PUMaC 2012 · 几何（B 组） · 第 4 题

PUMaC 2012 — Geometry (Division B) — Problem 4

[ 4 ] Three circles, ω , ω , and ω , are externally tangent to each other, with radii of 1 , 1 , and 1 2 3 2 respectively. Quadrilateral ABCD contains and is tangent to all three circles. Find the √ minimum possible area of ABCD . Your answer will be of the form a + b c where c is not divisible by any perfect square. Find a + b + c .

解析

[ 4 ] Three circles, ω , ω , and ω , are externally tangent to each other, with radii of 1 , 1 , and 1 2 3 2 respectively. Quadrilateral ABCD contains and is tangent to all three circles. Find the √ minimum possible area of ABCD . Your answer will be of the form a + b c where c is not divisible by any perfect square. Find a + b + c . Solution: In order to have the smallest quadrilateral, we want to make it tangent at as many points as possible. Consider the following quadrilateral. Note that 3 of the sides are fixed as 3 sides of a rectangle, so the area is proportional to the distance from the side tangential to the two small circles to the midpoint of the fourth side. This can be minimized by making the fourth side the fourth side of a rectangle, which is tangent to the large circle at its midpoint. To find the width of this rectangle, note that it is the sum of a radius of a small circle, a radius of the large circle, and the altitude of a isosceles triangle with side lengths of 2 , 3 , 3. Using the Pythagorean Theorem, we find the altitude to √ be 2 2. To calculate the whole area of the rectangle, we multiply its height, 4, by its width, √ √ 3 + 2 2, to get 12 + 8 2. So a + b + c = 12 + 8 + 2 = 22 .