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.
with the smaller base 
are constructed the squares 
and 
. Prove that the perpendicular bisector of passes through the midpoint of 
.
such that
for all real numbers 
satisfying the equation
has circumcenter 
. The lines 
and 
intersect at 
, while 
and 
intersect at 
and 
and 
intersect at 
. Show that if the triangles and 
are similar(with vertices in that order), than is equilateral.
candidates and 
judges, where 
is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most 
candidates. Prove that ![\[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]](http://latex.artofproblemsolving.com/3/8/7/387c8824b28f6422306e3c25b34fccb0a841f4cd.png)
be a triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of onto line 
. Show that 
.
have the property that 
is a perfect square for infinitely many positive integers 
?
functions such that
for all real numbers 
and 
with equal radii are given. Let 
,
be the intersection of the circles.
and 
are on and such that they are inside and respectively.
,
are on and respectively, such that 
and 
.Denote by 
as the other intersection of 
and 
. Prove that 
are concurrent.
days later (but perhaps earlier, as well). Hedvig received the trophy today. Determine all integers 
for which we can be absolutely certain that she cannot receive the trophy again in days, given the above information.
be real numbers such that : 
. Prove that : 
and some point on 
. The perpendicular from to intersects the circumcircle of 
at and the perpendicular from to 
intersects the circumcircle of 
at . Show that the line 
does not depend on the choice of .(Shining Sun, USA)
be a convex heptagon with all sides equal. It's known that 
and 
. Calculate the measure of angles 
, 
and 
.
?
such that 
such that 
is a regular pentagon. Since 
and 
,
is equilateral. Thus, 
.
.
is a rhombus with 
,
.
because 
.
.