HMMT 二月 2008 · 几何 · 第 3 题

HMMT February 2008 — Geometry — Problem 3

[ 4 ] Let ABC be a triangle with ∠ BAC = 90 . A circle is tangent to the sides AB and AC at X and Y respectively, such that the points on the circle diametrically opposite X and Y both lie on the side BC . Given that AB = 6, find the area of the portion of the circle that lies outside the triangle. C Y A X B

解析

[ 4 ] Let ABC be a triangle with ∠ BAC = 90 . A circle is tangent to the sides AB and AC at X and Y respectively, such that the points on the circle diametrically opposite X and Y both lie on the side BC . Given that AB = 6, find the area of the portion of the circle that lies outside the triangle. C Y A X B ′ ′ Answer: π − 2 Let O be the center of the circle, and r its radius, and let X and Y be the points ′ ′ ′ ′ ◦ diametrically opposite X and Y , respectively. We have OX = OY = r , and ∠ X OY = 90 . Since ′ ′ ′′ ′ triangles X OY and BAC are similar, we see that AB = AC . Let X be the projection of Y onto ′′ ′ ′′ ′ ′′ AB . Since X BY is similar to ABC , and X Y = r , we have X B = r . It follows that AB = 3 r , so r = 2. C ′ X ′ O Y Y ′′ A X B X 1 ′ ′ Then, the desired area is the area of the quarter circle minus that of the triangle X OY . And the 1 1 2 2 answer is πr − r = π − 2. 4 2