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p/2 < AC + BD < p

Source: South Africa 2001

October 1, 2005
geometryperimeterinequalitiestriangle inequality

Problem Statement

ABCDABCD is a convex quadrilateral with perimeter pp. Prove that 12p<AC+BD<p. \dfrac{1}{2}p < AC + BD < p. (A polygon is convex if all of its interior angles are less than 180∘180^\circ.)
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