AwesomeMath 2016
by djmathman, Dec 27, 2015, 12:14 AM
I guess I'm going to help out with Cornell and Puget Sound (western site) this year~
and now for some geo (darn why am I still doing geo i'm in college nobody cares about geo anymore)
Sorta complicated but w/e
and now for some geo (darn why am I still doing geo i'm in college nobody cares about geo anymore)
Quote:
Points
and
are the circumcenter and orthocenter of acute triangle
, respectively. The perpendicular bisector of segment
meets sides
and
at
and
, respectively. Prove that 









Sorta complicated but w/e
For convienence, denote by
the circumcircle of
. Let
be the antipode of
with respect to
and define
and
.
Suppose
and
. I claim
and
. To prove this, first remark that by Pascal on cyclic hexagon
, points
,
, and
are collinear. Now let
and
. Note that by angle chasing
This means
is isosceles with
. But remark that by the definition of antipode
, so as a result
is the midpoint of
. Similarly,
is the midpoint of
, so
. Combining this with earlier discoveries yields that
with ratio of similitude
as desired.
From here, the problem is simple. Remark that from well-known angle equalities
Finally, note that the homothety
sends
to
,
to
, and
to
. Since homotheties preserve angle equalities, we get
as desired. 







Suppose





















From here, the problem is simple. Remark that from well-known angle equalities
![\[\angle AA_0E_0\equiv\angle AA_0B'=\angle ABB'=\angle ACC'=\angle AA_0C\equiv \angle AA_0D_0.\]](http://latex.artofproblemsolving.com/2/7/b/27ba0f877b92a333f155c1f078503b3100ffd0d6.png)








