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2014 IMO Problems/Problem 3

Revision as of 04:25, 9 October 2014 by Timneh (talk | contribs) (Created page with "==Problem== Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}. Point </math>H<math> is the foot of the perpendicual from </math>A<math> to </mat...")
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Problem

Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}. Point$H$is the foot of the perpendicual from$A$to$BD$. Points$S$and$T$lie on sides$AB$and$AD$, respectively, such that$H$lies inside$\triangle{SCT} and \[\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.\]

Solution

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2014 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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