HMMT 二月 2004 · 几何 · 第 8 题

HMMT February 2004 — Geometry — Problem 8

A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle.

解析

A triangle has side lengths 18, 24, and 30. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle. Solution: 3 There are many solutions to this problem, which is straightforward. The given triangle is a right 3-4-5 triangle, so the circumcenter is the midpoint of the hypotenuse. Co- ordinatizing for convenience, put the vertex at (0 , 0) and the other vertices at (0 , 18) and (24 , 0). Then the circumcenter is (12 , 9). The centroid is at one-third the sum of the three vertices, which is (8 , 6). Finally, since the area equals the inradius times half the perimeter, we can see that the inradius is (18 · 24 / 2) / ([18 + 24 + 30] / 2) = 6. So the incenter of the triangle is (6 , 6). So the small triangle has a base of length 2 and a height of 3, hence its area is 3.