AoPS Academy
be a scalene oblique triangle with circumcenter 
and orthocenter 
, and 
(
, 
) a point in the plane.
, 
be the circumcevian triangles of , , respectively.
be the pedal triangle of with respect to .
be the reflection in 
of 
. Define 
, 
cyclically.
be the reflection in of 
. Define 
, 
cyclically.
, 
be the circumcenters of 
, 
, respectively., , are collinear if and only if lies on the Jerabek hyperbola of ., , are collinear if and only if lies on the Lemoine cubic (= 
) of .
is a triangle with incenter 
. The foot of the perpendicular from to 
is 
, and the foot of the perpendicular from to 
is . Prove that 
.
with an obtuse angle 
, let 
be feet of altitudes from 
on sides 
respectively. The tangents from to circumcircle of triangle intersect line 
at points 
respectively and we know that 
. Point 
lies on segment 
in such a way that 
and let 
be a point on line such that 
is between 
and 
. Prove that the circle with diameter 
is tangent to circumcircle of triangle .![\[
x^2 + 2|x| - 6x + 15 = 0.
\]](http://latex.artofproblemsolving.com/1/1/d/11dbf7a605e8c49bb730eb0bd3737019b277fa41.png)
(Note: the modulus of a complex number 
is 
.)
and 
.
and 
.
.
are the 256 roots (not necessarily distinct) of the equation 
, evaluate 
.
such that 
, 
, 
, and 
are real numbers that satisfy the system of equations
and 
be positive real numbers such that 
and 
. What is the value of 
?
be an acute triangle with circumcircle 
. The altitudes , 
and 
of the triangle intersect at point . A point is chosen on the line such that 
. Prove that the reflection of in 
lies on .
, let 
be the feet from 
. Let 
be the reflection of 
across the perpendicular bisector of . Let 
be the queue point. Then, 
.
to 
. 
is sent to 
, 
to 
so we need that is mapped to . as 
, 
is tangent to , so by properties of spiral simiarities, the lemma is proven.
be the intersection of the circle with radius 
with center with . Let be the point s.t. 
, with 
, and be the -queue point.
is tangent to as 
.
and 
we have that 
has to be tangent as well, which implies the conclusion.
, we have that is the center of 
and 
is tangent to .
is collinear.
, so 
as desired.
, where 
, but by the reflection of orthocenter lemma, is the midpoint of 
, so 
so as 
, is the perpendicular bisector of 
as desired.
if
,
,
?
in terms of 
.
.
to z because we already know the value of 
.
(in the equation 
)
.
.
and cross multiply.
.
.
![\[
\frac{ab + bc + ca}{a^2 + b^2 + c^2} + \frac{1}{6} \left( \frac{(a - b)^2}{a^2 + b^2} + \frac{(b - c)^2}{b^2 + c^2} + \frac{(c - a)^2}{c^2 + a^2} \right) \leq 1.
\]](http://latex.artofproblemsolving.com/d/8/b/d8b3869addff6715d86b73083c77395d64ab1995.png)
,
for the same reason.
and 
.
.
.
,
as it is also equal to 
and 
.
but once again substitute 
with 
with 
.
,
.
and 
Prove that
hence 
finishes
lies on the nine point circle of 
due to a homothety of factor 2 in 
be the foot from 
we find 
(remember that 
must be the diameter of this circle), we get 
.
.
is the diameter (since 
)
we now get 
and