[/quote]
,
such that for all 
,
.
such that, for all integers 
that satisfy 
,![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
,
be the 
-
and 
are placed on line 
such that 
is between 
,
and 
,
be the 
,
be the reflection of the orthocenter of triangle 
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
to the side 
and 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
Y by matol.kz, tapir1729, NO_SQUARES, Adventure10, Mango247, Rounak_iitr, ohiorizzler1434, Retemoeg
Let 
be a convex quadrilateral with 
. The internal angle bisectors of 
and 
meet 
at 
and 
respectively, and meet each other at point 
. Let 
be the midpoint of and let 
be the circumcircle of triangle 
. Segments 
and 
intersect again at 
and 
respectively. Denote by 
the intersection point of lines 
and 
. Prove that 
.
be a convex quadrilateral with 
. The internal angle bisectors of 
and 
meet 
at 
and 
respectively, and meet each other at point 
. Let 
be the midpoint of and let 
be the circumcircle of triangle 
. Segments 
and 
intersect again at 
and 
respectively. Denote by 
the intersection point of lines 
and 
. Prove that 
.
.
,![[asy] size(12cm); draw(unitcircle, red); pair P = dir(0); pair R = dir(180); pair K = dir(75); pair L = conj(K); pair B = dir(65); pair D = dir(-58); pair A = extension(K, D, B, L); pair C = extension(K, B, D, L); pair T = 2/(K+L); draw(A--B--C--D--cycle, orange); draw(P--B--K--A--L--D--P, orange); draw(K--T--L, red); draw(A--T, paleblue); pair E = extension(B, P, A, C); pair F = extension(D, P, A, C); draw(F--P, orange); pair M = midpoint(A--C); pair O = midpoint(P--R); pair X = -B+2*foot(O, B, M); pair Y = -D+2*foot(O, D, M); draw(B--M--D, yellow); draw(B--Y, yellow+dashed); draw(D--X, yellow+dashed); pair Q = extension(X, E, Y, F); draw(X--Q, brown+dotted); draw(Y--F, brown+dotted); pair Z = R*Q/B; draw(Z--K--X--L--cycle, paleblue); draw(Z--T, paleblue); draw(Z--B, dashed+paleblue); draw(T--R--Q, paleblue);
dot("$P$", P, dir(P)); dot("$R$", R, dir(R)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Q$", Q, dir(Q)); dot("$Z$", Z, dir(Z));
/* TSQ Source:
!size(12cm); unitcircle 0.1 yellow / red P = dir 0 R = dir 180 K = dir 75 L = conj K B = dir 65 D = dir -58 A = extension K D B L C = extension K B D L T = 2/(K+L) A--B--C--D--cycle 0.1 lightred / orange P--B--K--A--L--D--P orange K--T--L red A--T paleblue E = extension B P A C F = extension D P A C F--P orange M = midpoint A--C O := midpoint P--R X = -B+2*foot O B M Y = -D+2*foot O D M B--M--D yellow B--Y yellow dashed D--X yellow dashed Q = extension X E Y F X--Q brown dotted Y--F brown dotted Z = R*Q/B Z--K--X--L--cycle paleblue Z--T paleblue Z--B dashed paleblue T--R--Q paleblue
*/ [/asy]](http://latex.artofproblemsolving.com/6/0/0/600ee0ffef4cb9d1a841d4222e8515ba75e42977.png)
lies on 
lies on 
.
,
(not pictured). Then 
are collinear by Brokard Theorem on 
and 
.
,
,
and 
,
be the point on 
.
.
.
the antipode of 
)
,
,
are parallel. Indeed, this follows by Pascal on 
.
and 
.
,
,
,
.
such that 
.
which is obvious by reflection about 
;
and 
,
and 
and hence by analogy 
.
be the reflection of 
is cyclic on the 
as desired.
and 
where 
.
.
be a parallelogram; 
.
so 
are concyclic, just as we desired. 
lies on ray 
such that 
.
depends only on 
.
be midpoint of arc 
where projection is preceded by reflection of the pencil in line 
.
and 
;
as desired. 
Evidently 
It is well－known that 
the power of 
Let 
then 
are on the circle with diameter 
Likewise 
are on the circle with diameter 
,
.
be the circumcircle of 
the circumcircle of 
,
passes through 
,
passes through 
,
.
be on 
.
and 
.
