2014 IMO Problems/Problem 3

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Problem

Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular 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