[/quote]
,
board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average.
,
lie on a moving circle 
.
at 
and 
.
intersect at 
.
intersect at 
.
is the perpendicular image of 
.
and the orthocenter 
are collinear.
be a positive integer. Let 
satisfying 
for odd values of 
.
be the number of pairs of integers 
.
and 
be positive integers and let 
be a set of 
of 
![\[ k < \frac {1}{2} + \sqrt {2 \cdot n}
\]](http://latex.artofproblemsolving.com/d/a/5/da56146bf1ad38f257e203e4a0231bd604060a7a.png)
.
and 
. Prove that
small equilateral triangles of side 
.
vertices of the small triangles, such that for every integer 
,
,
is the center of the inscribed circle, point 
.
is equal to the diameter of the circle inscribed in triangle 
from the set of non-negative integers to itself such that 
,
,
.
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
.
.
for every 
or 
for every 
be an integer. Ion draws a regular 
Y by Adventure10, mathematicsy, HWenslawski, Mango247, ItsBesi, and 1 other user
Let 
be a triangle. Let 
and 
be the points in which the median and the angle bisector, respectively, at 
meet the side 
. Let 
and 
be the points in which the perpendicular at to 
meets 
and 
, respectively. And 
the point in which the perpendicular at to meets 
produced.
Prove that
is perpendicular to .
be a triangle. Let 
and 
be the points in which the median and the angle bisector, respectively, at 
meet the side 
. Let 
and 
be the points in which the perpendicular at to 
meets 
and 
, respectively. And 
the point in which the perpendicular at to meets 
produced.Prove that
is perpendicular to .
at the midpoint K of the arc BC opposite to the vertex A. Since 
,
and the isosceles triangles 
together with the points 
resp. 
on their bases are similar. Hence, the angles 
are equal, which means that the lines 
are parallel and 
is the perpendicular bisector of BC.
be the second intersection of 
and the circumcircle of triangle 
.
on 
and 
on 
such that 
,
.
is perpendicular to 
.
,
.
.
(since 
)
hit 
at 
.
tangent to 
and 
.
be the intersection of 
and the line through 
of 
be the intersection of 
and 
.
,
,
are harmonic conjugates, whence 
and 
.
cut the circumcirle of 
are collinear and 
.
cut the circumcircle again at 
cut 
.
,
,
corresponds to 
.
,
(since 
is cyclic
.
,
,
.
,
meet 
.
.
.
,
.
,
and 
.
are angles such that 
and 
.
and 
.
.
and 
.
as desired. 
is isosceles, which implies that 
is cyclic. Therefore 
and 
is also isosceles.![\[\dfrac{AB\sin\angle BAM}{AC\sin\angle MAC}=\dfrac{BM}{CM}=1\implies \dfrac{AB}{AC}=\dfrac{\sin\angle QAR}{\sin\angle PAR}.\]](http://latex.artofproblemsolving.com/4/b/1/4b1cb921d36d58c8f9f5fb3241f1146d9180d51c.png)
Next, applying the Ratio Lemma to 
and applying it to 
.![\[\dfrac{\sin\angle POQ}{\sin\angle QOR}=\dfrac{PQ}{QR}=\dfrac{\sin\angle PAR}{\sin\angle QAR}=\dfrac{AC}{AB}.\]](http://latex.artofproblemsolving.com/b/2/4/b24311a23e98d3cf8bf9e86525654bf87bd158c9.png)
In addition, by LoS we have 
,
.
is cyclic we have 
.
.
is cyclic, so 
and 
,
by the South Pole Theorem. Let 
and 
,
)
is the Simson line of point 
,
,
.
be the second intersection of AO with circumcircle of 
and 
.
and 
maps to 
.
maps to 
.
maps to 
.
collinear.So, 
. But this is trivial. Q.E.D. Please check my solution whether it is correct.
,
,
,
.
and 
.
is a projective transformation, and so lines 
and 
each have degree 1, therefore their intersection 
is fixed, 
has degree 1. Now to prove 
we only must check 4 points for 
where 
with 
.
.
.
and is the 
.
.
and hence 
and we know 
to meet 
at 
be the altitude from 
,
collinear, because 
.
is cyclic, so by Simson line wrt 
,
collinear; hence, 
.
be the homothety about 
,
and 
;
lies on both 
,
.
,
.
and 
,
,
and 
are cyclic.
.
,
note that 
.
is collinear and 
so there's a homothety now at A sending 
implying the desired result.
.
are collinear, call this line 
.
