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 real number such that 
Prove that
Let 
be real number . Prove that
, 
, 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 
. )
be a triangle. The points 
and 
lie on the segments 
and 
respectively, such that the lines 
and 
intersect in a common point. Prove that it is possible to choose two of the triangles 
and 
whose inradii sum up to at least the inradius of the triangle .
and 
be two positive integer numbers such that 
.Prove that 
divides 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a point interior to triangle (with 
). The lines 
, 
and 
meet again its circumcircle 
at 
, 
, respectively . The tangent line at 
to meets the line 
at 
. Show that from 
follows 
.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of .
of positive integers such that 
is an integer.
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist.
, its circumcircle, and its incenter are drawn on the plane. Construct the circumcenter of using only a ruler.
such that:
is strictly increasing,
such that 
for all 
,
, there exists 
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
are equidistant from , so must pass through the midpoint of 
. In the same way we conclude that must pass through the midpoint of , while 
pass through the midpoint of 
. If 
or 
we're done: or is the diagonal we're looking for; if, on the other hand, we assume that 
and 
, from the observation above it follows that is the common midpoint of the diagonals 
, contradicting .
or 
then the conclusion follows. Then assume the contrary and note that the equality of the areas of 
and 
implies that 
or equivalently 
. Thus 
which means that 
or 
and 
are collinear. In the same fashion we get that and 
are collinear and hence and are collinear which means 
. In such a way we get a contradiction.
and 
Prove that
satisfying
Prove
there is a point 
have equal areas. Prove that the area of