AoPS Academy
be a triangle with orthocenter 
and circumcenter 
. Let 
be a point in the plane such that 
. Let 
and 
be the reflections of in the lines 
and 
, respectively. Let 
be the orthogonal projection of onto . Let 
be the orthogonal projection of onto . Assume that 
and 
. Prove that 
.
satisfying
for all integers 
.
be a cyclic quadrilateral. A circle centered at passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
and 
be positive integers. For a connected simple graph 
on vertices and edges, we consider the number 
of orientations of (all of) its edges so that, in the resulting directed graph, every vertex has even outdegree. only depends on and , and determine its value.
be a triangle, 
its incenter and 
the touchpoints of the incircle with sides , 
, and , respectively. Let be the intersection of segment 
with line 
. Prove that 
, and the midpoint of are collinear.
be a triangle, its incenter and the touchpoints of the incircle with sides , , and , respectively. Let be the intersection of segment with line . Prove that , and the midpoint of are collinear.
as 
respectivily. Let 
be a point such that 
is diameter on the incircle. and now let the tangent from to the incircle meet 
at 
. By newton's theorem on 
we have 
concurrent at . Now let 
and 
, then by pascal on the degenerate hexagon 
we have 
colinear and now since is the ortocenter of 
we have 
. Now by brokard on 
(here we take all the polars w.r.t. incircle btw) we have that 
and now by la'hire we have 
and now by la'hire on definitions of 
and by the last fact we got, we have 
and 
and now by brokard on 
and the past facts we got, we have 
colinear and 
colinear. Now note that 
lies on the -symedian on 
and since 
is cyclic we have that is midpoint of 
now let 
be the point of infinity of 
then by projecting:
Thus we are done 
as respectivily. Let be a point such that is diameter on the incircle. and now let the tangent from to the incircle meet at . By newton's theorem on we have concurrent at . Now let and , then by pascal on the degenerate hexagon we have colinear and now since is the ortocenter of we have . Now by brokard on (here we take all the polars w.r.t. incircle btw) we have that and now by la'hire we have and now by la'hire on definitions of and by the last fact we got, we have and and now by brokard on and the past facts we got, we have colinear and colinear. Now note that lies on the -symedian on and since is cyclic we have that is midpoint of now let be the point of infinity of then by projecting:Thus we are done 