.![\[\angle ARX = 180 - \angle XAR - \angle AXR = 180-\angle DAC-\angle BXM = 180-\angle DBM - \angle BDM = 90\]](http://latex.artofproblemsolving.com/6/f/2/6f2c1973234f98a955657c55cb575f42baabe37e.png)
Thus, 
,
,
.
![\[\frac{AQ}{AM}=\frac{AP}{AX}=\frac{AO}{AD}\]](http://latex.artofproblemsolving.com/2/2/a/22addc01ea67bbd999ecdea6778f6216d425789f.png)
Thus, by SAS-ratio similarity, we have 
.
,
![[asy]
unitsize(0.8cm);
pair A, B, C, B1, C1, M, N, P, Q, O;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
M=(4,0);
N=(3.5,0);
P=(29/4,3sqrt(5)/8);
Q=(127/32,3sqrt(5)/64);
O=(127/32,-15sqrt(5)/16);
B1=(-31/16,-93sqrt(5)/32);
C1=(221/16,-93sqrt(5)/32);
draw(A--B1--C1--A--(-1/4,-3sqrt(5)/8)--P--A--B--C--P--O--A--M--Q--N--P--N);
draw(circle(O,63sqrt(5)/32));
label("$A$", A, dir(90));
label("$B$", B, W);
label("$C$", C, NE);
label("$B'$", B1, SW);
label("$C'$", C1, SE);
label("$M$", M, S);
label("$N$", N, SSW);
label("$O$", O, S);
label("$P$", P, NE);
label("$Q$", Q, NNE);
[/asy]](http://latex.artofproblemsolving.com/a/0/4/a04ca8e9e83f08c611c82090f8a22dd79589eed2.png)
be the line such that 
and 
,
,
,
and 
.
(with 
)
.
and 
be the Simson line of 
wrt 
at 
,
so 
.![[asy]
size(200);
pair A=dir(145),B=dir(-150),C=dir(-30),M=1/2*(B+C),Op=dir(-90),Pp=foot(Op,A,B),Rp=foot(Op,A,C),N=extension(A,Op,B,C),P=extension(A,B,N,N+M-Rp),R=extension(N,P,A,C),Q=extension(N,P,A,M),O=extension(Q,foot(Q,B,C),A,N);
draw(circumcircle(A,B,C),red);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$M$",M,dir(-60));
dot("$O'$",Op,dir(Op));
dot("$P'$",Pp,dir(Pp));
dot("$R'$",Rp,dir(-90));
dot("$N$",N,dir(N));
dot("$R$",R,dir(90));
dot("$P$",P,dir(-170));
dot("$Q$",Q,dir(90));
dot("$O$",O,dir(-140));
draw(Pp--A--C--B^^Op--A--M,red);
draw(P--O--R^^Q--O,blue);
draw(Pp--Op--Rp^^Op--M,blue);
draw(P--R^^Pp--Rp,green);
[/asy]](http://latex.artofproblemsolving.com/6/9/0/690bf116f9ceece2de14ad67f66c229faf77a993.png)
at 
.
and 
as 
and 
.
.
as 
Now since 
.
,
,
,
and 
,
.
.
and 
and 
for later use.
to be a cevian, s.t. 
and 
then 
then from sine rule we get 
and 
, dividing these we get 
and 
we get 
hence we get :
and hence 
and as 
,
and therefore points 
are concyclic which gives 
are concyclic which gives us 
and since 
applying 
and since 
we get 
and hence 
to be the median of 
and 
and let 
is cyclic so we get 
.![\[\measuredangle(OQ,KN)=\measuredangle(PQ,CN)\implies \measuredangle QOK=\measuredangle CNP.\]](http://latex.artofproblemsolving.com/b/f/4/bf443b239be31da8d62906c9d6688e993cb6f14d.png)
:![\[\measuredangle(OQ,BC)=\measuredangle(OQ,KN)-\measuredangle(KN,BC)=\measuredangle QOK-\measuredangle BNA\]](http://latex.artofproblemsolving.com/3/1/c/31c433a98e58ac5bf4e509d1affdc424c74a0d19.png)
![\[\measuredangle QOK-\measuredangle BNA=\measuredangle CNP-\measuredangle BNA=\measuredangle BNP-\measuredangle BNA=90^{\circ}\]](http://latex.artofproblemsolving.com/0/4/c/04cc697b4111272b313020cb4a2c69f4093c24b2.png)
done!
where 
.
.![\[-1=(\infty_{BC},M;B,C)\stackrel{A}{=}(R,AQ\cap (APO);P,T)\]](http://latex.artofproblemsolving.com/7/b/0/7b060e8faa263ae99069e1e559e10c7e344dceb8.png)
so 
and 
bisects 
.
