PUMaC 2013 · 几何（A 组） · 第 1 题

PUMaC 2013 — Geometry (Division A) — Problem 1

[ 3 ] Let O be a point with three other points A, B, C and ∠ AOB = ∠ BOC = ∠ AOC = 2 π/ 3. Consider the average area of the set of triangles ABC where OA, OB, OC ∈ { 3 , 4 , 5 } . The √ average area can be written in the form m n where m, n are integers and n is not divisible by a perfect square greater than 1. Find m + n .

解析

[ 3 ] Let O be a point with three other points A, B, C and \ AOB = \ BOC = \ AOC = 2 ⇡ / 3. Consider the average area of the set of triangles ABC where OA, OB, OC 2 { 3 , 4 , 5 } . The p average area can be written in the form m n where m, n are integers and n is not divisible by a perfect square greater than 1. Find m + n . Solution By symmetry, it su ces to solve for the average area of 4 OAB and multiply by 3. 1 | 4 OAB | = | OA || OB | sin(2 ⇡ / 3), so summing over all the possible combinations, we have the 2 p p 1 total area of OAB as (3 + 4 + 5)(3 + 4 + 5)( 3 / 2) = 36 3. Thus the average area of OAB 2 p p is 4 3. Therefore the average area of 4 ABC is 12 3, and the answer is 15.