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 
be given . On the extension of the segment 
beyond 
, a point 
is chosen arbitrarily. Let the line through and 
intersect the line 
at the point 
. Let 
be the intersection of the lines 
and 
. Prove that the line 
is tangent to the circumcircle of the triangle 
.
be any point. Let 
be projection point from to 
. Circumcircle 
cuts circumcircle 
at 
. Let 
be antipode of wrt . Prove that 
are cyclic.
, 
be real numbers, such that 
. Prove that 
.
, angle 
. Point lies on 
such that 
. Line parallel to 
passing trough intersects 
at 
. Prove that 
.
be a parallelogram of center 
. Prove that for any point 
, there exist unique points 
and 
such that is the center of mass of 
.
Compute -![\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}](http://latex.artofproblemsolving.com/8/6/f/86fb48db976012106bd30f74cb6ada8c6e6d7800.png)
For 
define 
Determine the open interval(s) (if any) where 
is decreasing and all the open interval(s) (if any) where is increasing.
Determine all the local minima of (if any) and all the local maxima of (if any) 
of positive integers such that
, then it must not contain edge 
.
with circumcenter and orthocenter 
, let be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line . Let be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line such that 
. Let and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
, such that 
is a perfect square.
, such that is a perfect square.
which can never be perfect square.
Hence 
is only solution.
and 
for 
, the expression for will be congruent to 
, which can never be a perfect square. This puts a very nice bound for 
, and it's easy to check the case of 
.
and 
Prove that
is a natural number, here is my solution:
we have 
so this number is not a perfect square.
If 
we have to find 
is a perfect square. By taking modulo 
we see that 
and 
)
and this implies 
and 
We can only have 
and 
so in this case 
we have to find 
is a perfect square. Let 
This leads to 
Since 
it follows that we can only have 
and 
so 
and 
so in this case 
If 
,
,
.
