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.
It is well known that 
. Prove that this is the only perfect square in the sequence.
such that 
and 
are both perfect squares and 
is a power of a prime number.
, points 
are chosen such that when going around the triangle, the points occur in the order 
. It is given that![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the lines 
and 
are equal.
be a triangle such that 
. Let the excircle opposite to A be tangent to the lines 
, and 
at points 
, and 
, respectively, and let 
be its centre. Let 
be a point on the side . The circumcircles of the triangles 
and 
intersect for the second time at 
. Let 
be the foot of the perpendicular from 
to the line 
. Prove that the points 
, and are collinear. opposite to is the circle that is tangent to the line segment , to the ray 
beyond 
, and to the ray 
beyond 
.)
, find the minimal integer 
, such that for any positive integer 
, in any subset with elements of the set 
there are at least 
relatively prime elements.
is polynomial with minimum degree such that 
for 
, find 
and 
such that 
and is to the left of , and replaces the pair 
by either 
or 
. Prove that she can perform only finitely many such iterations.
. We are varying this length. The base of the area overlapping can be determined as 
(the base length is 1 unit). Due to the congruency between the two triangles, it can be easily determined that the height of the area overlapping is 
. The area of the triangle is then 
Then diffrentiating gets 
, but verification by the second diffrential shows this is in fact the min (and this should be obvious anyway). Now, the pentagon can be split up into two trapeziums, such that the Area to be computed is now 
. Since the two triangles are congruent, it does not take much to get 
. By similarity, one can also show that 
and 
. Putting this together
and diffrentiating yields the solution x=1/3. Substitution then shows that 
and the second diffrential shows that this in fact the max.
) slide on a line as on the picture. What is the maximal area of overlap?
Find the value of ![$[100A] .$](http://latex.artofproblemsolving.com/f/5/c/f5cb78c5a80d7eb64d2887c195a92ba19375626f.png)
Prove that :
Where 
such that 
.
Where 
Prove that 
Where 
Prove that
Where 
