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Segment AB has no common point with the square R

Source: Germany Bundeswettbewerb Mathematik 2001, Round 2, Problem 4

January 17, 2004
geometryperimeterrotationalgebrafunctiondomaingeometry unsolved

Problem Statement

A square R R of sidelength 250 250 lies inside a square Q Q of sidelength 500 500. Prove that: One can always find two points A A and B B on the perimeter of Q Q such that the segment AB AB has no common point with the square R R, and the length of this segment AB AB is greater than 521 521.
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