HMMT 二月 2006 · 冲刺赛 · 第 24 题

英文解析

24. Two 18-24-30 triangles in the plane share the same circumcircle as well as the sameincircle. 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 areright. Thus, the midpoint of the hypotenuse of each is the center of their commoncircumcircle, and the inradius is (18 + 24 − 30) = 6. Let one of the triangles be ABC ,1
    °72
    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: Thehypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicularto the longer leg, of the other. Now it is not hard to see, as a result of these paralleland perpendicularisms, that the other triangle “cuts off” at each vertex of 4 ABC asmaller, 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 totalarea 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 theoriginal triangles is thus 216 − 84 = 132 .