HMMT 二月 2015 · 几何 · 第 5 题
HMMT February 2015 — Geometry — Problem 5
题目详情
- Let I be the set of points ( x, y ) in the Cartesian plane such that ( ) 1 / 4 4 y x > + 2015 9 2 2 2 Let f ( r ) denote the area of the intersection of I and the disk x + y ≤ r of radius r > 0 centered at 2 the origin (0 , 0). Determine the minimum possible real number L such that f ( r ) < Lr for all r > 0. √ ◦
解析
- Let I be the set of points ( x, y ) in the Cartesian plane such that ( ) 1 / 4 4 y x > + 2015 9 2 2 2 Let f ( r ) denote the area of the intersection of I and the disk x + y ≤ r of radius r > 0 centered at 2 the origin (0 , 0). Determine the minimum possible real number L such that f ( r ) < Lr for all r > 0. π Answer: Let B ( P, r ) be the (closed) disc centered at P with radius r . Note that for all 3 ( ) 1 / 4 √ 4 | y | y ′ ′ √ ( x, y ) ∈ I , x > 0, and x > + 2015 > . Let I = { ( x, y ) : x 3 > | y |} . Then I ⊆ I and the 9 3 ′ π 2 π 2 intersection of I with B ((0 , 0) , r ) is r , so the f ( r ) = the area of I ∩ B ((0 , 0) , r ) is also less than r . 3 3 π Thus L = works. 3 ( ) 1 / 4 ( ) ( ) 4 1 / 4 4 | y | | y | | y | y 4 √ √ On the other hand, if x > +7, then x > +7 > + 7 > + 2015 , which means 9 9 3 3 √ ′′ ′′ ′ ′′ that if I = { ( x, y ) : ( x − 7) 3 > | y |} , then I ⊆ I . However, for r > 7, the area of I ∩ B ((7 , 0) , r − 7) π π 2 ′′ 2 is ( r − 7) , and I ⊆ I , B ((7 , 0) , r − 7) ⊆ B ((0 , 0) , r ), which means that f ( r ) > ( r − 7) for all 3 3 π r > 7, from which it is not hard to see that L = is the minimum possible L . 3 √ 4 4 Remark: The lines y = ± 3 x are actually asymptotes for the graph of 9 x − y = 2015. The bulk of √ α α the problem generalizes to the curve | 3 x | − | y | = C (for a positive real α > 0 and any real C ); the case α = 0 is the most familiar case of a hyperbola. √ ◦