IMO 2010 #2
by Wolstenholme, Nov 1, 2014, 3:04 AM
Given a triangle 
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on the arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the meeting point of the lines 
and 
lies on .
Proof:
Let
meet 
again at 
and let 
meet again at 
It suffices to show that points 
are collinear. Now by Pascal's Theorem on degenerate cyclic hexagon 
we have that if 
then 
Therefore it suffices to show that points 
are collinear.
Now we proceed with barycentric coordinates. Let
and 
and 
Moreover let 
and 
and 
Let 
for some 
It is well-known that 
and that 
Therefore we easily find 
Now denote the point at infinity on line 
as 
Clearly 
Therefore line 
has equation 
Since line has equation 
we find that 
Now it suffices to show that the determinant 
which is a trivial computation (made even easier by the fact that 
and 
cancel out).
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on the arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the meeting point of the lines 
and 
lies on .Proof:
Let
meet 
again at 
and let 
meet again at 
It suffices to show that points 
are collinear. Now by Pascal's Theorem on degenerate cyclic hexagon 
we have that if 
then 
Therefore it suffices to show that points 
are collinear.Now we proceed with barycentric coordinates. Let
and 
and 
Moreover let 
and 
and 
Let 
for some 
It is well-known that 
and that 
Therefore we easily find 
Now denote the point at infinity on line 
as 
Clearly 
Therefore line 
has equation 
Since line has equation 
we find that 
Now it suffices to show that the determinant 
which is a trivial computation (made even easier by the fact that 
and 
cancel out).
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