HMMT 十一月 2013 · 冲刺赛 · 第 21 题
HMMT November 2013 — Guts Round — Problem 21
题目详情
- [ 11 ] Suppose A , B , C , and D are four circles of radius r > 0 centered about the points (0 , r ), ( r, 0), (0 , − r ), and ( − r, 0) in the plane. Let O be a circle centered at (0 , 0) with radius 2 r . In terms of r , what is the area of the union of circles A , B , C , and D subtracted by the area of circle O that is not contained in the union of A , B , C , and D ? (The union of two or more regions in the plane is the set of points lying in at least one of the regions.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND
解析
- [ 11 ] Suppose A , B , C , and D are four circles of radius r > 0 centered about the points (0 , r ), ( r, 0), (0 , − r ), and ( − r, 0) in the plane. Let O be a circle centered at (0 , 0) with radius 2 r . In terms of r , what is the area of the union of circles A , B , C , and D subtracted by the area of circle O that is not contained in the union of A , B , C , and D ? (The union of two or more regions in the plane is the set of points lying in at least one of the regions.) 2 Answer: 8 r Solution 1. Let U denote the union of the four circles, so we seek 1 2 2 2 2 U − ([ O ] − U ) = 2 U − [ O ] = 2[(2 r ) + 4 · πr ] − π (2 r ) = 8 r . 2 (Here we decompose U into the square S with vertices at ( ± r, ± r ) and the four semicircular regions of radius r bordering the four sides of U .) Solution 2. There are three different kinds of regions: let x be an area of a small circle that does not contain the two intersections with the other two small circles, y be an area of intersection of two small circles, and z be one of those four areas that is inside the big circle but outside all of the small circles. Then the key observation is y = z . Indeed, adopting the union U notation from the previous solution, 2 we have 4 z = [ O ] − U = π (2 r ) − U , and by the inclusion-exclusion principle, 4 y = [ A ] + [ B ] + [ C ] + 2 2 [ D ] − U = 4 π − U , so y = z . Now U = (4 x + 4 y ) − (4 z ) = 4 x . But the area of each x is simply 2 r by moving the curved outward parts to fit into the curved inward parts to get a r × 2 r rectangle. So 2 the answer is 8 r .