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 a degree 6 polynomial with integer coefficients. Prove that there exists a real number ![$a \in [0,3]$](http://latex.artofproblemsolving.com/c/e/0/ce048e201e2d842cb846be9a98a64da6bd47a364.png)
such that![\[
|P(a)| \geq \frac{45}{64}.
\]](http://latex.artofproblemsolving.com/1/7/5/175d3c5451574d61d0d4923d2b9bc240ec3935fd.png)
with 
, 
acute angles, 
, 
, 
midpoint of 
, 
. Prove that is isosceles.
be an acute triangle, and let 
be its incenter. Let 
be the incircle of triangle 
, which touches 
at points 
, respectively. Let 
intersect again at point 
, and let the circumcircle of triangle 
intersect again at point 
. The circumcircles of triangles 
and 
intersect again at a point 
. Let 
be the midpoint of arc 
. Prove that points 
are collinear.
board, we call a bar each segment on the border of the board that is adjacent to some cell. The 
bars are numbered from 
to (as shown in the figure). bars is formed. For example, in the figure below, the pairing is 
.
board, how many possible pairings are there?![\[
(1,10), (2,5), (3,4), (6,7), (8,9)
\quad \text{and} \quad
(2,5), (7,6), (9,8), (10,1), (3,4).
\]](http://latex.artofproblemsolving.com/5/3/f/53f2327bbb39da69ead41167d22589311b7cc5c4.png)
, let 
be the largest prime number that divides 
. For 
, define 
.
such that the numbers 
, 
, and 
are all less than 
.
, prove that there exists a positive integer such that the numbers , , and are all less than 
.
be a positive integer. Let 
, 
, 
, 
be a sequence of positive integers such that ![\[\operatorname{gcd}(a_1,a_2),\,\operatorname{gcd}(a_2,a_3),\,\dots,\,\operatorname{gcd}(a_{n-1},a_n)\]](http://latex.artofproblemsolving.com/a/d/0/ad0cf276c77f67a7565f695af477a24b9045f6e6.png)
is a strictly increasing sequence. Find, in terms of , the maximum possible value of ![\[\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\]](http://latex.artofproblemsolving.com/8/7/b/87be81cac06a437ab90a3e166dc27c793e890eb1.png)
over all such sequences.
be the circumcircle of triangle 
, and let be the midpoint of side 
. The line 
intersects at point 
(with 
). The circumcircles of triangles 
and 
intersect at point 
(with 
). The circumcircles of triangles 
and 
intersect at point 
(with 
). Prove that the circumcircle of triangle 
is tangent to .
be an altitude of right angled triangle (
). An excircle of triangle 
opposite to 
touches 
at point 
; a point 
is defined similarly. Prove that 
.
such that there exist pairwise distinct positive integers 
with 
for all integers 
.
is 
.
which satisfy:
. Prove that: 
, which is increasing function.
be an integer greater than 
, and let 
be non-negative real numbers with 
. Determine the minimum value of ![\[ \frac{a_1}{a_2^2 + 1}+ \frac{a_2}{a^2_3 + 1}+ \cdots + \frac{a_n}{a^2_1 + 1}.\]](http://latex.artofproblemsolving.com/1/3/7/1374198f13a8e0a8c3507c5f086c69ad3a719b3b.png)
with 
, we have![\[b-(a+1)<|a(b+1)-(a+1)b|<(b+1)-a.\]](http://latex.artofproblemsolving.com/b/0/c/b0c096aaf4306ae0a815983cfa3305ec41cada2d.png)
It suffices to show that there are at least 
pairs 
with the four numbers 
, 
, 
, 
in separate groups.
pairs of consecutive numbers that are located in different groups, and of the numbers that are in at least 1 such pair, 
are in the first group, 
in the second, and so on (We have 
). Now since there are 
pairs of consecutive integers, and at most 
in each group, then 
. with the four numbers , , , in separate groups is at least ![\[{k\choose2}-\left(\sum_{i=1}^n{{k_i\choose2}+k_i}\right)\geq \frac{k(k-3)}2-\sum_{i=1}^n{k_i\choose2}.\]](http://latex.artofproblemsolving.com/6/e/c/6ec49743e24a002ee5a0bd725e61dd0dcff1d1c9.png)
If 
, then we have![\[\frac{k(k-3)}2-\sum_{i=1}^n{k_i\choose2}\geq \frac{(n+2)(n-1)}2-n\cdot{4\choose2}>\frac{(n-6)^2}2\text{ for all }n>100.\]](http://latex.artofproblemsolving.com/3/6/9/3693df716a547097e447f2d1f7946fb3425a39e8.png)
Now if 
, notice that 
for non-negative integers 
with 
. So, we have 
as required.
with the four numbers , , , in separate groups is at least If , then we have
here
,
.
be an integer. The numbers 
are split into 
.
quadruples 
such that they are all in different groups, 
and 
.
quadruples for 
