[/quote]
,
positive real numbers such that 
. Prove that :
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
.
to 
meets 
at points 
,
to 
is 
.
intersects line 
at 
,
intersects line 
,
intersect at 
.
is tangent to 
.
be an acute triangle. Points 
,
lie on a line in this order and satisfy 
.
and 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
pairwise distinct complex numbers are written on a board. It turns out that there are exactly 760 ways to choose two numbers 
and 
from the board such that:![\[
a^2 + b^2 + 1 = 2ab
\]](http://latex.artofproblemsolving.com/8/f/c/8fc5df8dbc8e1647edfaa96dcfa8a0dfdac0c961.png)
Ways that differ by the order of selection are considered the same. Prove that there exist two numbers 
and 
from the board such that:![\[
c^2 + d^2 + 2025 = 2cd
\]](http://latex.artofproblemsolving.com/2/7/0/2706d72797c5e392b3f068a8174bd257e4d93fa1.png)
,
are even?
and 
be real numbers such that 
.
and 
.
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be an acute-angled, non-isosceles triangle with circumcenter 
and incenter 
,![\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]](http://latex.artofproblemsolving.com/7/8/1/78125759bf5c73149e4a56538e9b0c44309704a7.png)
,
,
.
,
,
be the excenters opposite to vertices 
,
,
,
,
,
be the Lemoine points of triangles 
,
,
,
,
,
all pass through a common point 
.
.
intersect at point 
.
and 
with their common external tangents lie on a circle 
.
and 
with their common external tangents lie on a circle 
.
Y by anantmudgal09, k12byda5h, MehmetBurak, Adventure10, Mango247, Rounak_iitr
Let 
be an acute triangle. Point 
lies on side 
. Let 
be the circumcenters of triangles 
and 
, respectively. Suppose that the points 
lies on a circle centered at 
. Let 
be the orthocenter of triangle . Prove that 
.
Zuming Feng.
be an acute triangle. Point 
lies on side 
. Let 
be the circumcenters of triangles 
and 
, respectively. Suppose that the points 
lies on a circle centered at 
. Let 
be the orthocenter of triangle . Prove that 
.Zuming Feng.
This post has been edited 1 time. Last edited by MellowMelon, Jul 19, 2009, 5:49 AM
.
and 
hence 
which means that point 
lies on the circumcircle of triangle 
which implies 
and point 
(2). From (1) and (2) we have 
- radical axis of circles 
and 
,
- radical axis of circles 
and 
.
and 
.
is radical axis of circles 
and 
hence 
and 
which implies that 
is cyclic quadrilateral and 
(3). We know that 
(4) (it was posted on the forum, I'll write proof later). A simple angle chase shows that 
is cyclic quadrilateral and 
hence 
is angle bisector of 
which combined with 
implies 
and 
(5). Let 
.
(because 
and 
are similar) hence 
is cyclic. Using this and (3) we get 
qed.
,
and 
are concurrent.
be foot of 
Let 
be midpoints of 
and 
for any 
Let 
be midpoints of 
and 
rectangles; 
are perpendicular bisectors of 
respectively. Since 
are A-altitude feet of 
these 2 rectangles are similar 
are similar by SAS and 
Consequently, foot 
of perpendicular 
to 
from 
is on midparallel 
of 
This means that the perpendicular bisector 
vertex tangent 
If 
at 
the parabola tangent 
Since 
being cyclic or not. But if it happens to be cyclic, the intersection 
of 
is its circumcenter.
is the intersection of 
and 
i
is the bisector of 
are collinear
which indicates 
are collinear
are on a circle where 
respectively?
and 
in terms of the angles of 
then realize that the construction is equivalent to the angle trisection.