,
,
.
,
cyclic hence 
.
be diametrically opposite 
.
.
.
parallelogram. Now invoking a little bit more trigonometric bashing we have 
so 
,
,
we have (considering signed length) 
.
we have 
.
,
.
we have 
.
,
.
since again it is not hard to verify that 
.
if it's nearer to 
otherwise). Therefore, 
,
and the two perpendicular lines coincide. (Sorry for the trigo-heavy proof).
,
.
is cyclic and denote its circumcircle by 
.
and 
both pass through the midpoint of the arc 
.
.
and 
the tangents from 
the midpoint of 
the midpoint of 
is the radical axis of 
is the polar of 
.
.
.
,
.
.
.
is enough, that is, it is not used at all that they are less that 
.
,
and 
.
and 
lie on 
.
,
,
.
inscribed in 
and 
.
the center of 
and 
is, actually, the external angle bisector of 
.
,
.
.
is cyclic, therefore 
.
is cyclic, hence 
.
,
.
,
![[asy]
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label("$X$", (-3.581983191970996,-5.0287202570899865), NE * labelscalefactor);
dot((7.617876917278405,1.98716542535431),linewidth(4pt) + dotstyle);
label("$Y$", (6.795046263553144,2.0143232353915335), NE * labelscalefactor);
dot((-6.265589433796959,-3.915676740913798),linewidth(4pt) + dotstyle);
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dot((22.765729294890875,10.16187829142016),linewidth(4pt) + dotstyle);
label("$G$", (22.21472858702275,10.891058524732147), NE * labelscalefactor);
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dot((9.90366577557506,7.7519661926992285),linewidth(4pt) + dotstyle);
label("$J$", (10.087356813699357,8.098846015878527), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/0/e/f0e805df5b2811c33aea18926a13b7cb50cda803.png)
and 
.
and 
are collinear. Note that![\[\angle PDI = \pi-\angle PBI = \angle PBC = \angle PDA,\]](http://latex.artofproblemsolving.com/d/f/5/df5193475342c7f0d4c2fdad3dd95cf5963ab38d.png)
so 
,
.
.
,
,
and 
are tangent to 
are the intersections of 
is on the polar of 
,
,
are collinear. Also, we see that (by Brokard on the same quadrilateral), that 
is on the polar of 
.
.
is tangent to 
(its on 
where 
is cyclic. Thus,
so 
,![[asy]
unitsize(0.45inches);
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[/asy]](http://latex.artofproblemsolving.com/7/7/3/773be7d03ebd41e0832787c4ecc6c42458e8d7c0.png)
passes through its center, we see 
.
,
,
and 
lie on 
similarly for any arbitrary point 
and 
are projective maps. Thus, it suffices to show our claim for three positions of 
,
,
and 
meet on 
.
.
.
,
,
.
and 
be the antipodes of 
are concurrent. Applying Pascal on 
,
lies on 
.
,
is the 
,
is cyclic. Thus, for proving 
,
is the 
,
and 
,
. Hence, done. 
post on HSO 
and 
as shown in the diagram below.![[asy]
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dot((-3.9607235462495836,1.9853181724732336),linewidth(2pt) + dotstyle);
label("$X$", (-3.8866004163876657,2.0724511294738717), NE * labelscalefactor);
dot((-2.9486521304059456,-2.0677614476898487),linewidth(2pt) + dotstyle);
label("$Y$", (-2.869750317276038,-1.994949266972681), NE * labelscalefactor);
dot((-2.9069134899946403,-1.4273208830994124),linewidth(2pt) + dotstyle);
label("$Q$", (-2.829076313311573,-1.3441652035412326), NE * labelscalefactor);
dot((-14.58832583745327,-0.7334908956054996),linewidth(2pt) + dotstyle);
label("$U$", (-14.502515451113059,-0.6527071361453185), NE * labelscalefactor);
dot((-11.433793812647506,3.8474892637385096),linewidth(2pt) + dotstyle);
label("$K$", (-11.350280143867012,3.9231183098570535), NE * labelscalefactor);
dot((-11.540362797522098,-6.525836664813006),linewidth(2pt) + dotstyle);
label("$L$", (-11.451965153778175,-6.4487527010816565), NE * labelscalefactor);
dot((-5.256493221081515,-6.058574568103326),linewidth(2pt) + dotstyle);
label("$R$", (-5.167831541268317,-5.981001655490303), NE * labelscalefactor);
dot((-3.09873970725935,0.12086294484481758),linewidth(2pt) + dotstyle);
label("$V$", (-3.312109331151666,0.20144694710845756), NE * labelscalefactor);
dot((-6.896106866781368,4.184907010995224),linewidth(2pt) + dotstyle);
label("$S$", (-6.8151287018291535,4.26884734355501), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/c/8/7/c876ac897c390ba14fb2a7230c2ab6f91d940202.png)
Let 
and the second intersection of 
with 
and 
be 
and 
are concurrent.
at 
By Brocard's Theorem, it is easy to see that 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.172652771186477, xmax = 10.198530855245242, ymin = -4.859595378688876, ymax = 5.831089427237665; /* image dimensions */
/* draw figures */
draw(circle((2.235675488430095,-0.6187792711653043), 3.3775654652468465), linewidth(0.4) + blue);
draw((-0.08,1.84)--(5.6,-0.32), linewidth(0.4));
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draw((1.276600275234065,5.179061053430356)--(-1.138688740973718,-0.7657980432597458), linewidth(0.4));
draw((1.276600275234065,5.179061053430356)--(5.6,-0.32), linewidth(0.4));
draw((4.845512684584302,1.5252119325475504)--(1.3536390081741951,3.9396046309159387), linewidth(0.4));
draw((1.5220063065799334,1.23078633411749)--(1.8454944779205704,-3.9737319162954723), linewidth(0.4));
/* dots and labels */
dot((-0.08,1.84),linewidth(2pt) + dotstyle);
label("$A$", (-0.026763442960632506,1.8952154190856052), NE * labelscalefactor);
dot((3.32,2.58),linewidth(2pt) + dotstyle);
label("$B$", (3.3772356992249386,2.6265433597895353), NE * labelscalefactor);
dot((5.6,-0.32),linewidth(2pt) + dotstyle);
label("$C$", (5.651000751231707,-0.27217465972786026), NE * labelscalefactor);
dot((-1.138688740973718,-0.7657980432597458),linewidth(2pt) + dotstyle);
label("$D$", (-1.0905131748936234,-0.7109714241502183), NE * labelscalefactor);
dot((1.276600275234065,5.179061053430356),linewidth(2pt) + dotstyle);
label("$X$", (1.329517465253931,5.232730203025358), NE * labelscalefactor);
dot((1.5220063065799334,1.23078633411749),linewidth(2pt) + dotstyle);
label("$Y$", (1.568861154938854,1.2835593232241365), NE * labelscalefactor);
dot((1.3536390081741951,3.9396046309159387),linewidth(2pt) + dotstyle);
label("$P$", (1.4092986951489053,3.9961211396532588), NE * labelscalefactor);
dot((-0.8973864599809147,0.6429145504281734),linewidth(2pt) + dotstyle);
label("$K$", (-0.8378726135595381,0.6984969706609924), NE * labelscalefactor);
dot((4.845512684584302,1.5252119325475504),linewidth(2pt) + dotstyle);
label("$L$", (4.893079067229451,1.5760904995057083), NE * labelscalefactor);
dot((1.4330467410096437,2.6620339764674266),linewidth(2pt) + dotstyle);
label("$I$", (1.4890799250438795,2.7196214613336718), NE * labelscalefactor);
dot((1.8454944779205704,-3.9737319162954723),linewidth(2pt) + dotstyle);
label("$J$", (1.9012829461679137,-3.915517491598348), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/4/7/b/47bbd0b8581bb951ff330438df63842310d71f3a.png)
we obtain that 
which implies that 
are concurrent, thus again by Brocard, 
Let 
and 
let 
and 
with 
then it follows that 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -22.680500494225463, xmax = 17.0747406351982, ymin = -14.444614950563711, ymax = 13.205224035644418; /* image dimensions */
/* draw figures */
draw(circle((-3.64,-1.16), 5.381152292957336), linewidth(0.4) + blue);
draw((-3.722093369329189,11.091310229368936)--(-8.790835128644087,-2.7174650806763028), linewidth(0.4));
draw((-8.790835128644087,-2.7174650806763028)--(-0.4303602250270546,3.1591448823710016), linewidth(0.4));
draw((-6.56,3.36)--(1.6846835554896908,-1.9375249410132471), linewidth(0.4));
draw((1.6846835554896908,-1.9375249410132471)--(-3.722093369329189,11.091310229368936), linewidth(0.4));
draw((-3.722093369329189,11.091310229368936)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4));
draw((-3.722093369329189,11.091310229368936)--(-5.1074720182720705,4.017192856711877), linewidth(0.4));
draw((-6.951948556963188,-5.401202277188562)--(-5.1074720182720705,4.017192856711877), linewidth(0.4));
draw((-6.56,3.36)--(-3.3947298101688306,4.79211902592437), linewidth(0.4));
draw((-8.790835128644087,-2.7174650806763028)--(-2.5295738376256147,-11.855374989077417), linewidth(0.4));
draw((-2.5295738376256147,-11.855374989077417)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4));
/* dots and labels */
dot((-6.56,3.36),linewidth(2pt) + dotstyle);
label("$A$", (-6.413866364279621,3.50714618227291), NE * labelscalefactor);
dot((-0.4303602250270546,3.1591448823710016),linewidth(2pt) + dotstyle);
label("$B$", (-0.2923845986763235,3.3008041002862822), NE * labelscalefactor);
dot((1.6846835554896908,-1.9375249410132471),linewidth(2pt) + dotstyle);
label("$C$", (1.8054265681877277,-1.788967255383871), NE * labelscalefactor);
dot((-8.790835128644087,-2.7174650806763028),linewidth(2pt) + dotstyle);
label("$D$", (-8.649238919134758,-2.5799452363326107), NE * labelscalefactor);
dot((-3.722093369329189,11.091310229368936),linewidth(2pt) + dotstyle);
label("$X$", (-3.5938579104623716,11.244974256771453), NE * labelscalefactor);
dot((-3.3947298101688306,4.79211902592437),linewidth(2pt) + dotstyle);
label("$P$", (-3.2499544404846583,4.9171504091815335), NE * labelscalefactor);
dot((-5.1074720182720705,4.017192856711877),linewidth(2pt) + dotstyle);
label("$K$", (-4.969471790373225,4.360562775230565), NE * labelscalefactor);
dot((-3.2083861761728762,1.2064597573855096),linewidth(2pt) + dotstyle);
label("$Y$", (-3.0780027054958015,1.340554321413318), NE * labelscalefactor);
dot((-6.951948556963188,-5.401202277188562),linewidth(2pt) + dotstyle);
label("$L$", (-6.826550528252877,-5.262392302158773), NE * labelscalefactor);
dot((-2.5295738376256147,-11.855374989077417),linewidth(2pt) + dotstyle);
label("$Z$", (-2.390195765540375,-11.727777537739778), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/7/d/a/7da084c8e98c307d7fc99af95bbd4b6cdea53493.png)
let 
Let 
is a trapezium and 
intersects 
Prove that 
"
we get 
and 
intersect on 
Now, letting 
and applying Pascal's Theorem on cyclic hexagon 
we get 
are collinear. Thus 
is cyclic.
Therefore, 
now since 
therefore, 
Hence, 
H
H
thus, ![\[\angle AXC=180^{\circ}-\angle ABC=180^{\circ}-ADC=\angle AYC\]](http://latex.artofproblemsolving.com/3/a/1/3a11d9b6275cbbebfa431a1da39e9aec18d9648b.png)
So, the claim follows. 
Let 
intersect 
and 
again by Brocard, we have 
also since 
so 
But we also have 
therefore, 
are collinear, which is a contradiction to the assumption that 
in 
in 
of triangle 
therefore the first part follows. Similarly, as the 
H
is the exterior angle bisector of 
of triangle 
we obtain 
and 
thus, it follows that 
So, it suffices to solve the following problem:
is perpendicular to the parallel sides."
Let the common perpendicular bisector of the parallel sides of the trapezium be 
.
be 
it is obvious that 
also midpoint of 
is 
so 
but 
is same as 
Thus, we are done. 
,
,
be the radius.
are concyclic. (Similarly 
are concyclic.)
,
.
by the power of point 
.
.
,
.
lie on 
and 
are collinear respectively, and 
,
,
.
,
.
,
.
,
(by properties of the Miquel point), so 
is the perpendicular bisector of 
.
,
.
is the midpoint of arc 
.
,
,
.
and lines 
respectively intersect on 
,
and ![\[\frac{AX}{CX} = \frac{\sin\angle XCA}{\sin\angle XAC} = \frac{\sin\angle MBC}{\sin\angle MBA} = \frac{AB}{CB} = \frac{AE}{CE}.\]](http://latex.artofproblemsolving.com/a/5/1/a51271185437e8823e0e8e2a91294da907e9daf6.png)
,
is the unique point on 
.
.
be an acute angled triangle 
.
be such that, 
are Internal and External angle bisectors of 
.
,
is cyclic. Let 
,
,
H
& 
are cyclic. Let 
is cyclic. Let 
is a parallelogram
Let 
a
,
is the Internal bisector WRT 
pass through midpoint of arcs 
)
)
cyclic. Now Let 
.
.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(16cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -23.523904823024644, xmax = 8.06162313100848, ymin = -12.224812830046798, ymax = 6.545275503292614; /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqccqq = rgb(0,0.8,0); pen ffxfqq = rgb(1,0.4980392156862745,0); pen qqqqcc = rgb(0,0,0.8); pen ffdxqq = rgb(1,0.8431372549019608,0);
draw((-0.13378578198266103,1.3093270004863191)--(-2.88,-4.51)--(7.66,-4.59)--cycle, linewidth(2) + rvwvcq);
draw((-3.7158381930057414,4.020679236309496)--(-0.13378578198266103,1.3093270004863191)--(-2.557364084571636,-7.802866403645595)--(-4.760942572202613,-8.495785167438727)--cycle, linewidth(0.8) + ubqqys);
/* draw figures */
draw((-0.13378578198266103,1.3093270004863191)--(-2.88,-4.51), linewidth(0.8) + rvwvcq);
draw((-2.88,-4.51)--(7.66,-4.59), linewidth(0.8) + rvwvcq);
draw((7.66,-4.59)--(-0.13378578198266103,1.3093270004863191), linewidth(0.8) + rvwvcq);
draw((xmin, -0.007590132827324485*xmin-4.531859582542695)--(xmax, -0.007590132827324485*xmax-4.531859582542695), linewidth(0.4)); /* line */
draw((-0.13378578198266103,1.3093270004863191)--(2.39,-4.55), linewidth(0.8) + dtsfsf);
draw(circle((2.3984000473006253,-3.4432937681426545), 5.385106242631295), linewidth(0.8));
draw(circle((2.3815999526993745,-5.656706231857345), 5.385106242631294), linewidth(0.8));
draw(circle((-10.94115381135354,-4.4488147718303335), 12.245627807195754), linewidth(0.4) + linetype("4 4"));
draw((-2.88,-4.51)--(-2.557364084571636,-7.802866403645595), linewidth(0.8) + rvwvcq);
draw((-2.557364084571636,-7.802866403645595)--(7.66,-4.59), linewidth(0.8) + rvwvcq);
draw((-2.557364084571636,-7.802866403645595)--(2.39,-4.55), linewidth(0.8) + dtsfsf);
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draw((-0.13378578198266103,1.3093270004863191)--(-2.557364084571636,-7.802866403645595), linewidth(0.8) + ubqqys);
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draw((-2.88,-4.51)--(-3.7158381930057414,4.020679236309496), linewidth(0.4));
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label("$B$", (-2.7861934368110934,-4.2507615105039855), NE * labelscalefactor);
dot((7.66,-4.59),dotstyle);
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dot((2.39,-4.55),dotstyle);
label("$M$", (2.495321073535429,-4.3025410645269915), NE * labelscalefactor);
dot((0.6677688333267355,-0.5515958374881434),dotstyle);
label("$K$", (0.7607060137647575,-0.2896256277440821), NE * labelscalefactor);
dot((-2.557364084571636,-7.802866403645595),dotstyle);
label("$L$", (-2.44962633566156,-7.53876319096482), NE * labelscalefactor);
dot((-1.5295710756849874,-7.127097794067173),dotstyle);
label("$F$", (-1.4140352552014577,-6.865628988665751), NE * labelscalefactor);
dot((-3.7158381930057414,4.020679236309496),dotstyle);
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dot((-0.3924891064001432,-4.52888053809184),dotstyle);
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dot((0.8421124628484566,-2.661766673288232),dotstyle);
label("$S$", (0.9419344528452754,-2.4125873426872984), NE * labelscalefactor);
dot((-11.947365553758662,-1.5938661023574054),dotstyle);
label("$T$", (-11.84761539083699,-1.3252167082041877), NE * labelscalefactor);
dot((4.913785781982661,-10.409327000486318),dotstyle);
label("$M'$", (5.006629443651177,-10.153630669126587), NE * labelscalefactor);
dot((0.8494807422243454,-1.6909958655147632),dotstyle);
label("$U$", (0.9419344528452754,-1.4287758162501982), NE * labelscalefactor);
dot((2.4392725416538292,1.9416573628920448),dotstyle);
label("$M_{BC}$", (2.5471006275584345,2.1957929653601713), NE * labelscalefactor);
dot((2.3407274583461697,-11.041657362892046),dotstyle);
label("$M_{BC}'$", (2.443541519512424,-10.774985317402649), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/b/6/0b6066a2cce6ef09618e744a46c1413c8a28d840.png)
)
be the circles 
respectively. Since 
hence 
so that the rays 
meet 
respectively.
are concyclic, say lie on 
then 
is the reflection of 
This is because the orthocenter of 
is concyclic with 
Then we have
Hence ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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[/asy]](http://latex.artofproblemsolving.com/b/3/5/b35f56dc8daf6a87c7dbadf92b084e95beef0a39.png)
It is easy to see that 
since they subtend equal arcs in their circles. Hence
Hence we are done here. 
respectively, so that 
lie in the interiors of 
respectively.
respectively.
This is because it lies on the 
Hence, by a well-known lemma, it lies on the 
so 
Since 
hence 
as 
are arc midpoints.
are both cyclic, due to right angles. Hence
Hence 
which is what we wanted. 
Let 
Our goal is to show 
Then
Thus 
It is not too hard to check now, say algebraically that since 
have the same midpoint, this is possible if and only if 
Now, 
are collinear happens only if 
(Proved formally in the comments). Even here, 
are "parallel", since they coincide. Hence we are done. 
(note that 
exactly twice, when 
Move 
while fixing the other points (in particular 
Then 
is projective and so 
with degree 
Also, 
is projective and so 
with degree 
(note that 
and 
Now assume on the contrary that 
are collinear for more than three positions of 
Then by Zack's lemma 
a contradiction.
there are exactly two positions of 
and hence we are home free!
and 
aren't parallel.
by the circumcircle of triangle 
,
.
,
,
.
and 
.
,
,
.
,
are concyclic. Similarly 
are concyclic too.
,
.
,
,
is parallel to 
,
,
lies on 
.
is the external angle bisector of 
.
,
is concyclic quadrilateral, and 
.
with 
.
.
be the reflections of 
with respect to 
.
lies on 
respectively.
.
which is equivalent to 
and since 
, we have 
; 
respectively. Therefore 
.
.
we have 
and 
are collinear. Therefore 
, so 
as claimed.
thus 
,
.
so 
and 
.
so 
so 
are parallel. Thus 
and 
.
as the intersection of the tangents to 
as the intersection of the tangents at 
.
.
,
,
implies that 
.
,
is a harmonic bundle on 
finishes. 
and 
.![\[ \measuredangle DSB = \measuredangle SCB + \measuredangle CBS = \measuredangle DCB + \measuredangle CBP + \measuredangle PDC = \measuredangle DPB \]](http://latex.artofproblemsolving.com/9/7/c/97cd043e15c6604831b441184ec0f2865d48d76c.png)
as desired. The proof for 
and let 
.
,
it follows that 
.![\[ (VW; AC) \overset{B}= (VW; ST) \overset{A}= (VW; BD) = -1 \]](http://latex.artofproblemsolving.com/7/d/4/7d4b89bb943088dec156bf5f85ab1feaf17ec7de.png)
It thus follows that the polar of 
lies on 
gives that the polar of 
is 
(from which it follows that 
are collinear), it follows that 
be a cyclic quadrilateral in 
and 
.
as originally. Let 
be the conjugate of 
,
be the harmonic conjugate of 
,
concur.
,
over 
is perpendicular to 
.
.
be 
respectively. Let 
and 
.
and 
.
has slope equal to 
,
.
is a positive real number, which is well known. 
