PUMaC 2012 · 团队赛 · 第 11 题

PUMaC 2012 — Team Round — Problem 11

(3 digits) Let √ 2 2 2 2 2 x − 4 x +4 x − 4 x +4 Ξ( x ) = 2012( x − 2) + 278( x − 2) 2012 + e + 139 + ( x − 4 x + 4) e find the area of the region in the xy -plane satisfying: √ { x ≥ 0 and x ≤ 4 and y ≥ 0 and y ≤ Ξ( x ) }

解析

Problem: (3 digits) Let √ 2 2 2 2 2 x − 4 x +4 x − 4 x +4 Ξ( x ) = 2012( x − 2) + 278( x − 2) 2012 + e + 139 + ( x − 4 x + 4) e find the area of the region in the xy -plane satisfying: √ { x ≥ 0 and x ≤ 4 and y ≥ 0 and y ≤ Ξ( x ) } Answer: 556 √ √ 2 x − 4 x +4 Solution: Let f ( x ) = Ξ( x ) and observe that f ( x ) = 139 + ( x − 2) 2012 + e . Let S √ 2 x be the set of points ( x, y ) which satisfy all four inequalities. Let g ( x ) = x 2012 + e . Notice ◦ that this function is odd. Thus, f ( x ) = g ( x − 2) + 139 has 180 rotational symmetry around the point ( x, y ) = (2 , 139). ◦ ′ If we rotate the region S by 180 around this (2 , 139), we get a new set S , with the following ′ ′ properties: (1) S and S are disjoint (except at the boundary). (2) The union S ∪ S forms a rectangle with base 4 and height f (0) + f (4) = 139 + g ( − 2) + 139 + g (2) = 2 · 139. 1 Thus, the area of S is · 4 · 2 · 139 = 556 . 2 Author: Alan 7