HMMT 二月 2015 · 几何 · 第 8 题
HMMT February 2015 — Geometry — Problem 8
题目详情
- Let S be the set of discs D contained completely in the set { ( x, y ) : y < 0 } (the region below the 3 2 x -axis) and centered (at some point) on the curve y = x − . What is the area of the union of the 4 elements of S ?
解析
- 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 the 4 elements of S ? √ 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 4 2 2 3 circle is at most the distance from ( x, y ) to the x -axis. Some manipulation yields x − 2 y ( x − ) − 0 4 2 2 2 xx + x + y < 0. 0 0 0 Observe that ( x , y ) ∈ S if and only if the optimal choice for x that minimizes the expression satisfies 0 0 x 0 the inequality. The minimum is achieved for x = . After substituting and simplifying, we obtain 1 − 2 y 0 2 − x 2 2 3 2 2 3 0 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 1 − 2 y 2 2 0 1 2 0 ⇐⇒ 1 > x + ( y + ) . 0 0 2 Geometry 1 S is therefore the intersection of the lower half-plane and a circle centered at (0 , − ) of radius 1. This 2 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 π 3 is + . 3 4 1 Solution 2. Let O = (0 , − ) and ℓ = { y = − 1 } be the focus and directrix of the given parabola. Let 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. It is not hard to see that this is precisely the intersection of the unit circle centered at O and the lower half-plane, so now we can proceed as in Solution 1.