Difference between revisions of "1996 IMO Problems/Problem 2"
Nxk12102008 (talk | contribs) (Created blank page) |
(→Problem) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | Let <math>P</math> be a point inside triangle <math>ABC</math> such that | ||
+ | |||
+ | <cmath>\angle APB-\angle ACB = \angle APC-\angle ABC</cmath> | ||
+ | |||
+ | Let <math>D</math>, <math>E</math> be the incenters of triangles <math>APB</math>, <math>APC</math>, respectively. Show that <math>AP</math>, <math>BD</math>, <math>CE</math> meet at a point. | ||
+ | |||
+ | ==Solution== | ||
+ | {{solution}} | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1996|num-b=1|num-a=3}} | ||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 12:08, 3 June 2024
Problem
Let be a point inside triangle such that
Let , be the incenters of triangles , , respectively. Show that , , meet at a point.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |