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.
. Prove that: 
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
are positive real numbers such that 
, Prove that ![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
such that 
holds for all integers 
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each , determine the largest number of dominoes that can be placed in this way.
or 
. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be a scalene triangle with orthocenter 
and circumcenter 
. Denote by 
, the midpoints of 
, 
. Suppose the circle 
with diameter meets the circumcircle of at 
, and meets line 
at a point 
. The tangent to at 
meets line 
at 
. Show that the circumcircles of 
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
are intersecting at the point 
.
and 
are the midpoints of 
and 
, respectively. Let the points 
on 
and 
on 
s.t.
and 
.Prove that 
.
)
and the points 
where 
is the midpoint of the segment 
are collinear, as well 
from the complete quadrilateral 
we have that the points 
where 
is the midpoint of the segment 
are collinear, as well.
bisects the segment 
with diameter we conclude that the line segment 
is the midperpendicular of the cord and the proof is completed.
be the origin, and the lowercase letters denote the position vectors of the respective points. Then we have to show that 
. Using 
and 
it remains to show that 
i.e. 
; here 
. But we have 
and 
. Therefore we are done.
be a cyclic quadrilateral whose diagonals intersect at . Let , and be the midpoints of and , respectively. Let and 
be the projections of onto , and 
, respecitvely. Prove that 
is the perpendicular bisector of 
.
be the reflection of in . Since 
is indirectly similar to 
, and is indirectly similar to 
, so is directly similar to . Note that 
. It follows that 
. But 
and 
share corresponding side so they are congruent. The conclusion follows.
are intersecting at the point . and are the midpoints of and , respectively. Let the points on and on s.t. and .Prove that .
and 
then prove 
and 
.
so EH=EI
and 
be the respective orthocenters of 
and 
,
and 
meet at 
meet 
at 
and 
and 
and 
be 
and 
respectively. Now, since 
,
,
.
.
(notice that since 
,
)
is cyclic, so 
from 
and 
.
,
from which we see that 
.
and 
.
,
is cyclic, so 
.
is perpendicular to the line connecting the centers of these circles, 
.
,
.