AoPS Academy
be a cyclic quadrilateral, and let 
, 
, 
, and 
be the midpoints of 
, 
, 
, and 
respectively. Let 
, 
, 
and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.
with real coefficients is the average of two monic polynomials of degree with real roots.
of positive integers, for which![$$(a, b) + 3[a, b] = a^3 - b^3$$](http://latex.artofproblemsolving.com/b/8/3/b83f52a0152dc39b6d966880a1d6ca824a0f3786.png)
denotes the greatest common divisor of , and ![$[a, b]$](http://latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
denotes the least common multiple of .
and power 
).
is the circumcenter of 
, 
is the intersection of and 
, 
is the tangent of the circumcircle of triangle 
at 
. We have to prove following results:
is parallel to 
is a square 
is collinear
reflection of 
in ; well known, 
is the bisector of 
, hence 
collinear; since 
is the image of , we have also 
. But 
is tangent to this circle, hence 
.
to 
and the circle 
to 'another' circle, tangent to , after the rule that the common point is sent to . AAs it is not possible to have to circle passing through 
and being tangent to , the answer is straight forward.
be the midpoint of 
and 
be the center of 
. Since is the center of the circumcircle of triangle 
are collinear. We have 
so triangles 
are isosceles which equal sides are the radii of the circumcircle of 
and . They have an equal common angle, so their rest angles are also equal. Now, 
so 
are collinear.
is similar to 
so 
or 
. From power of point of 
to we have 
. From the similarity of and 
we get 
. Now 
and radius 
and that results:
and are collinear.
. Let 
at point 
is fixed?
It is well known that 
bisects 
.Hence 
.
are collinear 
?