HMMT 二月 2006 · 冲刺赛 · 第 24 题

HMMT February 2006 — Guts Round — Problem 24

[9] Two 18-24-30 triangles in the plane share the same circumcircle as well as the same incircle. What’s the area of the region common to both the triangles? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

解析

Two 18-24-30 triangles in the plane share the same circumcircle as well as the same incircle. What’s the area of the region common to both the triangles? Answer: 132 Solution: Notice, first of all, that 18-24-30 is 6 times 3-4-5, so the triangles are right. Thus, the midpoint of the hypotenuse of each is the center of their common 1 circumcircle, and the inradius is (18 + 24 − 30) = 6. Let one of the triangles be ABC , 2 7 ◦ where ∠ A < ∠ B < ∠ C = 90 . Now the line joining the midpoints of sides AB and AC is tangent to the incircle, because it is the right distance (12) from line BC . So, the hypotenuse of the other triangle lies along . We may formulate this thus: The hypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicular to the longer leg, of the other. Now it is not hard to see, as a result of these parallel- and perpendicularisms, that the other triangle “cuts off” at each vertex of 4 ABC a smaller, similar right triangle. If we compute the dimensions of these smaller triangles, we find that they are as follows: 9-12-15 at A , 6-8-10 at B , and 3-4-5 at C . The total area chopped off of 4 ABC is thus 9 · 12 6 · 8 3 · 4

- = 54 + 24 + 6 = 84 . 2 2 2 The area of 4 ABC is 18 · 24 / 2 = 216. The area of the region common to both the original triangles is thus 216 − 84 = 132 .