:
and 
and 
with 
(If 
is A-altitude foot, pick 
so that 
)
and 
the intersection of 
and 
.
is the bisector of 
or 
are concurrent.
we have:
so 
then 
.
and 
are all issoceles triangles respectively with respect to vertexes 
.
such that 
.
and 
.
such that 
,
,
.
and 
.
and 
then we have 
and 
are two sets of concyclic points, which leads to the fact that 
.
is concyclic 
are collinear. Further, if 
then 
is the polar of 
with respect to 
,
are collinear and 
at 
.
.
.
,
,
,
bisects 
.
.
while 
,
is cyclic, 
and 
.
,
.
and 
.
,
and 
are cyclic.
so we're done by (*).
and of 
and 
now so 
also 
and that implies 
is similar to 
now also note 
.
then 
also 
, using this we get 
now apply ceva with 
and note 
and so obviously 
,and hence 
and which is indeed 
,so done!
.
.
is cyclic, it follows that 
is cyclic.
,
.![\[
\angle BO_BO_C = \angle BO_BD + \angle D_OBC = 2x+B \implies \angle BXC = 360 - 4x - 2B \implies \angle BAX = \angle BO_CX = 2x+B-90.
\]](http://latex.artofproblemsolving.com/7/a/8/7a83dea638a623d5e464870174655fa4321cf554.png)
On the other hand, 
.
,
is cyclic? Is it a property..
by simple angle chasing, so let 
and 
.
and 
so since 
are concyclic, we have by radical axis theorem that 
concur, so 
.
be 
and clearly 
(Let 
,
are concyclic). So, we must show 
bisects 
,
which follows since 
.
)
.
concylic by a little angle chasing, and thus 
(by 
.
.
is cyclic, and 
implies line 
is the bisector of 
.
and 
,
,
.
be the circumcenter of 
be the circumcenter of 
.
so 
lie on a circle.Also by some angle chasing we obtain that 
.
lie on the circumcircle of 
.
denote the circumradii of 
and 
and 
.
.
so that they meet at 
.
bisects 
and consequently 
.
and 
are both isoceles. Furthermore, since 
and 
are circumcenters, we know that 
where the angles are directed. It follows that 
with similar orientation. Therefore, 
to 
,
Hence, it is well-known (and can be proven using harmonic divisions) that 
with respect to 
Then since 
are collinear, under inversion in 
,
are concyclic. Then because 
and the points ![\[\measuredangle DAX = \measuredangle O_BAX - \measuredangle O_BAD = \measuredangle O_BCX - \left(90^{\circ} - \measuredangle DBA\right) = \measuredangle O_BCX + \measuredangle DBA - 90^{\circ}.\]](http://latex.artofproblemsolving.com/6/4/7/647b17206f8ac2ffe31da55f23b9ac9f6710ffff.png)
Meanwhile, since 
,
Therefore, ![\[\measuredangle DAX = \measuredangle HAD \iff \measuredangle O_BCX + \measuredangle DBA + \measuredangle ADB= 180^{\circ} = 0.\]](http://latex.artofproblemsolving.com/e/0/0/e00593dd5e209d11eeeb1a243e236b4a71a8ce2f.png)
Since the sum of the directed angles in 
,
But because 
,![\[\measuredangle O_BCX = 90^{\circ} - \measuredangle CBO_B = 90^{\circ} - \measuredangle DBO_B = \measuredangle BAD.\]](http://latex.artofproblemsolving.com/b/5/f/b5fddbd341d49090284c462765e1dbed324a1bb5.png)
be the circumcircle of triangle 
b
.
be the circumcircle of 
.
By the known lemma we know that lines 
are concurrent. Let ![\[\measuredangle O_bBA = \measuredangle O_cCA\]](http://latex.artofproblemsolving.com/a/1/8/a185e112515fdc52362e9c835b1763b22f4dcba3.png)
.
so we find that 
.
be the circumcenter of 
we find that
is cyclice.
and likewise 
.
.
is cyclic. So 
Also, 
,
.
and the result follows. 
.
and set 
so 
Consequently, 
.
we conclude that 
is tangent to the circle 
Finally, we have 
establishing the result.
,
and 
concur at the Miquel Point of the quadrilateral, where 
.
Consequently, 
be the foot of the perpendicular from 
,
.
.
,
and 
are isosceles. Hence, there 
a Spiral Similarit
.
.
we get that 
are concurrent at a point 
are concurrent and also 
.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
size(8cm); defaultpen(fontsize(10pt));
draw((-1.110459431125061,4.058413889003155)--(-1.62,1.75)--(2.8,1.75)--cycle);
/* draw figures */
draw((-1.110459431125061,4.058413889003155)--(-1.62,1.75));
draw((-1.62,1.75)--(2.8,1.75));
draw((2.8,1.75)--(-1.110459431125061,4.058413889003155));
draw(circle((-1.0413773597967102,2.832722400774602), 1.2276367365368883));
draw(circle((1.1686226402032898,3.452813719451886), 2.3581701917399545));
draw(circle((0.59,1.2649171606442673), 2.262610298093205));
draw((1.1686226402032898,3.452813719451886)--(-4.900190463027996,1.75));
draw((-4.900190463027996,1.75)--(-1.62,1.75));
draw((-4.900190463027996,1.75)--(-1.110459431125061,4.058413889003155));
draw((-1.110459431125061,4.058413889003155)--(-0.46275471959342035,1.75));
draw((-1.110459431125061,4.058413889003155)--(0.59,1.2649171606442673));
draw((-1.110459431125061,4.058413889003155)--(-1.110459431125061,1.75));
/* dots and labels */
dot("$A$",(-1.110459431125061,4.058413889003155),NW);
dot("$B$",(-1.62,1.75),SW);
dot("$C$",(2.8,1.75),E);
dot("$D$",(-0.46275471959342035,1.75),S);
dot("$O_B$",(-1.0413773597967102,2.832722400774602),dir(150));
dot("$O_C$",(1.1686226402032898,3.452813719451886),N);
dot("$T$",(-4.900190463027996,1.75),SW);
dot("$M$",(-0.7866070753592407,2.9042069445015777),dir(-30));
dot("$X$",(0.59,1.2649171606442673),NE);
dot((-1.110459431125061,1.75));
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/4/4/f444c6a997258b0a406e9bdd80aadc70d596299b.png)
,
,
,
and 
,
.
,
.
is the perpendicular bisector of 
,
,
isosceles. Hence 
,
and 
,
.
be the intersection of 
and 
,
,
.
,
is a tangent to 
.
.
.
are isogonal in 
.
by properties of Miquel points and 
is perpendicular to line 
.
.
.
from the Miquel configuration and properties of the circumcenter of 
,
is acute and 
is obtuse; other cases are analogous. Thus angle 
.
are directly similar. Therefore 
to 
,
are cyclic. Let 
,
.
are congruent. Now angle chasing we get 
.
.
.
,
.
.
.
.
,
,
.
is the perpendicular bisector of 
is isosceles and 
.
and 
are perpendicular since the points 
.
,
and 
.
is isosceles.
as desired. 
