has an incenter 
.
be the midpoint of 
.
.
meets 
again at point 
.
and 
.
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
.
.
be a triangle. Let 
and 
be the intersections of the 
and 
angle bisectors with the opposite sides. Let 
.
where 
is the major 
arc midpiont.
and 
again at 
or 
and 
concur on 
be a positive integer. Alex plays on a row of 9 squares as follows. Initially, all squares are empty. In each turn, Alex must perform exactly one of the following moves:
Choose a number of the form 
,
a non-negative integer, and place it in an empty square.
Choose two (not necessarily consecutive) squares containing the same number, say 
and erase the number in the other square.
,
and circumcentre 
.
and 
lie on 
and 
.
meet 
at 
meet 
and 
have an intersection lie on line 
.
be collinear points (in order) and 
a point in plane. Consider the disc 
of center 
,
.![$\mathcal{D}\cap [AC]$](http://latex.artofproblemsolving.com/0/5/6/0569d4130943000380a87b72be211965b5ff02ad.png)
is either the empty set or a segment of length at most 
.![$P(X)\in\mathbb{C}[X]$](http://latex.artofproblemsolving.com/9/9/e/99e8cde28b0b9bc9414425ebc30fa2dc0386fb37.png)
be a polynomial of degree ![\[\sup_{x\in[0,1]}|P(x)|\le(2n+1)^{n+1}\int\limits_{0}^{1}|P(x)|\mathrm{d}x.\]](http://latex.artofproblemsolving.com/9/1/6/91647cd75888c5e9d291280c8ae3f199cc6d1bfd.png)
![$f \colon [0,1] \to \mathbb{R} $](http://latex.artofproblemsolving.com/2/1/8/21859d37ed035abd07d0be112f49d52598fa477e.png)
be a differentiable function such that its derivative is an integrable function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
,
.![\[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]](http://latex.artofproblemsolving.com/2/3/e/23e33eed39fc1e20486bbb86261d3566e8fd2017.png)
be an odd prime number, and 
be an odd number not divisible by 
be a field with 
elements, and 
be the set of elements of 
,
.
has at least 
.
and 
be a continuous function for which there exists an antiderivative 
,
,
,
.
for all 
be a positive integer, ![$g \in \mathbb{R}[X]$](http://latex.artofproblemsolving.com/0/1/e/01e7ebcebcb9f16935716ee859f7acfaa66c29d4.png)
,
be a polynomial with all of its roots being real, and 
for any 
for all 
be two matrices such that 
.
;
,
statements are equivalent:
is differentiable, with continuous first derivative.
and for any two sequences 
,
,
for any positive integer 
is convergent.
matrices which has only 
and 
,
:![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
![$f\colon[0,1]\rightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/f/d/b/fdb1fb0216ab9d17007b1b7b03d20d1060b88531.png)
be a continuous function. Suppose that for each 
,![\[f_t\colon[0,1-t]\rightarrow\mathbb{R}, f_t(x)=f(x+t)-f(x)\]](http://latex.artofproblemsolving.com/4/f/c/4fc6a66d0600a999c000136a26ea23dfcc269cc3.png)
has an unique global minimum point, which we will denote by 
.
,
is constant zero.
Y by LordofAngaband, hwl0304, Davi-8191, anantmudgal09, Ankoganit, Wizard_32, magicarrow, The_Maitreyo1, OlympusHero, myh2910, mathleticguyyy, Lcz, megarnie, squareroot12621, ihatemath123, Deinoncyus_Ban, Adventure10, Mango247, MathLuis, axsolers_24, zhenghua, oVlad, and 11 other users
Let 
be a finite set of at least two points in the plane. Assume that no three points of 
are collinear. A windmill is a process that starts with a line 
going through a single point 
. The line rotates clockwise about the pivot 
until the first time that the line meets some other point belonging to . This point, 
, takes over as the new pivot, and the line now rotates clockwise about , until it next meets a point of . This process continues indefinitely.
Show that we can choose a point in and a line going through such that the resulting windmill uses each point of as a pivot infinitely many times.
Proposed by Geoffrey Smith, United Kingdom
be a finite set of at least two points in the plane. Assume that no three points of 
are collinear. A windmill is a process that starts with a line 
going through a single point 
. The line rotates clockwise about the pivot 
until the first time that the line meets some other point belonging to . This point, 
, takes over as the new pivot, and the line now rotates clockwise about , until it next meets a point of . This process continues indefinitely.Show that we can choose a point
in and a line going through such that the resulting windmill uses each point of as a pivot infinitely many times.Proposed by Geoffrey Smith, United Kingdom
points in the plane, no three collinear, prove that two of them determine a line that separates 
out of 
,
and 0, 1 or 2 points left. Now we choose the line 
such that it contains points inside 
.
then the line l will always have points inside that polygon because the only way to "get out" of the polygon ethis meeting two consecutive points 
and 
of the polygon and leaving the polygon but if it touches 
such that the line will meet 
and 
at the same time infinitely many times (after the second time it meets them the same movements from the first time start again leading to a cicle) and in that cicle every point is met at least once (that's not very hard to prove)
will be met at least once in the process. So we start with the process until one of the 1 or 2 points inside 
and 
where the line has the same position every point inside 
of points from 
)
that to each point 
(called anchor) corresponds some distinguished line 
;
pair; for a square containing a point (other than its centre) we can use a 
pair. However, a pair known to exist for all points of 
,
pair of number of points situated on their right, respectively left sides. Therefore, since a windmill reaches (infinitely often) all orientations for its lines, and its lines are always around some pivot, the windmill line having the orientation of any of the distinguished lines must have as pivot that line's anchor point.
be the smallest. After each operation, the angle made by the new line with the original increases by 
there exists a state of the line 
and 
denote two lines
.
points on its side and 
,
for all 
onto two sets, whose cardinalities differ by at most 
Fix starting position 
of 
there are moments, when it in white half-plane and in black one, therefore 
.
that 
into two sets/regions B and W colored black and white, s.t. 
,
When both of the points are on the windmill line, the points on one side decreases by 1, and stays the same on the other side of the line. When 
if the pivot point changes from 
in the cycle, there is a line which is a windmill, and is parallel to 
there will be a windmill through 
