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 real numbers such that 
. Show that
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 an acute-angled, non-isosceles triangle with circumcenter 
and incenter 
, such that![\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]](http://latex.artofproblemsolving.com/7/8/1/78125759bf5c73149e4a56538e9b0c44309704a7.png)
, 
, and 
.
, 
, and 
be the excenters opposite to vertices 
, 
, and 
, respectively, and let 
, 
, and 
be the Lemoine points of triangles 
, 
, and 
, respectively.
, 
, and 
all pass through a common point 
. Moreover, the isogonal conjugate of with respect to lies on the line 
.
intersect at point 
. The points of tangency of the circumcircles of triangles 
and 
with their common external tangents lie on a circle 
. The points of tangency of the circumcircles of triangles 
and 
with their common external tangents lie on a circle 
. Prove that the centers of circles and coincide.
grid, taking turns. Petya goes first. On his turn, a player writes an uppercase Russian letter in an empty cell (each cell can contain only one letter). When all cells are filled, Petya is declared the winner if there are four consecutive cells horizontally spelling the word ``ПЕТЯ'' (PETYA) from left to right, or four consecutive cells vertically spelling ``ПЕТЯ'' from top to bottom. Can Petya guarantee a win regardless of Vasya's moves?
, point is marked. Point 
is on segment 
, and point 
is on segment 
such that the circumcircles of triangles 
and 
are tangent to line 
. Lines are drawn through points and passing through the center of the circumcircle of triangle 
, and through points and passing through the center of the circumcircle of triangle 
. Prove that there exists a circle tangent to all four drawn lines.
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than 
. For every group of 
marked points, he calculated the sum of the numbers on all 
connecting segments. It turned out that the absolute value of each such sum is at least , and there are both positive and negative such sums. What is the maximum possible value of ?
pairwise distinct complex numbers are written on a board. It turns out that there are exactly 760 ways to choose two numbers and 
from the board such that:![\[
a^2 + b^2 + 1 = 2ab
\]](http://latex.artofproblemsolving.com/8/f/c/8fc5df8dbc8e1647edfaa96dcfa8a0dfdac0c961.png)
Ways that differ by the order of selection are considered the same. Prove that there exist two numbers 
and 
from the board such that:![\[
c^2 + d^2 + 2025 = 2cd
\]](http://latex.artofproblemsolving.com/2/7/0/2706d72797c5e392b3f068a8174bd257e4d93fa1.png)
is given. It is known that triangles 
, 
, 
, and are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle , lie on the same sphere.
and 
with integer coefficients is called 
if from the divisibility of both differences 
and 
by 
, it follows that both 
and 
are divisible by 100. Does there exist such an important pair of polynomials 
, 
, such that the pair 
and 
is also important?
and 
of triangle 
intersect sides 
and 
in points 
and 
, respectively. Let points 
and 
be foots of perpendiculars from point 
on lines 
and 
, respectively. Prove that 
and of triangle intersect sides and in points and , respectively. Let points and be foots of perpendiculars from point on lines and , respectively. Prove that 
be perpendicular to CI through I. 
. Thus is tangent to 
. Note 
is diameter of 
. Thus is tangent to . By Reim, we are done
and of triangle intersect sides and in points and , respectively. Let points and be foots of perpendiculars from point on lines and , respectively. Prove that 
extended meet at 
. Now, 
become the mid points of 
. Hence we are done by mid point theorem.
to 
, now its enough to prove 
is cyclic. we have 
is cyclic (where 
is the incenter of 
)therefore,
. Thus we only need to prove 
is cyclic (where 
is the intersection point of and 
), which will be true if we have 
which is completely clear. 
and 
and 
.
