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)
is the altitude, the point lies on the side, and 
. Prove that the center of the circumscribed circle of triangle 
coincides with the center of an excircle of triangle 
.
is the altitude, the point lies on the side, and . Prove that the center of the circumscribed circle of triangle coincides with the center of an excircle of triangle .
![[asy]
size(8cm);
defaultpen(fontsize(8pt));
pen pri=green;
pen sec=springgreen;
pen tri=chartreuse;
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pair A, B, C, H, K, Ic, X, Y;
A=(0,0);
B=(1,1);
C=(3,0);
H=foot(B, A, C);
K=(3-sqrt(5),0);
Ic=circumcenter(A,B,K);
X=intersectionpoint(Ic--C,B--K);
Y=intersectionpoint(Ic--H,A--B);
draw(A--B--C--cycle,pri);
draw(B--H,sec);
draw(B--K,sec);
draw(Ic--C,tri+dashed);
draw(Ic--H,tri+dashed);
dot("$A$",A,W);
dot("$B$",B,N);
dot("$C$",C,E);
dot("$K$",K,S);
dot("$H$",H,S);
dot("$I_C$",Ic,N);
dot("$X$",X,NW);
dot("$Y$",Y,S);
[/asy]](http://latex.artofproblemsolving.com/e/0/d/e0db40b5602f25aeed102a4f52dd78a14ce8797e.png)
and 
, 
. Thus 
is an isosceles right triangle.
and the external angle bisector of 
. These will intersect at the -excenter of 
. Our goal is to show that both of these lines are also perpendicular bisectors of the sides of 
.
and 
. It is clear that is the midpoint of 
, since 
is the external angle bisector of and thus the internal angle bisector of 
. Since the internal angle bisector of an isosceles triangle is its altitude and its median, 
is a perpendicular bisector of . A similar result follows for point 
(it is the foot of the altitude from to 
, and we are given that .) Thus, both the internal angle bisector of and the external angle bisector of are perpendicular bisectors of , and it follows that the center of the circumscribed circle of coincides with the center of an excircle of 
. 
.
and 
Prove that
