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2003 OIM Problems/Problem 5

Revision as of 04:32, 14 December 2023 by Tomasdiaz (talk | contribs) (Created page with "== Problem == In the square <math>ABCD</math>, let <math>P</math> and <math>Q</math> be points belonging to the sides <math>BC</math> and <math>CD</math> respectively, differe...")
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Problem

In the square $ABCD$, let $P$ and $Q$ be points belonging to the sides $BC$ and $CD$ respectively, different from the ends, such that $BP = CQ$. Points $X$ and $Y$ are considered, belonging to the segments $AP$ and $AQ$ respectively. Show that, whatever $X$ and $Y$, there exists a triangle whose sides have the lengths of the segments $BX$, $XY$, and $DY$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

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