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HMMT 二月 2015 · 几何 · 第 8 题

HMMT February 2015 — Geometry — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Let S be the set of discs D contained completely in the set { ( x, y ) : y < 0 } (the region below thex -axis) and centered (at some point) on the curve y = x − . What is the area of the union of the 23
    elements of S ?4
解析

英文解析

  1. Let S be the set of discs D contained completely in the set { ( x, y ) : y < 0 } (the region below the
    2 3
    x -axis) and centered (at some point) on the curve y = x − . What is the area of the union of theelements of S ?4
    √
    2 π 3
    Answer: + Solution 1. An arbitrary point ( x , y ) is contained in S if and only if there
    0 0
    3 4
    2 3 2 2 2
    exists some ( x, y ) on the curve ( x, x − ) such that ( x − x ) + ( y − y ) < y , since the radius of the
    0 0
    2 2 34
    circle is at most the distance from ( x, y ) to the x -axis. Some manipulation yields x − 2 y ( x − ) −
    2 240
    2 xx + x + y < 0.
    0 00
    Observe that ( x , y ) ∈ S if and only if the optimal choice for x that minimizes the expression satisfies
    0 0
    0 xthe inequality. The minimum is achieved for x = . After substituting and simplifying, we obtain
    1 − 2 y − x 20
    2 2 3 2 2 3
    y ( + x + y + y ) < 0. Since y < 0 and 1 − 2 y > 0, we find that we need − 2 x − 2 y + − 2 y >0
    0 0 0 0 0
    0 0 0 0
    1 − 2 y 2 2
    0 ⇐⇒ 1 > x + ( y + ) .210
    0 0
    Geometry 2
    S is therefore the intersection of the lower half-plane and a circle centered at (0 , − ) of radius 1. This 1
    is a circle of sector angle 4 π/ 3 and an isosceles triangle with vertex angle 2 π/ 3. The sum of these areas 2
    √
    2 π 3
    is + .
    3 4
    Solution 2. Let O = (0 , − ) and ℓ = { y = − 1 } be the focus and directrix of the given parabola. Let 1
    ′ ′2
    ℓ denote the x -axis. Note that a point P is in S iff there exists a point P on the parabola in the lower
    ′ ′ ′
    half-plane for which d ( P, P ) < d ( P, ℓ ). However, for all such P , d ( P, ℓ ) = 1 − d ( P, ℓ ) = 1 − d ( P, O ),
    ′ ′
    which means that P is in S iff there exists a P on the parabola for which d ( P , P ) + d ( P, O ) < 1. Itis not hard to see that this is precisely the intersection of the unit circle centered at O and the lowerhalf-plane, so now we can proceed as in Solution 1.