[/quote]
,
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
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 
.
on 
and 
on 
such that 
Interior points.
)
.
. Prove that: 
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
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 
.
and 
and 
then prove 
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
and 
.
,![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](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 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each 
or 
.
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 
.
,
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Y by laegolas, kk108, Adventure10, Chiaquinha
Find the largest number 
that for which there exists positive integers such that non of them divides another one, but between every three of them, one divides the sum of the other two.
Proposed by Morteza Saghafian
that for which there exists positive integers such that non of them divides another one, but between every three of them, one divides the sum of the other two.Proposed by Morteza Saghafian
This post has been edited 1 time. Last edited by bgn, Apr 23, 2017, 1:18 PM
.
elements.We'll prove a lemma.
are in these set of numbers, then 
.
be 4 numbers in that set. Then, 
and 
, also 
as desired. 
be the element such that the max power of 
that divides 
,
and 
)
where 
is odd. Consider another element 
with 
where 
is odd. Also, 
as 
show that 
.
is an example for 
.
?
.
.
.
,
,
.
s owe can get that 
.
,
,
if and only if 
.
so 
and 
,
.
since there are only two evens, which are 
since an even cannot divide an odd, but by Lemma 2 this implies 
,
.
for some 
in our set, we still have 
.
.
be the smaller even in our old set. Note that 
and 
so 
.
is even and 
is odd so 
.
,
.
.
where 
,
,
.
is not the number that divides the other two because 
divides at most one of these triples because if it divided both 
and 
,
which is a contradiction since 
.
divide 10-1=9 triples in total, so by pigeonhole one of them divides three of them. Say this number is 
.
.
modulo 
as well. Since 
.
,
,
.
.
,
which would lead to us getting that 
,
after using Lemma 2. Then we have 
,
so 
.
