AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a trapezoid with parallel sides 
. Points 
and 
lie on the line segments 
and 
, respectively, so that 
. Suppose that there are points 
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points , , 
and 
are concyclic.
, with 
, and the circle 
with center 
and radius 
. Denote by 
the midpoint of side . The line 
intersects the circle for the second time in 
. Let 
be a point on the circle such that 
and 
. Show that 
.
. Let and be the feet of the altitudes from to 
and from 
to respectively.
and 
as the reflections of across lines 
and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
![\[\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}\]](http://latex.artofproblemsolving.com/5/d/6/5d6881be29327bcdc3e7bd76c221968578c7f50b.png)
is isosceles.
be all the divisors of 
. It is known, that 
are all divisors for some 
(except 
). Find all such 
be a sequence of positive real numbers that satisfies 
and 
for every natural number . Prove that 
for every natural number .
be positive real numbers such that : 
. Prove that : 
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and .
(ans- 
)
satisfying 
Prove that 
When does equality hold?
be the center of 
. Let 
be the plane that contains 
and let be the point where touches .
is the intersection of and 
. Let 
be the point where touches .
because 
and are congruent. We also know that 
because they're both tangent lines to the sphere.
are collinear, so that 
.
as the points where touches the planes containing 
and 
, then 
and 
. But 
because they're all tangent lines, so 
which means is the circumcenter of 
. are collinear, I don't have the proof myself, but from many2 times visualizing, I almost believe that the following is correct:
is the plane containing 
then 
and 
.
are collinear.
the planes containing the the faces of the tetrahedron 
opposite to 
Since 
it follows by obvious specular symmetry that the angle bisector (plane) of the dihedral angle 
which pass through the S-excenter of 
is also the perpendicular bisector of 
Likewise, the angle bisector of the dihedral angle 
is the perpendicular bisector of 
lies on the perpendicular bisector of 
Therefore, orthogonal projection of on 
lies on the intersection 
i.e. the perpendicular bisector (line) of 
in 
Similarly, 
is the circumcenter of 
such that 
and 
.
?
