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.
marbles in 
colors such that there are exactly 
marbles of each color. She wants to arrange them on 
shelves, 
marbles on each shelf such that for any 
colors there is a shelf that has marbles of those colors.
and 
Find the maximum value of 
be triangle with 
and 
on the segment 
such that 
. Consider point 
on 
such that 
. Point 
is the intersection of tangent line through point to circle 
respectively . If the tangent through point to 
intersect 
at 
. Line 
intersect 
at 
. Prove that 
bisect 
.
and 
Find the maximum value of
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
Prove that
with 
, point 
lies on the circumcircle of such that 
. The line through parallel to 
intersects 
in 
and in 
. Prove that the center of the circumcircle of triangle 
lies on the circumcircle of triangle 
.
be a rearrangement of the integers from 1 to 2025.
, define 
to be the number of integers 
such that 
and 
.
are given such that 
and 
for all .
satisfying this condition. be a acute triangle with 
. 
are the orthocenter and circumcenter. Let be a point on the same side of 
as 
, such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be the midpoint of . Prove that 
are collinear.
, show that![\[\left(1+\frac{a}{b}\right)\left(1+\frac{a}{b}+\frac{b}{c}\right)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 16\]](http://latex.artofproblemsolving.com/c/1/7/c17cc42f0feafca5ecbeacde7076daa5b2615c60.png)
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction.
is coprime to 
. We know that 
is coprime with 
and 
for all 
, so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but 
cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
,
Prove that
Let 
Prove that
Prove that
Where 
such that for any positive integer 
there exists a positive integer 
are equal to 
