AoPS Academy
+4 w
on the board gets replaced by 
, where 
is the largest prime divisor of .
be triangle with incenter 
and circumcircle 
. Let 
and 
, the line parallel to 
through cuts 
, 
in 
and 
. Prove that the circumradius of 
and 
are equal.
, with 
, is inscribed in the circle 
. Let be a variable point on the arc 
that does not contain 
, and let 
and 
denote the incenters of triangles 
and 
, respectively. varies, the circumcircle of triangle 
passes through a fixed point.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to , and let be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
for which the sum ![\[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\]](http://latex.artofproblemsolving.com/6/b/0/6b010978a55642b7f7aabb2df2f7cf45c9db937a.png)
is an integer strictly between 
and 
. For that unique , find 
.
denotes the greatest integer that is less than or equal to 
.)
denotes the sum of the positive integer divisors of 
Let be the least positive integer such that 
is divisible by 
for all positive integers Find the sum of the prime factors in the prime factorization of 
to deduce all of them are not divisible by 
denote the k-th symmetric sum and 
that 
and by newton sums we have 
and we consider the polynomial
which in 
is 
and the roots are 
so we get 
but 
and we are done
we get 
so as 
we have 
as desired.
, consider the polynomial 
, clearly 
and ![$P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/0/5/a05997bd7ba17001c62bc17ae35c76dd78e24789.png)
. And therefore we can let 
where all 
's are integers, so by taking newton sums modulo we have 
which gives that 
therefore by putting 
in this we get that 
for any 
which proves that all 's are congrent modulo and therefore we are done 
.
. Suppose 
of the 's are nonzero and WLOG they are 
. Let 
be a generator of 
. Let 
where 
is an integer for 
. Then the condition becomes 
for all 
. It follows that the elements of are roots of ![$P(x):=x^{\alpha_1}+\cdots+x^{\alpha_r}\in\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/8/b/0/8b0a5b5937b0aff17391fefd8df1603fb6c521a8.png)
. Thus 
is divisible by![\[
\prod_{a\in\mathbb{F}_p^\times}(x-a)=x^{p-1}-1.
\]](http://latex.artofproblemsolving.com/c/9/d/c9d1ccf7462e298113392f4798239e5f502ae2f4.png)
Since 
if is nonzero, 
and thus either 
in which case we are done, or 
and all the 
's are equal, as desired. 
be a prime number and 
be integers for which 
is divisible by 
.
is divisible by 
)
denote the 
-
,
.
(interpreted in ![$\mathbb F_p[t]$](http://latex.artofproblemsolving.com/d/1/f/d1fa992d9fe693e162a54b2121d646af3dd9f0e2.png)
)
,
.
.
But polynomials in 
modulo