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.
are on the blackboard. Amelie and Boris take turns removing one of those until only two numbers remain on the board. Amelie starts. If the sum of the last two numbers is divisible by 
, then Amelie wins. Else Boris wins. Who can force a victory?
and 
for which there are three consecutive integers at which the polynomial ![\[ P(n) = \frac{n^5+a}{b} \]](http://latex.artofproblemsolving.com/9/7/5/975d942c43c8962888ebe4739c23492f0c3488ba.png)
takes integer values.
be an integer greater than or equal to 2. Consider the sequence such that its first term is , and for 
, the 
-th term of the sequence, we have![\[
a_{n+1} = \frac{a_n}{p_k^{e_k - 1}} + 1,
\]](http://latex.artofproblemsolving.com/3/1/b/31b670eef4898a72b4f0de3bc766e8db83fbcc0e.png)
where 
is the prime factorization of , with 
, and 
positive integers.
, the next two terms of the sequence are![\[
a_2 = \frac{a_1}{23^{1-1}} + 1 = \frac{2024}{1} + 1 = 2025 = 3^4 \cdot 5^2;
\]](http://latex.artofproblemsolving.com/0/7/2/072dd6b8030ad5d6f880150ce81fb31ed4ca2845.png)
![\[
a_3 = \frac{a_2}{5^{2-1}} + 1 = \frac{2025}{5} + 1 = 406.
\]](http://latex.artofproblemsolving.com/d/2/a/d2a9d7fcb1818ee4a4f95e16c9297312f724a1d8.png)
the sequence is eventually periodic and what all the possible periods are.
be a positive integer. A sequence 
is eventually periodic with period if is the smallest positive integer such that there exists an 
satisfying 
for all 
.
. Find all 
such that there exist non-negative integer 
satisfied
P/s:
such that 
(
is sum of digit of in base 10)
with its incircle 
. Let 
and 
be the points of tangency of with 
and 
respectively. On the edges 
and 
, take points 
and 
such that 
is tangent to at 
. Suppose 
is the intersection point of the lines 
and 
. Prove that two lines 
and 
are parallel or coincide.
be a triangle with 
. A circle 
is tangent to sides and , and 
is the midpoint of 
. Points and lie on sides and , respectively, such that segment 
is tangent to circle at point . Let and 
be the orthocenters of triangles 
and 
, respectively. Prove that line 
bisects segment 
.
, construct a graph 
on 
, where 
are adjacent if and only if divides 
. Prove that is disconnected for infinitely many
be a triangle with points and lying on rays and , respectively, such that 
. On the line , choose points 
and 
such that 
and 
. Let and 
be the feet of the perpendiculars from and to line , respectively. Let 
be the intersection point of lines 
and 
. Prove that the reflection of over line lies on the line connecting 
and the circumcenter of triangle .
with 
and point 
on side 
then 
is equivalent to 
.
be the midpoint of arc not containing 
. Then 
are collinear, the triangle 
is isosceles and from the Ptolemy Theorem we have: 
(since it is well known that 
). Now 
is equivalent to 
or 
or 
or 
.
.
Where 
are one of such 
such that![\[
f(x^4 + 5y^4 + 10z^4) = f(x)^4 + 5f(y)^4 + 10f(z)^4
\]](http://latex.artofproblemsolving.com/3/0/8/308079e75e1acd9a9f896885387b67a7b4020c16.png)
for all 
.
the following equation :
the incenter and by 
the circumcenter of the triangle 
holds if and only if 
.
![\[ \angle AIO\le\frac{\pi}{2}\Leftrightarrow\cos AIO\ge 0\Leftrightarrow IA^{2}+IO^{2}-OA^{2}\le 0\Leftrightarrow \frac{r^{2}}{\sin^{2}\frac{A}{2}}+R^{2}-2Rr-R^{2}\ge 0\Leftrightarrow \sin^{2}\frac{A}{2}\le \frac{r}{2R}\Leftrightarrow \frac{(s-b)(s-c)}{bc}\le \frac{2S^{2}}{sabc}=\frac{2s(s-a)(s-b)(s-c)}{sabc}\Leftrightarrow 2(s-a)\ge a\Leftrightarrow b+c\ge 2a \]](http://latex.artofproblemsolving.com/6/c/2/6c2f7d1d6dc325489d4d1bf6ed9a828236530d55.png)
and 