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)
and make it the reference triangle for all further operations. Then the cevian triangle of the isotomic conjugate of the orthocenter of should be degenerate. This implies that the orthic triangle of should be degenerate too (using Menelaus' theorem if the line doesn't coincide with a cevian or a side, and using plain angles otherwise); this happens when one of the angles of the orthic triangle is either 
or 
, and in the only non-degenerate case for the triangle is when one of the angles is 
, i.e. is right. For the other direction, note that all the midpoints of altitudes lie on a midline of the triangle if it is right.
is right, then the midpoint of altitudes are clearly collinear. Now suppose none of the angles of 
is right and the midpoint of altitudes are collinear. Fix 
and move 
linearly on a fixed line such that 
. Now midpoint of 
altitude is fixed, midpoint of 
altitude varies with degree 2 and midpoint of 
altitude varies on a fixed line, so has degree 1. For 
, 
, 
and when 
we get 4 cases where the midpoint of altitudes are collinear. Therefore we have 
cases, so for all points on the line, the midpoint of altitudes are collinear. This is clearly false if we choose 
, so we get a contradiction.
.
and 
Prove that
,
and 
.
,
,
.
,
,
.
,
,
.
,
,
.
is
The slope of 
is
These slopes must be equal, so:
This shows that either 
,
or 
.
.
,
.
must be a right triangle.