HMMT 二月 2014 · 团队赛 · 第 1 题

英文解析

1. [ 10 ] Let ω be a circle, and let A and B be two points in its interior. Prove that there exists a circlepassing through A and B that is contained in the interior of ω .
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   Answer: N/A WLOG, suppose OA ≥ OB . Let ω be the circle of radius OA centered at O . We
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   have that B lies inside ω . Thus, it is possible to scale ω down about the point A to get a circle ω
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   passing through both A and B . Since ω lies inside ω and ω lies inside ω , ω lies inside ω .
   Alternative solution 1 : WLOG, suppose OA ≥ OB . Since OA ≥ OB , the perpendicular bisector of
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   AB intersects segment OA at some point C . We claim that the circle ω passing through A and Band centered at C lies entirely in ω . Let x = OA and y = AC = BC . Note that y is the length of the
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   radius of ω . By definition, any point P contained in ω is of distance at most y from C . Applying the triangle inequality to OCP , we see that OP ≤ OC + CP ≤ ( x − y ) + y = x , so P lies in ω . Since P
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   was arbitrary, it follows that ω lies entirely in ω .
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   Alternative solution 2 : Draw line AB , and let it intersect ω at A and B , where A and A are on the
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   same side of B . Choose X inside the segment AB so that A X/AX = B X/BX ; such a point exists
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   by the intermediate value theorem. Notice that X is the center of a dilation taking A B to AB - the
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   same dilation carries ω to ω which goes through A and B . Since ω is ω dilated with respect to a
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   point in its interior, it’s clear that ω must be contained entirely within ω , and so we are done.