[/quote]
,
satisfies
for some 
and for all 
,
of the functional equation 
for all 
for all 
.
be distinct polynomials of degree 
with non-zero coefficients. Suppose that they have 
common real roots counting multiplicity and 
common coefficients. Determine the maximum possible value of 
.
such that:
is strictly increasing,
such that 
for all 
,
,
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
Prove that
Prove that
Where 
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
be real numbers such that 
and 
Prove that
Where 
divides 
.
workers stand in a row. Every contiguous group of at least 
workers is a 
.
.
be a fixed acute-angled triangle. Consider some points 
and 
,
be the midpoint of 
. Let the perpendicular bisector of 
at 
,
intersect the lines 
and 
, respectively. We call the pair 
,
is cyclic.
and 
are interesting. Prove that 
and 
Prove that the sum of the four prime numbers is divisible by 
.
and 
and one starting number from the set 
used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number 
is reached or exceeded.
Show that Bob cannot win.
Y by Adventure10, Mango247
Let 
be an acute angled triangle and let 
be its circumscribed circle. A point 
is given in the interior of the triangle, such that 
, where 
and 
are on and lies on 
while lies on 
. Prove that is on the angle bisector of 
if and only if 
.
be an acute angled triangle and let 
be its circumscribed circle. A point 
is given in the interior of the triangle, such that 
, where 
and 
are on and lies on 
while lies on 
. Prove that is on the angle bisector of 
if and only if 
.
since 
is bisector of 
and is easy to see that: 
since 
since 
and 
