Difference between revisions of "1996 IMO Problems/Problem 2"

Latest revision as of 12:08, 3 June 2024

Let $P$ be a point inside triangle $ABC$ such that

$\angle APB-\angle ACB = \angle APC-\angle ABC$

Let $D$ , $E$ be the incenters of triangles $APB$ , $APC$ , respectively. Show that $AP$ , $BD$ , $CE$ meet at a point.

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ 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]]

1996 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions