[/quote]
,
positive real numbers such that 
. Prove that :
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
.
at points 
,
to 
.
intersects line 
,
intersects line 
,
intersect at 
.
is tangent to 
.
be an acute triangle. Points 
,
lie on a line in this order and satisfy 
.
and 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
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)
,
are even?
and 
be real numbers such that 
.
and 
.
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 
,![\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]](http://latex.artofproblemsolving.com/7/8/1/78125759bf5c73149e4a56538e9b0c44309704a7.png)
,
,
.
,
,
be the excenters opposite to vertices 
,
,
,
,
,
be the Lemoine points of triangles 
,
,
,
,
,
all pass through a common point 
.
.
intersect at point 
.
and 
with their common external tangents lie on a circle 
.
and 
with their common external tangents lie on a circle 
.
Y by
Let 
be an acute triangle with circumcircle 
. A line passing through 
perpendicular to 
meets again at 
. Draw two circles 
, 
with 
as centers and 
, 
as radii, respectively, and they intersect 
, 
at 
respectively. Let 
be the second intersection of 
and , and let , intersect 
, 
at 
, respectively. The circumcenter of triangle 
is 
, prove that 
are collinear.
proposed by Li4
be an acute triangle with circumcircle 
. A line passing through 
perpendicular to 
meets again at 
. Draw two circles 
, 
with 
as centers and 
, 
as radii, respectively, and they intersect 
, 
at 
respectively. Let 
be the second intersection of 
and , and let , intersect 
, 
at 
, respectively. The circumcenter of triangle 
is 
, prove that 
are collinear.proposed by Li4
This post has been edited 2 times. Last edited by parmenides51, Dec 15, 2022, 7:32 PM
is the Miquel point of 
and thus 
and also
so 
if 
then 
and 
is the midpoint of 
and thus the triangles 
are perspective
are collinear. QED
