[/quote]
,
be an odd prime and 
be integers so that the integers 
are divisible by 
Prove that 
be real numbers such that 
Prove that 
Where 
.
and 
respectively denote the midpoints of the arcs 
and 
that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
)
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
friends, 
people are seated around a circular table in a fixed order. Ali places 
is the circumcenter of 
.
lie on 
,
is tangent to circumcircle of 
.
is tangent to circumcircle of triangle 
.
be the sequence defined as follows: 
,
,
is the smallest integer greater than 
such that in 
there are no arithmetic progressions of length 
.
with 
,
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
positive integers 
such that for every index 
,
,
is a multiple of 
.
and 
be positive integers satisfying the equation ![$(a, b) + [a, b]=2021^c$](http://latex.artofproblemsolving.com/2/4/6/246ccbb22acecb7adc72219afb8afb34c9b05101.png)
.
is a prime number, prove that the number 
is composite.
Y by mathenthusiastic, Mathuzb, itslumi, tiendung2006, Adventure10, Mango247, sami1618
Let 
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let 
be a point such that 
and 
. Prove that 
passes through the midpoint of 
.
Proposed by Iman Maghsoudi, Hooman Fattahi
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let 
be a point such that 
and 
. Prove that 
passes through the midpoint of 
.Proposed by Iman Maghsoudi, Hooman Fattahi
.
;
and 
be the 
.
.
bisects the arc 
of 
since 
is a symmedian in 
and 
.
.
hence 
.
.
so 
hence 
in 
.
;
bisects the side 
.
be the A-Humpty point and 
be the second intersection of the circumcircle and the A-symmedian. Let 
be the perpendicular from 
to 
.
and 
.
.
.
be the 
point. 
.
.
are conyclic.
are collinear.
is the angle bisector of 
.
is a harmonic quadrilateral
and 
are collinear). By Angle bisector theorem, the conclusion follows.
is the angle bisector of 
.
intersect the circumcircle of 
.
is a parallelogram. Since 
is the midpoint of 
.
is cyclic, so 
so 
which is 
.
and 
.
,
intersect 
in 
,
is the diameter of the circle 
.
.
and the perpendicular in 
be the orthocenter of 
be the intersection of 
is cyclic.
,
passes through the midpoint of arc 
;
.
and 
,
.
so 
,
is cyclic. It follows that 
and we're done.
be the midpoint of 
and let 
be the midpoint of arc 
arc 
respectively. Note that 
so 
On the other hand, 
are tangent to 
so by 
we get 
and 
are homothetic with center 
hence we conclude that 
be the foot of perpendiculars from 
.
so 
is the angle bisector of 
and this means that 
are collinear,so we are done by lemma 2.
.
is isosceles, and since 
is cyclic with 
the antipode of 
,
,
.
is also harmonic (say, after an inversion at 
)
is the orthocenter of the triangle, then we can see that 
,
and 
intersect at 
,
is a diameter of 
,
is the midpoint of BC.
![[asy]size(7cm);
pair A=(1.5,7),B=(0,0),C=(10,0),Ee,F,M,Nn,P,X,H,Xa,D;
Ee=foot(B,A,C);F=foot(C,A,B);
M=extension(Ee,F,A,bisectorpoint(B,A,C));
Nn=extension(B,C,A,bisectorpoint(B,A,C));
P=extension(Nn,Nn+(B-Nn)*(0,1),M,M+(Ee-M)*(0,1));
H=extension(B,Ee,C,F);D=(B+C)/2;
Xa=foot(H,A,D);X=extension(Ee,F,B,C);
draw(circumcircle(X,M,Nn)^^circumcircle(A,Ee,F),green);
D(MP("A",A,N)--MP("F",F,W)--MP("B",B,S)--MP("N",Nn,S)--MP("D",D,S)--MP("C",C,S)--MP("E",Ee,NE)--A--MP("M",M,NW)--F--MP("H",H,S)--C);
D(B--H--Ee--M--Nn--MP("P",P,E)--M);
D(A--MP("X_A",Xa,NE)--P--D,dotted);
D(F--MP("X",X,S)--B);
dot(A^^B^^C^^Ee^^F^^M^^Nn^^P^^Xa^^H^^D^^X);
[/asy]](http://latex.artofproblemsolving.com/d/6/1/d6130cd9e9eb731fe908bd6406e0d5e5e31bf281.png)
and 
,
,
as well. It's well-known that 
A
,
.
are collinear. Assume the configuration shown above. We have 
and also 
which proves our claim. But since 
,
and the line through 
cuts 
at 
is perpendicular to 
is parallel to 
then 
.
