[/quote]
,
and 
with real coefficients satisfying
be reals such that 
Show that
Let 
and 
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
.
and 
of 
are said to be friends if the following conditions are true:
elements.[/quote]
be a circle on the plane. Let 
and 
be circles which are internally tangent to 
and 
respectively and let the intersection points of 
and 
.
.
and 
respectively. Let the second intersection point of the line 
and 
and let the second intersection point of the circumcircle of 
and 
be 
and let the intersection points of 
and 
be 
.
be a scalene triangle with incenter 
and incircle 
be 
respectively. Let the line 
intersect the circumcircle of 
.
lies between the points 
and 
.
be a circle that passes through 
and that is tangent to 
which lies on different semi-planes with 
with respect to the line 
at points 
.
and 
lies on the circumcircle of 
and 
lies on 
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 
.
represent the number of unit squares that have exactly 
edges on the boundary of 
such that the inequality 
holds for each polygon constructed with these conditions.
such that, for all 
,
is defined as, 
and for all 
,
then 
.
.
,
is also an arithmetic sequence, show that the sequence 
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 
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 
