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 an acute triangle with incenter 
and circumcenter 
. The incircle touches sides 
and 
at 
and 
respectively, and 
is the reflection of 
over . The circumcircles of 
and 
meet at 
, and the circumcircles of 
and meet at a point 
, where 
is the midpoint of 
. Prove that if 
and meet at 
, then 
.
and 
Prove that
such that divides 
but 
does not divide 
.
with right angles at and 
has an inscribed circle with center and radius 
. Let the lengths of the parallel sides 
and 
be 
and 
, and the intersection of the diagonals 
and 
be 
. to one of the trapezoid sides has the length . and as a function of and .
be a point on the Neuberg cubic. Show that as varies, the Nine Point Circle of the antipedal triangle of goes through a fixed point.
be positive integers and 
be prime. We know that 
and 
. Prove that 
is a perfect square.
and such that:![\[ 3n^2+3n+7=m^3 \]](http://latex.artofproblemsolving.com/6/f/8/6f84f81f857094b3ea432284a7881697167b2418.png)
for which there exists a positive integer 
with the property that is divisible by and 
is divisible by 
.
be the circumcircle of a non-isosceles triangle and 
be a tangent line to (
). Let 
be midpoint of 
and 
. Prove that 
and 
be coprime positive integers such that:![\[ \dfrac{p}{q}=1-\frac12+\frac13-\frac14 \cdots -\dfrac{1}{1334}+\dfrac{1}{1335} \]](http://latex.artofproblemsolving.com/b/d/3/bd374388fe6b84fbcd66e23a7956f2fbeebe4c28.png)
is divisible by 2003.
and all positive integers 
and 
satisfying 
be midpoints of 
, and 
is the orthocenter of . Then the feet of perpendiculars from to the sides of 
are collinear, so lies on the circumcircle of . But note that the midpoint of 
, 
, is the orthocenter of 
, and lies on the circumcircle of 
. Similarly define 
, then 
lie on a circle thus 
lie on a circle. Then now it is easy to show that must be right. is right, 
.
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
,
and the midpoints of the altitudes of the triangle are collinear, find all possible values of the length of 
.