IMO 2012, Problem 1
by v_Enhance, Jul 10, 2012, 9:31 PM
Given triangle 
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
.
,
.
.![\[ G = \left( -a: b : \frac{-as + (s-c)b}{s-b} \right) \]](http://latex.artofproblemsolving.com/4/4/9/4495dc725c460e59a446472e08964c357e71f832.png)
It follows immediately that ![\[ T = \left( 0 : b : \frac{-as + (s-c)}{b}{s-b} \right) = ( 0 : b(s-b) : b(s-c) - as) \]](http://latex.artofproblemsolving.com/c/e/6/ce6802403daedaba220a139e4bd6dbd6a6381a39.png)
Normalizing, we see that 
,
.
,
D:
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