![[asy]import geometry; size(10cm); pen c1=orange+white+white; pen c2=orange+white; pen c3=brown+red+white; pair A,B,C,D,H,O_B,O_C,E,F,X; A=(-2,5); B=(-3,-3); C=(7,-3); D=(0,-3); H=orthocenter(A,B,C); O_B=circumcenter(A,B,D); O_C=circumcenter(A,C,D); E=extension(B,O_B,C,O_C); F=extension(B,C,O_B,O_C); X=circumcenter(B,C,O_B); fill(B--O_B--O_C--C--cycle,c1); draw(B--O_B--O_C--C--cycle,c2); draw(B--F,c2); draw(O_B--F,c2); draw(O_B--E,c2); draw(O_C--E,c2); draw(A--H); draw(A--O_B); draw(A--D); draw(A--X); draw(A--O_C); draw(E--F,dotted); draw(circle(B,C,O_B),c2); draw(circle(A,B,D)); draw(circle(A,C,D)); draw(A--B,c3); draw(A--C,c3); draw(arc(intersectionpoint(bisector(A,C),bisector(A,F)),C,F),dotted); dot(A^^B^^C^^D^^H^^O_B^^O_C^^E^^F^^X); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$H$",H,0.6S); label("$O_B$",O_B,SE); label("$O_C$",O_C,NE); label("$E$",E,NE); label("$F$",F,SW); label("$X$",X,NE);[/asy]](http://latex.artofproblemsolving.com/8/e/1/8e1548778475fc9027ba4628886433956f4ef55a.png)
Let 
and 
We see that 
and 
so 
Hence, the Miquel point of 
and since 
Notice that 
![[asy]
import olympiad;
size(230);
/** Coordinates from Geogebra **/
pair A = (-0.36 , 4.12);
pair B = (-3 , -2);
pair C = (4.64 , -1.82);
draw(A--B--C--cycle , green);
pair D = (0.27 , -1.92);
draw(A--D);
draw(circumcircle(A,B,D) , purple);
pair O_B = circumcenter(A,B,D);
draw(circumcircle(A,C,D) , heavycyan);
pair O_C = circumcenter(A,C,D);
draw(circumcircle(B , O_C , C) , orange);
pair X = circumcenter(B , O_C , C);
pair H = orthocenter(A,B,C);
draw(circumcircle(A,B,C) , blue);
pair O = circumcenter(A,B,C);
/** Dots and Labels **/
dot("$A$" , A , N);
dot("$B$" , B , SW);
dot("$C$" , C , SE);
dot("$D$" , D , S);
dot("$X$" , X , S);
dot("$H$" , H , S);
dot("$O$" , O , NE);
dot("$O_B$" , O_B , NW);
dot("$O_C$" , O_C , NE);
[/asy]](http://latex.artofproblemsolving.com/1/e/6/1e6f5863a7168eca58bea97c0a2d275c86f69727.png)
and 
.
and 
concur.
.
be the midpoint of 
Also 
.
bisects 
which is true since 
.
,
.
.
is isosceles. So 
.
inversion to get the following problem:
be a point on the circumcircle of triangle 
and 
be the reflections of 
and 
,
has circumcenter 
are isogonal.
and 
.
of 
,
![[asy]
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -0.004, xmax = 0.004, ymin = -0.0055, ymax = 0.0035; /* image dimensions */
draw(arc((-0.0015446764438354107,0.0017601067586798715),0.0005941504392848956,-101.31153334263735,-72.96771295549291)--(-0.0015446764438354107,0.0017601067586798715)--cycle, linewidth(0.75) + blue);
draw(arc((-0.0015446764438354107,0.0017601067586798715),0.00047532035142791644,-72.9677129554929,-44.653596847594095)--(-0.0015446764438354107,0.0017601067586798715)--cycle, linewidth(0.75) + blue);
draw(arc((-0.0019739622178513794,-0.000386010299822956),0.0005941504392848956,-51.24139176033055,-22.945299768886326)--(-0.0019739622178513794,-0.000386010299822956)--cycle, linewidth(0.75) + blue);
/* draw figures */
draw((-0.0015446764438354107,0.0017601067586798715)--(-0.003011434824851063,-0.002313120907518996), linewidth(0.5));
draw((-0.003011434824851063,-0.002313120907518996)--(0.0025781038703707464,-0.002313120907518996), linewidth(0.5));
draw((0.0025781038703707464,-0.002313120907518996)--(-0.0015446764438354107,0.0017601067586798715), linewidth(0.5));
