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1997 Poland - Second Round
2
P inside ABC such that 3 < ABP = 3 < ACP = < ABC + < ACB
P inside ABC such that 3 < ABP = 3 < ACP = < ABC + < ACB
Source: Poland 97
January 9, 2005
geometry proposed
geometry
Problem Statement
Let P be a point inside triangle ABC such that 3
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