PUMaC 2012 · 几何（A 组） · 第 2 题

PUMaC 2012 — Geometry (Division A) — Problem 2

[ 3 ] Two circles centered at O and P have radii of length 5 and 6 respectively. Circle O passes through point P . Let the intersection points of circles O and P be M and N . The area of triangle 4 M N P can be written in simplest form as a/b . Find a + b .

解析

[ 3 ] Two circles centered at O and P have radii of length 5 and 6 respectively. Circle O passes through point P . Let the intersection points of circles O and P be M and N . The area of triangle 4 M N P can be written in simplest form as a/b . Find a + b . Solution: Draw 4 M OP and 4 N OP . They are each triangles with side lengths of 5 , 5 , 6. Let V be the point of intersection of M N and OP . Note that M V and N V , the altitudes from OP to vertices M and N , are each equal to twice the area of each triangle divided by the length of side OP . To find the area of 4 M OP , simply take the altitude to M P and call the intersection at the base X . Since this triangle is isosceles, the altitude splits M P into two equal segments, so XP has a length of 3., so 4 OXP forms a 3-4-5 right triangle, and the altitude has a length of 4. As a result, M V has a length of 24 / 5, and so M V = 48 / 5. Using the Pythagorean Theorem, we have that the length of V P is 18 / 5, and the total area of 4 N OP is 432 / 25, so a + b = 432 + 25 = 457 .