draw(circle((-0.0002166654772401582,-0.0025748828877237124), 0.002807001075993056), linewidth(0.5) + red);
draw((xmin, -6.697329579162294*xmin-0.00858510047885425)--(xmax, -6.697329579162294*xmax-0.00858510047885425), linewidth(0.5) + blue); /* line */
draw((-0.0019739622178513794,-0.000386010299822956)--(-0.0015446764438354107,0.0017601067586798715), linewidth(0.5));
draw((-0.0015446764438354107,0.0017601067586798715)--(0.0008208071297595251,0.000031285775362950524), linewidth(0.5));
draw((xmin, -3.264272476241373*xmin-0.003282138041630463)--(xmax, -3.264272476241373*xmax-0.003282138041630463), linewidth(0.5) + blue); /* line */
draw(circle((0.0009335560702811858,0.00014839489907061385), 0.0029562225072598644), linewidth(0.5) + linetype("4 4") + blue);
draw(circle((-0.001439392660103263,-0.0005785507672779155), 0.0023410262063539478), linewidth(0.5) + linetype("4 4") + blue);
draw((-0.0002166654772401582,-0.0025748828877237124)--(-0.0019739622178513794,-0.000386010299822956), linewidth(0.5));
draw((-0.0019739622178513794,-0.000386010299822956)--(0.0025781038703707464,-0.002313120907518996), linewidth(0.5));
/* dots and labels */
dot((-0.0015446764438354107,0.0017601067586798715),linewidth(3pt) + dotstyle);
label("$A$", (-0.0015962278670139036,0.001969223957289919), NE * labelscalefactor);
dot((-0.003011434824851063,-0.002313120907518996),linewidth(3pt) + dotstyle);
label("$B$", (-0.0034737432551541735,-0.002296776196775628), NE * labelscalefactor);
dot((0.0025781038703707464,-0.002313120907518996),linewidth(3pt) + dotstyle);
label("$C$", (0.0026341232606945527,-0.0023443082319184196), NE * labelscalefactor);
dot((-0.0009364896108516965,-0.002313120907518996),linewidth(3pt) + dotstyle);
label("$D$", (-0.001406099726442737,-0.0026057344252037734), NE * labelscalefactor);
dot((-0.0019739622178513794,-0.000386010299822956),linewidth(3pt) + dotstyle);
label("$O_1$", (-0.0023923894556556638,-0.0002647816944212866), NE * labelscalefactor);
dot((0.0008208071297595251,0.000031285775362950524),linewidth(3pt) + dotstyle);
label("$O_2$", (0.0007566078725542828,0.0001748896306495358), NE * labelscalefactor);
dot((-0.0002166654772401582,-0.0025748828877237124),linewidth(3pt) + dotstyle);
label("$X$", (-0.00013461778637306054,-0.0026532664603465647), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/c/5/d/c5d0d5eac74bffa8a67eb50a01c55189b3e4e491.png)
.
.
.
.
is cyclic, we thus get the 
and 
.
and 
are also cyclic.
.
(I felt that this was really messy IDK why lol (maybe due to my nubness 
))
and we also have 
.
,![\[\measuredangle DAH = \measuredangle O_BPD = \measuredangle APO_B = \measuredangle XAD. \quad \blacksquare\]](http://latex.artofproblemsolving.com/6/0/9/60920d9fb111d27bf81fa3890eef60a7a39b29f1.png)
and 
.
,
,
.
and 
are cyclic.
Now by our first claim, 
,
.
.
bisects 
as 
is cyclic. From the following 
and 
,![\[\angle HAD = 90^{\circ} - \angle ADB = \frac{1}{2}(180^{\circ} - \angle AO_BB) = \frac{1}{2} \angle AZB,\]](http://latex.artofproblemsolving.com/c/3/b/c3b6dd302facb63695458ad536bc247a9cf679ba.png)
and ![\[\angle DAX = \angle AZO_B = \frac{1}{2} \angle AZB,\]](http://latex.artofproblemsolving.com/a/a/a/aaae9ef38fd07e6fe397e3c2795fd4f51cfe364c.png)
so that 
,
and let 
be on line 
cyclic. Since 
so 
.
are collinear.
since Miquel inversion fixes 
.
are cyclic we have 
,
is the external bisector of 
.
we get the desired result.
.
.
.
,
.
,
.
.
