HMMT 二月 2010 · GEN1 赛 · 第 9 题

英文解析

9. [ 7 ] Three unit circles ω , ω , and ω in the plane have the property that each circle passes through the
   1 2 3
   centers of the other two. A square S surrounds the three circles in such a way that each of its foursides is tangent to at least one of ω , ω and ω . Find the side length of the square S .
   1 2 3
   √ √
   6+ 2+8
   Answer:
   B C4
   2ω
   3ωQω
   1R
   A M N DEPF
   By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say ω . Since Ssurrounds the circles, these two sides must be adjacent, so we can let A denote the common vertex 1
   of the two sides tangent to ω . Let B , C , and D be the other vertices of S in clockwise order, andlet P , Q , and R be the centers of ω , ω , and ω respectively, and suppose WLOG that they are also 1
   1 2 3
   in clockwise order. Then AC passes through the center of ω , and by symmetry (since AB = AD ) it
   General Test, Part 11
   must also pass through the other intersection point of ω and ω . That is, AC is the radical axis of ω
   2 3 2
   and ω .
   Now, let M and N be the feet of the perpendiculars from P and R , respectively, to side AD . Let E3
   and F be the feet of the perpendiculars from P to AB and from R to DC , respectively. Then P EAMand N RF D are rectangles, and P E and RF are radii of ω and ω respectively. Thus AM = EP = 1
   1 2
   and N D = RF = 1. Finally, we have
   °
   M N = P R · cos(180 − ∠ EP R )
   °
   = cos(180 − EP Q − RP Q )
   ° ° °
   = − cos ((270 − 60 ) / 2 + 60 )
   °
   = − cos(165 )
   °
   = cos(15 )
   √ √
   6 + 2 = .
   √ √ √ √4
   6+ 2 6+ 2+8
   Thus AD = AM + M N + N D = 1 + + 1 = as claimed.
   4 4