[/quote]
,
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 at a party. Prove that there are two people such that, of the remaining 
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 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
,
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these 
.
and 
and 
respectively such that 
.
and 
meet at 
.
is the incenter of 
and 
is the orthocenter of 
.
and 
are circles with diameter 
and 
,
and 
.
.
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers 
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 
Prove that
,
inequalities, these given comparisons satisfy the following three constraints:
and 
,
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
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 
