_{2} \cup S_{1}$ are similar. This gives us that the radius of $S_{2}$ is $\sqrt{10}$ .

To compute the answer, we define $V$ to be either intersection of the arcs of $S_{2}$ and $S_{3}$ . By the Pythagorean theorem, $PQ = \sqrt{PV^{2} - VQ^{2}} = \sqrt{100 - 10} = \sqrt{90}$ . By the Pythagorean theorem again, $PC = \sqrt{PQ^{2} - QC^{2}} = \sqrt{90 - 10} = \sqrt{80}$ . Thus $AC \cdot CB = (10 + PC)(10 - PC) = 100 - PC^{2} = \left[20\right]$ .

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