HMMT 十一月 2013 · 冲刺赛 · 第 21 题

英文解析

21. [ 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 notcontained 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.)
    Answer: 8 r Solution 1. Let U denote the union of the four circles, so we seek 2
    2 2 2 21
    U − ([ O ] − U ) = 2 U − [ O ] = 2[(2 r ) + 4 · πr ] − π (2 r ) = 8 r .
    (Here we decompose U into the square S with vertices at ( ± r, ± r ) and the four semicircular regions of 2
    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 notcontain the two intersections with the other two small circles, y be an area of intersection of two smallcircles, 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,
    we have 4 z = [ O ] − U = π (2 r ) − U , and by the inclusion-exclusion principle, 4 y = [ A ] + [ B ] + [ C ] +2
    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 rby moving the curved outward parts to fit into the curved inward parts to get a r × 2 r rectangle. Sothe answer is 8 r .2