[/quote]
,
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
is selected inside acute 
so that 
and 
.
is chosen on ray 
so that 
.
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
and 
intersect at 
.
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
,
.
intersects 
.
cuts segment 
at 
respectively. If 
and 
,
equals 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
are points on a circle 
,
,
.
at 
.
is on 
such that 
.
intersects 
.
is the reflection of 
wrt 
lies on the same circle.
in its exterio
so that 
,
,
.
and
such that 
and 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
has a right angle at 
.
such that 
and 
lies between 
.
and 
.
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and 
are tangent to the circumcircle of 
be the reflection of 
be the reflection of 
is the circumcenter of 
,
is perpendicular to 
be a positive integer. Lexi has a dictionary 
consisting of some 
grid so that each column contains a string from 
such that if 
Y by Generic_Username, artsolver, anantmudgal09, DaniyalQazi2, Davi-8191, Wizard_32, rashah76, OlympusHero, itslumi, v4913, HamstPan38825, hamburgerwhizz, samrocksnature, suvamkonar, Jc426, megarnie, diegoca1, Adventure10
Let 
, 
, 
, 
be 
distinct points on the unit circle 
, other than 
. Each point is colored either red or blue, with exactly 
red points and blue points. Let 
, 
, , 
be any ordering of the red points. Let 
be the nearest blue point to traveling counterclockwise around the circle starting from . Then let 
be the nearest of the remaining blue points to travelling counterclockwise around the circle from , and so on, until we have labeled all of the blue points 
. Show that the number of counterclockwise arcs of the form 
that contain the point is independent of the way we chose the ordering 
of the red points.
, 
, 
, 
be 
distinct points on the unit circle 
, other than 
. Each point is colored either red or blue, with exactly 
red points and blue points. Let 
, 
, , 
be any ordering of the red points. Let 
be the nearest blue point to traveling counterclockwise around the circle starting from . Then let 
be the nearest of the remaining blue points to travelling counterclockwise around the circle from , and so on, until we have labeled all of the blue points 
. Show that the number of counterclockwise arcs of the form 
that contain the point is independent of the way we chose the ordering 
of the red points.This post has been edited 3 times. Last edited by DPatrick, Apr 20, 2017, 11:13 PM
Reason: minor LaTeX fix
Reason: minor LaTeX fix
that pass through 
and 
,
can be obtained by composition of transpositions in Part 1 (a.k.a. the transpositions of the form 
generate 
)
![[asy]size(6cm); pair A(int i) { return dir(22.5+45*i); }
draw(unitcircle, grey); dot("$(1,0)$", dir(0), dir(0));
dotfactor *= 2; draw(A(7)--A(0), EndArrow, Margins); draw(A(1)--A(2), EndArrow, Margins); draw(A(3)--A(5), EndArrow, Margins); draw(A(4)--A(6), EndArrow, Margins);
dot("$-1$", A(0), A(0), blue); dot("$0$", A(1), A(1), red); dot("$-1$", A(2), A(2), blue); dot("$0$", A(3), A(3), red); dot("$+1$", A(4), A(4), red); dot("$0$", A(5), A(5), blue); dot("$-1$", A(6), A(6), blue); dot("$+1$", A(7), A(7), red); [/asy]](http://latex.artofproblemsolving.com/6/2/f/62f9f83d7ffa45090f2c183bc1ab126ddfecc088.png)
;
and decrement by 
at each blue point and increment by 
be the lowest number ever recorded. Then:
(and hence independent of the choice of chords).
;
.
.
,
generate 
.
,
and 
keeps the number of 
such that 
is crossing constant.
and 
are determined, 
are all determined by the respective 
is crossing or not is independent of where 
are, so it suffices to prove it for flipping 
and 
.
and 
,
is crossing iff 
is, and similarly 
is crossing iff 
is, so in this case flipping them keeps the number of 
and 
,
such that 
is crossing as is 
but 
and 
are not, so the number of crossing arcs is constant after flipping.
is crossing, then so are 
is crossing, then 
is crossing, then none of the four aforementioned arcs are, and the number of crossing arcs remains constant.
this is true, and for 
,
and 
,
,
,
remains the same, and one case where the B's swap.
is implied by the claim for 
.
points remaining also works for the original 
points! (When we trace clockwise from ![[asy]
picture bdot;
filldraw(bdot, circle((0,0),.15), lightblue);
picture rdot;
filldraw(rdot, circle((0,0),.15), red);
int [] reds = {0, 1, 3, 6};
int [] blues = {2,4,5,7};
for(int i = 0; i<4; ++i){
add(shift(dir(reds[i]*45+22.5))*rdot);
}
for(int i = 0; i<4; ++i){
add(shift(dir(blues[i]*45+22.5))*bdot);
}
draw(arc((0,0),1,-100, -80));
label("$X$", dir(-112.5));
[/asy]](http://latex.artofproblemsolving.com/d/3/a/d3a7a0b78c310af075196ab61a275bc42cf1070a.png)
arc intersects the smallest arc with 
,
,
with 
and the 
s
if 
.
:
for each red point.
-
is a red point between 
since 
.
.
the number of arcs that contain 
case(with the indices shifted up by 1) and by our inductive hypothesis, this is constant, regardless of whether 
contains 
from both diagrams because they cannot be used anymore. Set 
case in both diagrams, thus this is constant.
from both diagrams because they cannot be used anymore. This becomes the 
we choose. Therefore, we can make 
they are constant in both orderings, and both give the 
and move along the circle counterclockwise one full circle. During this, we will keep a counter 
that starts at 
instead, 
.
,
and 
be 
for any 
,
such that the number of counterclockwise arcs is 
.
the number of counterclockwise arcs through 
is purely dependent on the number of points between 
as the point adjacent to 
as 
.
if the point is blue.
for all 
.
points. Assume that it's true for 
points where 
.
points. Start at any red point 
such that 
.
points, and we're done. 
and 
.
(finishes).
does not contain 
,
.
.
and 
,
is unchanged. Then we just need to show that the goodness of 
are unchanged.
)
does not change number of good arcs.
cases: 0,1,2 of blue points lies on arc 
.
or 
that occupies the circle does not change, hence the conclusion. ( Note that there are no extra blue points in these arcs), hence conclusion. 
red dots you see, and the number 
blue dots you see. I claim the answer is the maximum value of 
.
is less than 
blue points between this point and 
red points from this point to 
appear in this order, notice that upon swapping 
,
.
.
are in counterclockwise order by first swapping 
at point 
there are 
at 
is trivial.
(our intent is to induct downwards 
)
)
has 
,
with 
and 
for each ordering intersecting 
