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.
, determine the maximal constant 
satisfying the following condition: for any partition of the set 
into two -element subsets 
and 
, there exist labellings 
and 
of and , respectively, such that
be a positive integer and 
be positive real numbers such that 
Prove that 
+1 w
be prime numbers such that their sum isn't divisible by 
. Find the all 
positive integer quadruples satisfy:
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and -median of 
are concurrent.
+1 w
such that 
divides 
. be a triangle with circumcenter 
and circumcircle 
. Let 
be the foot of the altitude from to 
. Let 
and 
be points on the circumcircles of triangles 
and 
, respectively, such that , , and are collinear. Prove that if the circumcircle of triangle 
is tangent to at 
, then 
.
. Find all functions 
such that for all 
,![\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]](http://latex.artofproblemsolving.com/f/9/3/f93b7526ff4338eea89a1c59bc009aa001ffc15d.png)
Prove that: 
is the sum of square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
be triangle with point 
and 
on 
and 
are on the same side of 
be midpoint of segment and 
be midpoint of segment be intersection of 
with 
and 
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 .
with incenter 
, is a point on ,the bisector of 
meet 
at .Prove that 
Also great title.
is odd. 
is even and 
is not even. Thus is even. If 
, its clear that 
, by second and thirth condition 
, contradiction. So 
. If 
so 
. And by second condition 
is a divisor too. If 
by firs t condition we get 
and 
, if 
is not a divisor and 
is, we get 
and , if 
and doesnt, 
and 
but is not a divisor. if and both are not divisors, we get 
and 
and we know that 
is a prime thus it must be 
. 
. If 
Let 
If 
. By Last two conditions we get 
gives a contradiction. Thus 
so 
are primes. And by is even its 
thus we get 
and ![$\Delta=[2p(p-4)]^2-4\cdot 4p^2+4$](http://latex.artofproblemsolving.com/c/e/2/ce28fa3d1cff5202692b30ff64666425d190e3e2.png)
and it can't be perfect square eqsily. Thus theres no solution more. Thus 
.
.
works. The divisors of are 
which indeed satisfy
is a multiple of
is odd, then all of its divisors will be odd, so 
, a contradiction. Thus, is even. Since 
is an odd perfect square, it must satisfy 
, the claim follows. A corollary is that 
.
and .
![\[
(d_6-d_7)^2=n+1-\frac{4n}{d_3}
\]](http://latex.artofproblemsolving.com/d/4/b/d4b5c12a7adf13da6f8271b47e5649f684038353.png)
If 
then the right hand side would be strictly smaller than 
, a contradiction. Thus 
. As 
we must have that . Thus 
and hence . A consequence of the fact that is that 
.
and into the first equation yields![\[
3 + d_4 + d_5 = d_6.
\]](http://latex.artofproblemsolving.com/2/0/7/207a22d94b04115b009f326e57abe5d8d10afa43.png)
We proceed with casework on 
.
(and thus a prime smaller than )
, skipping .
. But 
.
, but .
is even
, but then 
, a multiple of
is odd
. As is the first odd divisor, it must be prime so either 
or 
. The only possibility is 
giving 
. Then 
, as desired.
. We start with the following observation. and . were odd then 
and are all odd. But then,![\[d_7 = 2d_2+d_4+d_5 \equiv 0 + 1 + 1 \equiv 0 \pmod{2}\]](http://latex.artofproblemsolving.com/b/e/a/bea61628b07c20ceed6747b2e236443a7a26c702.png)
which is a contradiction. Thus, is even which implies . Further, since is even,![\[(d_6+d_7)^2 = n+1 \equiv 1 \pmod{2}\]](http://latex.artofproblemsolving.com/2/a/a/2aa0ae016da690d73ede18945bd8b7bcc71fcba2.png)
so 
is even. Since it is well known that the square of an odd integer is always 
it follows that is divisible by . Thus, 
so or . Note that if ,
But by AM-GM we have, 
which is a clear contradiction. Thus, which proves the claim. holds for any satisfying the given conditions., letting 
we have
which implies as claimed.
and the square of an integer is always 
or 
, since the only possibility is 
so 
which implies 
.
, is a factor of . Further since , 
. Thus, is one of 
, , 
and . These will be our four cases.
. We then require 
, 
, 
and , which is a contradiction since .
. But this is a contradiction since 
.
. Then we require, or 
(since 
). If , then 
so . This immediately gives 
which is a contradiction since this is not divisible by . If , then we have 
and which is a contradiction since .. Here we have two possibilities. If , then 
and which is divisible by 3 which is impossible. If there exists some odd prime 
, 
so 
. Since 
. So 
or 
., 
which is impossible since .
, 
so 
. Since no multiple of proceeds we must have . So and 
. Then, 
. It now remains to write out the divisors of and check that all the conditions are met, which is easy enough.
is even it's contradiction. I mean 
and .
and 
,we can see the only solution is
and it is contradiction.
and 
. So 
.İt means 
.
but 
.İt is contradiction. we have 2 option. Firstly 
but 
is contradiction. Secondly 
but is contradiction. I mean 
. So 
. If 
but is contradiction. I mean 
. The only solution is .
![\[\sum_{k=1}^n\frac{\phi(k)}k\ln\frac{2^k}{2^k-1}\ge 1-\frac 1{2^n}.\]](http://latex.artofproblemsolving.com/3/d/3/3d32207efff8052214e65aca2505c9e4e7bb581c.png)
and 
Find the minimum 
denote all positive divisors of 
,
even, which means 
,
.
is odd, and 
is odd, which means 
.
,
,
,
,
.
.
is 
.
,
fails by checking. If 
,
fails by checking.
,
,
.
,
,
,
,
,
,
.
,
,
,
,
,
,
.
,
s. Then, 
is odd and a perfect square, it must be in the form 
hence 
.
,
.
is 
but 
,
,
so 
but 
,
so 
but 
also divides 
but 
.
where 
is a prime number greater than 
is an odd number so 
is even. Then, 
.
so 
.
is odd. But 
.
.
then 
and 
then 
so 
that gives us 
and 
but 
.
when 
that means 
but 
when 
and 
we can get 
and 
.
and 
then 
and 
.
.
.
.
then 
when 
then 
and 
.
therefore
That means 
