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 scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let 
intersect 
and 
at 
and 
respectively and 
intersect 
and 
at 
and 
respectively .
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
be a positive integer such that 
.
.
be lines in the plane such that no two of them are parallel and no three of them are concurrent. Let 
be the intersection point of lines 
. We call an "Interior Point" if there are points 
on 
and 
on 
such that is between and . Prove that there are at least 
Interior points.(
)
.
. 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 the side lengths of a triangle. Prove that![\[ (ab+bc+ca)\left( \frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{1}{a^2+b^2}\right) \le \frac{9}{2}.\]](http://latex.artofproblemsolving.com/3/c/0/3c0127a42c46fcbc0236dbee617b7dc18cb3011f.png)
or 
and its permutations. 
, 
, . (
isn't true).
. 
, , , you would only have a line, but I understand this is a degenerative case and as such, may be considered a triangle.
a triangle, the statement is still not true - take any triple 
and all of its permutations and the equality will obviously still hold since the inequality is homogeneous.
, , , you would only have a line, but I understand this is a degenerative case and as such, may be considered a triangle. a triangle, the statement is still not true - take any triple and all of its permutations and the equality will obviously still hold since the inequality is homogeneous.
but I just want to write it as . I think it is nothing at all since the main important thing is my inequality. By the way, my inequality is true indeed. I am sure of it.
and are not completely the same (they are in this problem because it's homogeneous... but you get the picture). I was also a little confused because is not a triangle in the most obvious sense of the word, but I'm sure we got everything straightened out.
, 
, 
and 
, 
, the inequality becomes
.
for 
. Since 
is convex, because![$ f"(x)=\frac{24(p-x)[(x-1)^2+p^2+1]}{(x^2-2px-1)^4}>0$](http://latex.artofproblemsolving.com/6/9/2/6923f02a2f9e3eae6a85b50aa5fc6c31a5ff4416.png)
,
and 
. These cases are equivalent to 
and 
, respectively.
We put 
and 
are sidelenghts of a triangle, by Valenticu-Shur inequality.
, 
and 
it is sufficient to show that for 
and this is true being equivalent at 
we find that 
which is not least of zero as 
!!!! is not equal at
We put 
and 
are sidelenghts of a triangle, by Valenticu-Shur inequality.
, 
and 
it is sufficient to show that for and this is true being equivalent at 
we find that 
which is not least of zero as !!!! is not equal at
My solution for it used the old way-SOS. Here is it
and 
then prove 
and 
.
and it's nothing. 