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 set of Natural numbers from 
to 
.
(
) distinct elements from this set.
. If 
and 
, show that the area of 
is not less than the area of 
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by 
the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than 
) of the line 
and the circle . Prove that 
must hold.
the length of the line segment .)
such that 
(where 
), and a functional equation 
that satisfies 
for all , has a finite number of solutions. Find such .
is the function obtained by composing 
times, that is, 
)
be an acute triangle with 
, and let 
be the projection from onto 
. Let be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
sastifying 
such that the relation ![\[|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1 \]](http://latex.artofproblemsolving.com/1/2/2/1229a609403904449c447225094b3d97e063efd4.png)
holds for all positive integers .
, 
, where is a positive integer.
of positive integers for which there exist injective functions 
such that 
.
denotes component-wise function multiplication)
, the equilateral triangle 
is constructed. Let 
be an interior point of the triangle such that 
, 
, 
and 
. Find the area of the square .
. Now construct in the exteror of 
the triangle 
.
is equilateral and hence 
. From the reciprocal of Phytagorean theorem it follows now that 
, so after summing we get 
. Now, by the Cosine theorem it follows that 
and this also represents the area to find. inside it such that 
, 
, 
using carteaian coordinates, with 
, 
, 
, 
and 
, solving the system formed by the equations of distances we find that there exists only two squares, with 
, and that squares don’t verify the condition 
.
square we find no solution to the problem. But, analysing their solution we see that the only condition about used was that it is a rectangle. Therefore, we can state the problem in the following way: be a rectangle with 
and an equilateral triangle 
inside it is constructed. It is known that there exists a point inside such that , 
, , . Find the lengths of and . will be found in the same way, so in this problem also 
. Now, will be different. And hence the area of the rectangle will be different.
.
and 
Prove that
