HMMT 二月 1998 · CALC 赛 · 第 10 题

英文解析

10. Problem: Let S be the locus of all points ( x, y ) in the first quadrant such that + = 1 for somet 1 − tt with 0 < t < 1. Find the area of S .
    Solution: Solving for t in the given equation, we get t + ( y − x − 1) t + x = 0. Using the quadratic 2
    √
    ( x +1 − y ) ± ( y − x − 1) − 4 x 2
    equation, we get t = . For all valid combinations of ( x, y ), t is positive and less than 1
    (this is easy to see by inspection). All valid combinations of ( x, y ) are those that make ( y − x − 1) − 4 x ≥ 0.22
    √
    2 2 2
    Solving for y in the equation ( y − x − 1) − 4 x = 0 yields y − (2 x +2) y +( x − 1) ≥ 0 ⇒ y = ( x +1) ± 2 x .
    In the original equation, it is given that + = 1, and 0 < t < 1. This implies that x, y < 1. Then thexyt 1 − t
    √
    only possible y < 1 that satisfies ( y − x − 1) − 4 x = 0 is y = x + 1 − 2 x .2
    √
    Then to satisfy the inequality ( y − x − 1) − 4 x ≥ 0, we must have y ≤ x + 1 − 2 x . Recall that this is 2
    √
    when 0 < y < 1. Hence we integrate in the interval [0 , 1]: ∈ x + 1 − 2 x = .11
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