HMMT 二月 2011 · 冲刺赛 · 第 17 题

英文解析

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 .
    √
    √
    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.
    The area of triangle ABC is equal to · 1 · 2 = 1. But the area is also equal to the sum of the areas of 1
    √ √ √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
    2 2 2 2 2 2 223
    √
    xy + yz + zx , we obtain x + y + z = 5 − . Therefore ( x + y + z ) = ( x + y + z )+2( xy + yz + zx ) =
    √3
    5 + 2 3.