Difference between revisions of "2014 IMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}. Point < | + | Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}</math>. Point <math>H</math> is the foot of the perpendicular from <math>A</math> to <math>BD</math>. Points <math>S</math> and <math>T</math> lie on sides <math>AB</math> and <math>AD</math>, respectively, such that <math>H</math> lies inside <math>\triangle{SCT}</math> and |
<cmath>\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.</cmath> | <cmath>\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.</cmath> | ||
Revision as of 04:26, 9 October 2014
Problem
Convex quadrilateral 
has 
. Point 
is the foot of the perpendicular from 
to 
. Points 
and 
lie on sides 
and 
, respectively, such that lies inside 
and
![\[\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.\]](http://latex.artofproblemsolving.com/8/f/7/8f7fdb3d9a89e2d7864c895d94972778862131a4.png)
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 | ||