.![\[\angle XBA+\angle XDA=360 ^{\circ}- \angle BAD+\angle BQD= 360^{\circ} - \angle BAD-\angle BPD=\angle ABP+\angle ADP=2x\]](http://latex.artofproblemsolving.com/6/a/5/6a5defd6e4b861a19ec5910a52fc044145ea8512.png)
Thus 
.
then. However, it is also the only point that works since 
is fixed, and any point 
is on one of two lines, the first of which contains 
so 
and similarly 
,
concur, say at 
.
.
gives 
are collinear. This requires 
,
collinear and 
.
.
This finishes as if 
,
so 
as desired.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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draw(circle((-2.4858330282581638,0.9332447656807272), 6.0999977721559455), linewidth(0.8) + blue);
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dot((-0.7714483795412509,6.787376477262837),dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/d/1/b/d1bd13d41a74b5039972e3a08aa067f0a062aec7.png)
is cyclic
Similarly, 
.
be the reflections of 
.![\[ \angle BPD = \angle VPU = 180^\circ - \angle DVZ - \angle BUW = 180^\circ - \angle DVZ - \angle ZVS = 180^\circ - \angle DVS = 180^\circ - \angle DX'S = \angle BX'D, \]](http://latex.artofproblemsolving.com/1/3/6/136da0b43e44be7b68c6ee208807ca9772e28cbc.png)
and similarly 
,
is cyclic, and 
.
and 
.
,
,
,
lie on one line, which must be 
,
,
.
over 
,
,
(the point at infinity on line 
lie on one line, which is 
,
,
,
.![\[ \angle HPE = 180^\circ - \angle BPQ = 180^\circ - \angle BXQ = \angle SXZ = \angle SVZ, \]](http://latex.artofproblemsolving.com/5/1/a/51aef445f6e83909ee87041c305035ffd8d1b6e4.png)
and ![\[ \angle HEP = \angle BEA = 180^\circ - \angle ABE - \angle EAB = 180^\circ - \frac12 \angle AB'C - \angle B'AC = 180^\circ - \angle VAC - \angle B'AC = 180^\circ - \angle VAB' = \angle VZS. \]](http://latex.artofproblemsolving.com/0/a/3/0a3079263f49a2cfca0b54bc81391e0c05da9ca5.png)
Thus 
,
,
.
Hence 
.
and 
we have 
we have 
.
Assume that 
.
,
when 
.
and 
denote the circumcircles of triangles 
,
is tangent to line 
.
yields 
,
.
and 
be the midpoints of the minor and the major arc 
,
similarly. Note that 
and 
.
.
must cut 
and 
.
is the line passing through ![\[-1=(N_b;N_d;M;\infty)\overset{Q}{=}(F;E;M;l_q\cap (AC))\]](http://latex.artofproblemsolving.com/2/c/a/2ca359dc735277eb8f22bfbbbe2c84b55aa0118a.png)
Defining 
similarly and projecting from 
also yields 
,
months back and it's tiring to do all the work again. This is a kind of detailed outline from my notes.Some details might be extraneous because I found both the humpty points and some projective observations.
and 
.
,
and 
are all collinear, We get that by converse of pacal's on 
points uniquely determine a conic and 
c
.
.
,
and 
intersect 
and 
meet at 
.
.
.
And do the same for 
.
be a quadrilateral inscribed in a circle 
and 
.
,
.
,
,
.
)
and 
.
and hence 
lies on 
to get that 
are collinear, as desired.
.
and 
.
.
be a quadrilateral inscribed in circle 
.
with 
and define 
.
with 
and define 
.
.
are collinear.
to the center of 
is a rectangle, 
are reflections over 
,
so 
so 
are aligned to the sides of 
,
lies on 
with 
and ![\[\frac{\sin(\angle ADT)}{\sin(\angle ABT)}=\frac{\sin(\angle BAC)}{\sin(\angle DAC)}\]](http://latex.artofproblemsolving.com/1/3/8/138856d179883b7d73008ac5a98f9d91b03ef179.png)
Also note that since 
we must have 
.![\[\frac{\sin(\angle BAC)}{\sin(\angle DAC)}=\frac{DC}{BC}=\frac{\sin(\angle DBC)}{\sin(\angle BDC)}\]](http://latex.artofproblemsolving.com/6/6/e/66ebee1c71b2a136c4239724613d4090e4e81736.png)
This implies that 
and 
so the conclusion follows. 
and 
.
and 
respectively so we are done. 
let 
.
