HMMT 二月 2011 · 冲刺赛 · 第 17 题

HMMT February 2011 — Guts Round — Problem 17

[ 10 ] Given positive real numbers x , y , and z that satisfy the following system of equations: 2 2 x + y + xy = 1 , 2 2 y + z + yz = 4 , 2 2 z + x + zx = 5 , find x + y + z .

解析

[ 10 ] Given positive real numbers x , y , and z that satisfy the following system of equations: 2 2 x + y + xy = 1 , 2 2 y + z + yz = 4 , 2 2 z + x + zx = 5 , find x + y + z . √ √ Answer: 5 + 2 3 Let O denote the origin. Construct vectors OA , OB , and OC as follows: The lengths of OA , OB , and ◦ OC are x , y , and z , respectively, and the angle between any two vectors is 120 . By the Law of Cosines, √ we have AB = 1, BC = 2, and AC = 5. Thus ABC is a right triangle, and O is its Fermat point. 1 The area of triangle ABC is equal to · 1 · 2 = 1. But the area is also equal to the sum of the areas of 2 √ √ √ 1 3 1 3 1 3 triangles AOB , BOC , and COA , which is equal to · xy · + · yz · + · zx · . We thus obtain 2 2 2 2 2 2 4 √ xy + yz + zx = . Adding the three equations given in the problem and subtracting both sides by 3 2 2 2 2 2 2 2 2 √ xy + yz + zx , we obtain x + y + z = 5 − . Therefore ( x + y + z ) = ( x + y + z )+2( xy + yz + zx ) = 3 √ 5 + 2 3.