.![\[QO \parallel ME \perp BC. \quad \blacksquare\]](http://latex.artofproblemsolving.com/7/2/2/7220da6d45592c30db51b1b4fb22990e5857b9e4.png)
,
.
.
,
and 
.
.![[asy]
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 = -6.141728024042078, xmax = 3.174575507137483, ymin = -5.81445529676934, ymax = 3.2614274981217113; /* image dimensions */
/* draw figures *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */
draw((-4.88,-1.88)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8));
draw((-5.211598643768283,-3.0982645825399944)--(-1.494317431054321,-4.110068936296458), linewidth(0.8));
draw((-2.468833572833071,-1.892791333831124)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8));
draw((-2.468833572833071,-1.892791333831124)--(-3.96,1.5), linewidth(0.8));
draw((-3.96,1.5)--(-1.11,-1.9), linewidth(0.8));
draw((-1.480249684031839,-1.4582986225585077)--(-2.468833572833071,-1.892791333831124), linewidth(0.8));
draw((-2.468833572833071,-1.892791333831124)--(-1.494317431054321,-4.110068936296458), linewidth(0.8));
draw((-1.480249684031839,-1.4582986225585077)--(-1.494317431054321,-4.110068936296458), linewidth(0.8));
draw(circle((-1.1057141289867563,-1.0921133140036046), 3.8556451167217896), linewidth(0.8)); /* special point */
draw((-1.494317431054321,-4.110068936296458)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8));
draw((-1.1057141289867563,-1.0921133140036046)--(-1.11,-1.9), linewidth(0.8));
draw((-1.11,-1.9)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8));
draw((-1.1261681919167021,-4.947704176298286)--(2.66,-1.92), linewidth(0.8));
draw((-5.398473910768285,-3.784828063474779)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8));
draw((-5.398473910768285,-3.784828063474779)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); /* special point */
draw((0.9060966758985711,-1.0139049292406515)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8));
draw((0.9060966758985711,-1.0139049292406515)--(-5.398473910768285,-3.784828063474779), linewidth(0.8)); /* special point */
draw((-3.96,1.5)--(-4.88,-1.88), linewidth(0.8));
draw((-4.88,-1.88)--(2.66,-1.92), linewidth(0.8));
draw((2.66,-1.92)--(-3.96,1.5), linewidth(0.8));
/* dots and labels */
dot((-3.96,1.5),linewidth(3pt) + dotstyle);
label("$A$", (-4.443756574004514,1.578482344102179), NE * labelscalefactor);
label("$B$", (-5.345334335086407,-1.8925920360631068), NE * labelscalefactor);
label("$C$", (2.7087603305785044,-1.9677235161532645), NE * labelscalefactor);
label("$M$", (-1.0778662659654463,-2.238196844477832), NE * labelscalefactor);
label("$N$", (-3.1162584522915163,-1.7122764838467284), NE * labelscalefactor);
dot((-5.211598643768283,-3.0982645825399944),linewidth(3pt) + dotstyle);
label("$P$", (-5.630833959429006,-3.0946957175056298), NE * labelscalefactor);
label("$Q$", (-2.0944703230653709,-1.5418031555221607), NE * labelscalefactor);
label("$O$", (-1.9193388429752132,-4.4719308790383105), NE * labelscalefactor);
dot((-1.1057141289867563,-1.0921133140036046),linewidth(3pt) + dotstyle);
label("$D$", (-1.363365890308046,-0.9008564988730251), NE * labelscalefactor);
label("$E$", (-1.228129226145762,-5.318587528174298), NE * labelscalefactor);
dot((-5.398473910768285,-3.784828063474779),linewidth(3pt) + dotstyle);
label("$F$", (-5.8562283996994795,-3.966220886551459), NE * labelscalefactor);
label("$G$", (0.8304733283245604,-0.8858302028549936), NE * labelscalefactor);
label("$H$", (-2.595522163786633,-2.9242223891810623), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/f/3/8/f389a5ecdf332b8779738c21ab756f447ca7c575.png)
are collinear.
,
.
.
.
and 
are homothetic with center 
.
.
and 
as points 
is a Simson line of 
because 
is an angle bisector. Therefore, 
.
,
,
to 
.
.
.
be the point on the angle bisector such that if you apply to it the problem's statement (instead of to 
are collinear. Then, 
,
and draw perpendiculars from 
to get 
,
are collinear.
which intersects passes through 
on 
,
and as 
,
,![\[X_A:QO\mapsto ML\]](http://latex.artofproblemsolving.com/5/6/a/56a9dbb485018afbfb9756de6b8f6f3d36d33ca0.png)
as desired.
