[/quote]
,
such that, for all 
,
,
is defined as, 
and for all 
,
then 
.
.
,
is also an arithmetic sequence, show that the sequence 
has to be constant.
,
are coprime positive integers. Find all 
such that:
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
in term of a function 
such that 
vertices, each edge has one of the colors 
,
,
.
,
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
be given pairwise coprime positive integers where 
.
be positive integers. We call 
to be a grandson of 
for all functions 
,
for all non-negative real numbers 
and 
.
and 
be sequences of real numbers for which 
and
for all positive integers 
for which 
for all 
)
and 
be two concentric circles. Let 
and 
be any two equilateral triangles inscribed in 
and 
are any two points on ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
Prove that
,
,
,
and 
.
to meet the circumcircle of triangle 
at 
.
.
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
Y by kiyoras_2001
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point 
such that the reflection of each of these four points about lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
such that the reflection of each of these four points about lies on the circle passing through three remaining points.Proposed by:A Kuznetsov
This post has been edited 1 time. Last edited by Gengar_in_Galar, Mar 11, 2025, 9:49 AM
be the midpoint of 
,
.
so![\[
\measuredangle M_{AD}E - \measuredangle EM_{AB} = \measuredangle M_{AD}E - \measuredangle EM_{BD} + \measuredangle EM_{BD} - \measuredangle EM_{AB} = \measuredangle M_{AD}M_{CD} - \measuredangle M_{CD}M_{BD} + \measuredangle M_{BD}M_{BC} - \measuredangle M_{AB}M_{BC} = \measuredangle AC - \measuredangle BC + \measuredangle CD - \measuredangle AC = \measuredangle DCB = \measuredangle M_{AD}M_{AC}M_{AB}
\]](http://latex.artofproblemsolving.com/2/8/d/28d53fb59825d330e4ee9932929be2f947be0cdd.png)
so 
and we are done by symmetry.
the “vertex–circle”
![\[
\omega_A=(BCD),\quad \omega_B=(ACD),\quad \omega_C=(ABD),\quad \omega_D=(ABC).
\]](http://latex.artofproblemsolving.com/e/6/9/e69f5f0e3be32f018dbd090c3ed13b0680ebb779.png)
Next, for each vertex 
with center 
.
to the midpoint of 
.![\[
\omega'_X = h_X(\omega_X).
\]](http://latex.artofproblemsolving.com/6/a/9/6a902f425152ae4195c6e8243636fae8272a04a8.png)
Thus, for example,![\[
\omega'_A \text{ is the circle through } M_{AB},\; M_{AC},\; M_{AD},
\]](http://latex.artofproblemsolving.com/e/1/9/e19ded44149a9c61598b78401ba0bbdd8f777ef1.png)
where 
is the midpoint of 
,
passes through 
;
.
contains the midpoint 
(since 
)
contains 
)
.
lies on 
for some 
;
.
lies on ![\[
X' = 2Z-X
\]](http://latex.artofproblemsolving.com/3/6/8/3688438a9b5c8e73838b6986edf1c6780831a65b.png)
lies on 
then for each vertex 
,
,
and 
.
and 
and in a second point, say 
.
;
and in a second point 
.![\[
h_A(\omega'_A)=\omega_A,\quad h_B(\omega'_B)=\omega_B,\quad h_C(\omega'_C)=\omega_C,
\]](http://latex.artofproblemsolving.com/9/a/b/9abc3b73607f48c2194c5d074cb550997fe29728.png)
and the standard fact about Miquel points in the complete quadrilateral 
(One way to see this is to “lift” the configuration via the homothety of ratio 2 from each vertex. For example, applying the homothety with center 
and a circle through the image of ![\[
Z\in\omega'_A\cap\omega'_B\cap\omega'_C\cap\omega'_D.
\]](http://latex.artofproblemsolving.com/a/2/7/a277556188061b2b23281e61233e4779d075bd08.png)
![\[
Z\in\omega'_X
\]](http://latex.artofproblemsolving.com/3/7/3/37329ce61c4718348d714a26d977e9a5946fa9eb.png)
implies that the point![\[
X' = 2Z - X,
\]](http://latex.artofproblemsolving.com/4/e/b/4eb563151229c7ecea890b50f11d8949904fae72.png)
i.e. the reflection of 
,
,
in 
