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 an acute triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
positive real numbers such that 
. Prove that :
and 
, there cannot exist infinitely many 
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
, we have:![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
, 
must be an integer, which is not possible..
be a sequence of positive integers with 
, 
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
. Show that if 
is a prime less than 
for some positive integer 
, then there exists 
such that 
.
and be the incircle and incenter of triangle . Let 
, 
, 
be the tangency points of to 
, 
, 
and let 
be the reflection of about . Assume 
intersects the tangents to at and at points 
and 
. Show that 
.
, find all functions 
such that![\[f(x+y+f(y))=f(x)+f(ay)\]](http://latex.artofproblemsolving.com/7/5/d/75d1ac71fa0c2e53fdfc8e198b109d18e935fe74.png)
holds for all 
.
of the integers with the following property: for any integer 
there is exactly one solution of 
with 
.
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained? be a finite set of cells in a square grid. On each cell of we place a grasshopper. Each grasshopper can face up, down, left or right. A grasshopper arrangement is Asturian if, when each grasshopper moves one cell forward in the direction in which it faces, each cell of still contains one grasshopper., the number of Asturian arrangements is a perfect square. is the following set:
for all x, y 
0, where is the prime number.
satisfying
for all integers 
.
, and for any positive integer , 
means 
applied times to , and 
means 
applied times to .
the centroid is 
and is the midpoint of 
. The line through parallel to 
meets 
at . Prove that 
if, and only if, 
.
be the intersection of 
and 
, M midpoint of .
.
, or 
cyclic.
, equivalent to isosceles trapezoid, or 
. (Note 
, 
are not parallel.)
, thus equivalent to 
.
, by definition of M. Clearly, 
if and only if 
.
be a midpoint of .If we have 
. As then 
which implies 
thus 
is cyclic . As is cyclic and 
parallel to we have 
thus 
so 
. Second case is just working backwards.
