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.
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)
. Let the perpendicular line from 
to 
meets 
at points 
, respectively, and the foot from 
to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at 
.
is tangent to 
.
+1 w
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
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?
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.
, and express these only in 
, that is simple because they are only medians and the result follows.
be the centroid of triangle 
, 
the midpoint of 
and 
the midpoint of 
.
, so 
. This implies that 
if and only if 
is orthocenter of trianble 
. This is equivalent to 
. On the other hand, is orthocenter and is centroid of triangle 
. So, if and only if 
is median and altitude from in triangle . This is equivalent to 
, and to 
.
as the origin of the complex plane, and let 
be the coordinates of points 
respectively. Then we have ![\[d=\frac{a+b}{2} \text{ and } e=\frac{a+c+d}{3}=\frac{3a+b+2c}{6}.\]](http://latex.artofproblemsolving.com/8/9/1/8912f2c3e6900cc680fe60c6d132b803830a99fe.png)
Using the real product of complex numbers, if 
is the circumradius of triangle , then we have ![\[a\cdot a=b\cdot b=c\cdot c = R^2.\]](http://latex.artofproblemsolving.com/c/8/9/c899af10e64c453722ecc78b0c941e66cae26454.png)
Also note that 
, i.e. ![\[(a+b-2c)\cdot (3a+b+2c)=0.\]](http://latex.artofproblemsolving.com/f/2/8/f28b3b49b839c2b1ff1fc1b249382c42929499fe.png)
The last relation is equivalent to ![\[3a\cdot a+a\cdot b+2a\cdot c+3a\cdot b+b\cdot b+2b\cdot c-6a\cdot c-2b\cdot c - 4c\cdot c=0,\]](http://latex.artofproblemsolving.com/1/7/8/178640cb748434659678f80cb81cbb9edeb7a853.png)
which implies ![\[a\cdot b=a\cdot c (*)\]](http://latex.artofproblemsolving.com/e/2/8/e283699086fda0be1832ae8f3ce4aefaf91972db.png)
Now, observe that ![\[AB=AC \iff |b-a|^2=|c-a|^2.\]](http://latex.artofproblemsolving.com/0/4/a/04ab5ae490e38a0f82c4c0fc5749a96282b06541.png)
This yields ![\[(b-a)\cdot(b-a)=(c-a)\cdot(c-a)\]](http://latex.artofproblemsolving.com/5/9/e/59e147df80214cb700ad2b6e4d7509b62ec055de.png)
or ![\[b\cdot b-2a\cdot b+a\cdot a=c\cdot c-2a\cdot c+a\cdot c.\]](http://latex.artofproblemsolving.com/f/9/a/f9a8df43ac573f47f9e7b35fd608d6e60d387600.png)
Hence, we have ![\[a\cdot b=a\cdot c (**)\]](http://latex.artofproblemsolving.com/1/4/5/145b86d56e3780ff477becf3e8da3222989ca292.png)
From 
and 
we have 
. 
case, the other one can be done similarly (I swear I didn't read #2 )
, is its circumcenter, 
is the midpoint of and 
is the centroid of the triangle 
. Show that the lines 
and 
are perpendicular if and only if .
, where , let 
be the midpoints of 
. Let 
be a point on 
, such, 
. Let be the orthocenter of . Then, 
Back to the Problem: Let 
be midpoints of 
, then 
and 
are parallelogram. If 
and 
, then, 
are midpoints of , . 
and 
are concurrent, say at 
, which is centroid of and 
. 
is cyclic 
is orthocenter of . Now apply (Lemma)
and 
are perpendicular, then 
and write Pythagoras.
be the midpoint of and be the midpoint of .
therefore 
Let 
.
, as desired. 
and 
and 
.
for reals 
.
,
.
so 
.
is just 
.
is 
.
.
.
where 
.
.
.
and the slope of 
.
,
or
.
or 
.
which is equivalent to 
Notice that 
is pure imaginary. Or
Notice that 
,
and 
.
only if 
,
Now we prove the iff statement. Note that if
Then
If 
,
, or 
.
.
will imply perpendicularity. 
.
.
So we are done