[/quote]
,
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
,
,
,
such that 
and 
for 
.
,
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
,
,
,
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be a function such that
for all 
.
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
.
is defined as follows: 
if 
and 
if 
.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
and 
such that 
.
is a benefactor of 
because 
.
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
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?
Y by
Let 
be an acute scalene triangle with circumcircle 
and incenter 
. Suppose the orthocenter 
of 
lies inside . Let 
be the midpoint of the longer arc 
of . Let 
be the midpoint of the shorter arc 
of .
Prove that there exists a circle tangent to at and tangent to the circumcircles of 
and 
.
be an acute scalene triangle with circumcircle 
and incenter 
. Suppose the orthocenter 
of 
lies inside . Let 
be the midpoint of the longer arc 
of . Let 
be the midpoint of the shorter arc 
of .Prove that there exists a circle tangent to
at and tangent to the circumcircles of 
and 
.
resp. 
.
be the intersection of 
.
be the intersection of 
.
and 
be the centers of 
be the intersection of 
and 
.
.
and 
are congruent.
and 
.
,
i.e. we need to show that 
which is another (final) easy angle chase.
and 
implies 
since 
is the orthocenter of 
be midpoint of small arc 
.
are rhombus.
.
.
is a parallel transport of 
.
,
colinear and 
.
because there are a lot of half angles so directing is kind of annoying. Let 
and 
meet 
and 
respectively. For shorthand, let 
Observe that
so it follows that 
and 
lie on 
and 
respectively. Combined with 
this implies that 
Thus,![\[\angle HPL=\angle HRL=\angle CBL=\angle PLH\]](http://latex.artofproblemsolving.com/c/0/9/c0985e5987ef2d91d55a17d754a12f34bb434ad3.png)
and a very similar process also implies that 
so 
and 
share a perpendicular bisector 
and therefore are the parallel sides of an isosceles trapezoid. Finally, since
we find that 
This implies that 
so 
as well and we're done by homothety (the desired circle circumscribes the midpoints of arcs 
on their respective circles). 
,
,
,
be 
,
= 
,
= 
,
= 
,
= 
and 
= 
.
= 
= 
= 
= 
,
,
.
,
because of sine rule. 
= 
= 
=
= 
because 
is the center of 
= 
.
= 
= 
= 
= 
- 
- 
= 
be the point on 
and 
lie on 
and 
are parallelograms because 
= 
and let 
be the circle with center 
.
= 
.
= 
.
= 
and 
at 
and 
.
be midpoint of arc 
circles 
so circles are congruent and since 
and 
intersect on Radical Axis. Let the intersection point be 
is Rhombus.
and 
and 
.
where 
so 
and since 
then 
so 
so 
and we only need to prove that 
are collinear to prove that 
so 
are collinear as wanted.