meet ![\[-1=(V,U;L,K)\overset{D}{=}(A,C;L,K)=(A,C;F,S)\]](http://latex.artofproblemsolving.com/b/5/4/b54a35c184363113053583d3ce1386864f2dfc93.png)
Now since ![\[MA^2=ML\cdot MK=MF\cdot MS=MY\cdot MD\]](http://latex.artofproblemsolving.com/e/8/9/e895cb4ca5fa63171d574dcafc8c70fb52773a45.png)
Therefore, 
is cyclic so by Reim's we get 
and since 
and lines 
and ![\[\measuredangle PDS = \measuredangle PBA\]](http://latex.artofproblemsolving.com/4/9/1/491077d33abdd82b1d916d9536b4cb80cbc42564.png)
which implies that quadrilateral 
is cyclic. Similarly, quadrilateral 
is also cyclic, which implies the claim.
.
to a rectangle while preserving its circumcircle (in particular we treat ![[asy]
size(6 cm);
import geometry;
pair foot(pair P, pair A, pair B) {
return foot(triangle(A,B,P).VC);
}
pair S = dir(150);
pair B = dir(210);
pair T = dir(330);
pair D = dir(30);
pair X = dir(100);
pair Y = dir(280);
pair P = dir(20);
pair Q = dir(130);
draw(S--B--T--D--cycle , blue+1.3);
draw(Q--B,red);
draw(P--X,red);
draw(B--P,blue);
draw(Q--X,blue);
draw(Q--Y,blue);
draw(D--P,blue);
draw(Q--D,green+dashed);
draw(P--Y,green+dashed);
draw(circumcircle(D, B, S), blue);
dot("$S$", S, dir(190));
dot("$B$", B, dir(B));
dot("$T$", T, dir(T));
dot("$D$", D, dir(20));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
[/asy]](http://latex.artofproblemsolving.com/f/8/f/f8f0cf8bcea998e8a739f7d20e166e7d221dac63.png)
is a rectangle. Further, 
is an isosceles trapezoid so 
.
and 
as 
.
so 
is an isoceles trapezoid. Thus, 
and indeed points 
is cyclic. 
.![\[ME\cdot MR = MC^2\]](http://latex.artofproblemsolving.com/5/a/d/5ad8a852cfd3a1477e5355abf77a88aa69ccb745.png)
Further, it is well known that (and easy to check using linearity of power of point)![\[MC^2 = \text{Pow}_{\omega}(M)=MB \cdot MX\]](http://latex.artofproblemsolving.com/2/6/6/266f728ab5933433eb9afe05afd337fc3b2be4a1.png)
Thus,![\[ME \cdot MR = MB \cdot MX\]](http://latex.artofproblemsolving.com/5/d/8/5d846b9764788f6ecfb0ca0fd6af6d18560f6f5b.png)
and 
must be cyclic. Denote by ![\[\measuredangle ZPB = \measuredangle QPB = \measuredangle QXB = \measuredangle EXB = \measuredangle ERB = \measuredangle ZRB\]](http://latex.artofproblemsolving.com/2/5/6/256e75e5ddd3d19502037c01f3e6122634ffc339.png)
Thus 
is cyclic. Finally we note,![\[\measuredangle (PQ;AC) = \measuredangle PZE = \measuredangle PZR = \measuredangle PBR = \frac{\pi}{2}\]](http://latex.artofproblemsolving.com/e/0/1/e01788de5c6aa83a2621cd6063762efaaabf79a6.png)
which finishes the problem.
.
and 
meet on 
.
.
.
is the center of spiral homothety sending 
to 
.
which is true.
.
.
concur.
sufficies. Let 
meet 
.
which is followed by 
.
.
hence 
to get that 
are concurrent. Pascal at 
gives 
are collinear. Pascal at 
yields 
are collinear thus, 
gives that 
hence 
.
and 
are concyclic.
.
.
hence 
pass through 
.
by inversion at 
gives 
are collinear thus, 
concur. Pascal at 
yields 
are collinear hence 
concur which is equavilent to the collinearity of 
.
implies 
are collinear hence 
.
which shows that 
by 
intersect 
intersect 
is cyclic.
.
Thus, 
is cyclic. That is 
is cyclic, which mean 
is cyclic and orthogonal to 
By similar consequences, 
intersect 
at 
intersect 
at 
lie on 
is an isosceles trapezoid and 
,
