IMO 2012 #1
by Wolstenholme, Oct 28, 2014, 5:52 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 
Proof 1 (synthetic):
Note that
Also, since quadrilateral 
is cyclic note that 
Looking at 
we have that 
by our earlier calculations. Since 
as well we have that quadrilateral 
is cyclic so 
which implies that 
. This means that 
This means that 
and similarly we have that 
so it suffices to show that 
. But it is well known that 
and 
so we are done.
Proof 2 (bary):
We proceed with barycentric coordinates. Let
and 
and 
Moreover denote 
Also let 
.
It is clear that
. Moreover since 
we have that 
and 
. Also note that 
. Now the line 
has equation 
and by computing a simple determinant we have that line 
has equation 
. Intersecting these lines we have that 
. Then it is easy to find that 
. Similarly 
. Now it is easy to see that the midpoint of segment 
is 
as desired.
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 
Proof 1 (synthetic):
Note that
Also, since quadrilateral 
is cyclic note that 
Looking at 
we have that 
by our earlier calculations. Since 
as well we have that quadrilateral 
is cyclic so 
which implies that 
. This means that 
This means that 
and similarly we have that 
so it suffices to show that 
. But it is well known that 
and 
so we are done.Proof 2 (bary):
We proceed with barycentric coordinates. Let
and 
and 
Moreover denote 
Also let 
.It is clear that
. Moreover since 
we have that 
and 
. Also note that 
. Now the line 
has equation 
and by computing a simple determinant we have that line 
has equation 
. Intersecting these lines we have that 
. Then it is easy to find that 
. Similarly 
. Now it is easy to see that the midpoint of segment 
is 
as desired.
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