HMMT 二月 2025 · 几何 · 第 4 题

HMMT February 2025 — Geometry — Problem 4

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 is inscribed in a semicircle S , which is inscribed in another semicircle S . The radii of 1 2 3 S and S are 1 and 10, respectively, and the diameters of S and S are parallel. The endpoints of 1 3 1 3 the diameter of S are A and B , and S ’s arc is tangent to AB at C . Compute AC · CB . 3 2 S 3 S 2 S 1 A B C

解析

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 is inscribed in a semicircle S , which is inscribed in another semicircle S . The radii of 1 2 3 S and S are 1 and 10, respectively, and the diameters of S and S are parallel. The endpoints of 1 3 1 3 the diameter of S are A and B , and S ’s arc is tangent to AB at C . Compute AC · CB . 3 2 S 3 S 2 S 1 A B C Proposed by: Karthik Venkata Vedula Answer: 20 Solution: V S 3 S 2 Q S 1 R A B P C Let P , Q , and R be the midpoints of the diameters (i.e., the center of the circular arcs) of S , S , and 3 2 S , respectively. Observe that if one fixes S , the location of S is uniquely determined by the angle 1 3 2 between the diameters of S and S . The same holds for S and S . Thus, the figures S ∪ S and 2 3 2 1 3 2 √ S ∪ S are similar. This gives us that the radius of S is 10. 2 1 2 To compute the answer, we define V to be either intersection of the arcs of S and S . By the 2 3 p √ √ 2 2 Pythagorean theorem, P Q = P V − V Q = 100 − 10 = 90. By the Pythagorean theorem again, p √ √ 2 2 2 P C = P Q − QC = 90 − 10 = 80. Thus AC · CB = (10 + P C )(10 − P C ) = 100 − P C = 20 .