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.
such that for all 
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 
.
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.
, construct a graph 
on 
, where 
are adjacent if and only if divides 
. Prove that is disconnected for infinitely many
(
) are positive reals satisfying
be positive real numbers such that
inequality and 
for all real numbers 
and 
,
and 
to get: 
Multiply by the denominator to get:
But 
, from which the desired inequality follows
(QM-AM),
(Consequence of AM-GM).
.
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 :
Prove that
simply 
(?). Which is equal to 
,as, 
.
Now to finish notice that 
.![\[\frac{xy+yz+zx}{xyz}\leq \frac{x^2+y^2+z^2}{xyz} = \frac{2(a^2+b^2+c^2)}{xyz}=\frac{1}{2xyz}.\]](http://latex.artofproblemsolving.com/e/4/5/e45703e33db02b679950ae36f1b44fa0c1c49299.png)
We now wish to prove that 
,
and its cyclic forms.
,
.![\[\sum_{cyc} \frac{1}{\sqrt{a^2+b^2}} \le \sum_{cyc} \frac{\sqrt{2}}{a+b}.\]](http://latex.artofproblemsolving.com/4/4/f/44f5c5dea38147d48b8a2d808d5803050c705f44.png)
Thus, it suffices to show that ![\[\sum_{cyc} \frac{1}{a+b} \le \frac{1}{(a+b)(b+c)(b+c)}.\]](http://latex.artofproblemsolving.com/3/1/5/31501a383f9cdf2b14d3022f081ac35f8d0e4b68.png)
Clearing the fractions and expanding, this is equivalent to 
.
,
,
and likewise for the others. The desired inequality then turns into 
We also have 
by AMQM again, so the desired conclusion follows.
