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 positive integer. Show that if there exists a sequence 
of integers satisfying the condition ![\[a_n=\frac{a_{n-1}+n^k}{n}\text{ for all } n\geq 1,\]](http://latex.artofproblemsolving.com/9/7/3/973a8947bb62b9356852e9170699460216652c92.png)
then 
is divisible by 
.
such that 
.
is selected on arc 
not containing 
on 
. 
and 
are the reflections of point 
and 
with respect to 
, respectively. Circumcircles of 
and 
intersect at 
. Prove that 
is colinear
has 
vertices and 
edges, where 
, show that there exists a path with 
vertices. is good if 
for some positive integer 
. Here 
is the sum of all digits of . Prove that:
be a triangle with circumcenter 
, angle bisector with 
and altitude 
with 
. The lines 
and meet at 
. The circumcircle of 
meets 
at 
and 
meets at 
. The circumcircles of triangles 
and meet at 
. Show that 
is a symmedian in 
so that no two exist that their difference equals to 
.
, is a point on such that is the internal bisector of 
. Now Suppose 
=
and 
. Prove that 
.
? (The triangles may not overlap and may not exceed the margins of the square.)
can be tiled with 
congruent right-angled triangles, whose sides are of lengths 
, 
, and 
.![\[ \frac{b^2 \sqrt{3}}{2} \]](http://latex.artofproblemsolving.com/0/5/3/0534d2d3c3c9a4255ad911f638395ac8bbdbad27.png)
![\[
S_{\text{sq}} = k \cdot \frac{b^2 \sqrt{3}}{2}. \tag{1}
\]](http://latex.artofproblemsolving.com/c/d/8/cd85edeb4f81a5d97aff694c7b08606b151e0b98.png)
, is obtained by the contribution of an integer number of lengths , , and , hence![\[
a = m b \sqrt{3} + n b,
\]](http://latex.artofproblemsolving.com/3/d/1/3d1848972199602a32d9a6a8903e6e91b521a842.png)
where 
, and at least one of the numbers 
and 
is different from zero.![\[
S_{\text{sq}} = a^2 = \left(m b \sqrt{3} + n b\right)^2 = b^2 \left(3m^2 + n^2 + 2 \sqrt{3} m n\right). \tag{2}
\]](http://latex.artofproblemsolving.com/0/7/b/07b4a34ff9d1c9cbace32383f66c8ed22b6d40f6.png)
![\[
3m^2 + n^2 + 2 \sqrt{3} m n = k \cdot \frac{\sqrt{3}}{2},
\]](http://latex.artofproblemsolving.com/f/d/8/fd8cfc1180c70e32577631eab0b11511a5d1e5b3.png)
i.e.,![\[
6m^2 + 2n^2 = (k - 4mn) \sqrt{3}. \tag{3}
\]](http://latex.artofproblemsolving.com/c/4/d/c4dcb8e4abb426ce0ca323c336f6e62915df1b95.png)
and from equality (3), it follows that 
.![\[
\sqrt{3} = \frac{6m^2 + 2n^2}{k - 4mn},
\]](http://latex.artofproblemsolving.com/f/c/4/fc47449a7d0ec56473243683c71d56946a491565.png)
which contradicts the fact that 
is irrational, because 
is a rational number.
be an increasing sequence of positive integer such that 
.
there exists 
Prove that
Where 
and 
Prove that
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