.
.
,
;
and 
.
is isosceles, and since 
in 
,
.
with 
,
and 
;
via Simson's theorem, and so by spiral similarity we obtain that 
and hence 
.
through 
,
,
forms a harmonic bundle, and so projecting through 
be the points on 
s.t 
.
be midpoint of arcs opposite to 
,
So 
.
,
the intersection point of the line passing through 
with 
resp.
.
isosceles. So redefine 
(denote by 
)
.
be 
= 
= 
is internal bisector of 
= 
+ 
= 
= 
or 
)
= 
or 
and 
is 
.
is the 
and 
,
,
is rectangle
is cyclic. Let 
and 
point.
.
. Call this 
.
concyclic.Also since the images of 
and 
.Let 
where 
. Clearly 
lies on polar of 
is the polar of 
is the orthocentre w.r.t 
.
.
we can easily notice 
.
![[asy]
size(6cm); defaultpen(fontsize(10pt));
pair A,B,C,O,EE,F,NN,M,SS,T,P,H;
A=dir(120);
B=dir(210);
C=dir(330);
O=(B+C)/2;
EE=foot(B,C,A);
F=foot(C,A,B);
NN=extension(A,incenter(A,B,C),B,C);
M=extension(A,NN,EE,F);
SS=extension(B,C,EE,F);
T=foot(SS,A,O);
P=2*circumcenter(SS,M,NN)-SS;
H=A+B+C;
draw(circumcircle(SS,M,NN));
draw(A--NN,gray);
draw(B--EE,gray);
draw(C--F,gray);
draw(circumcircle(A,EE,F),gray);
draw(arc(reflect(B,C)*(0,0),1,-10,190),gray);
draw(B--A--C--SS);
draw(SS--EE);
draw(A--O);
dot("$A$",A,N);
dot("$B$",B,SE);
dot("$C$",C,E);
dot("$O$",O,S);
dot("$E$",EE,NE);
dot("$F$",F,dir(110));
dot("$N$",NN,dir(220));
dot("$M$",M,dir(250));
dot("$S$",SS,dir(210));
dot("$T$",T,dir(75));
dot("$P$",P,E);
dot("$H$",H,S);
[/asy]](http://latex.artofproblemsolving.com/5/c/b/5cb05103beb05e1cd84bc862501b7254cb36af58.png)
.
and 
.
and 
,
.
be the spiral similarity at ![\[\frac{NB}{NC}=\frac{AB}{AC}=\frac{AE}{AF}=\frac{ME}{MF},\]](http://latex.artofproblemsolving.com/5/a/d/5adfd1e7d64b94ca7bc2ab8a4ae83c85cc4d307b.png)
so 
is cyclic, say with circumcircle 
,
is a diameter of 
,
,
and 
,
.
is isosceles. 
and let 
be an altitude. Thus 
is a kite inscribed in some circle 
.![[asy]size(8cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair E = foot(B, C, A); pair F = foot(C, A, B); pair I = incenter(A, B, C); pair M = extension(E, F, A, I); pair N = extension(B, C, A, I); pair Z = N+dir(270); pair P = extension(A, midpoint(B--C), N, Z); draw(M--P--N, red); filldraw(unitcircle, invisible, blue); draw(A--B--C--cycle, blue); draw(A--P, blue); draw(A--N, blue); draw(E--F, blue); pair T = extension(E, F, B, C); draw(F--T--B, lightblue); draw(circumcircle(P, M, N), deepgreen); pair Q = foot(T, A, midpoint(B--C)); draw(T--Q, deepgreen); pair D = foot(A, B, C); pair S = extension(Q, T, M, N); draw(A--D, blue); draw(B--E, blue+dotted); draw(C--F, blue+dotted); pair X = extension(E, F, A, D);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(45)); dot("$F$", F, dir(F)); dot("$M$", M, dir(190)); dot("$N$", N, dir(N)); dot("$P$", P, dir(45)); dot("$T$", T, dir(T)); dot("$Q$", Q, dir(45)); dot("$D$", D, dir(D)); dot("$S$", S, dir(260)); dot("$X$", X, dir(X));
/* TSQ Source:
A = dir 130 B = dir 210 C = dir 330 E = foot B C A R45 F = foot C A B I := incenter A B C M = extension E F A I R190 N = extension B C A I Z := N+dir(270) P = extension A midpoint B--C N Z R45 M--P--N red unitcircle 0.1 lightcyan / blue A--B--C--cycle blue A--P blue A--N blue E--F blue T = extension E F B C F--T--B lightblue circumcircle P M N deepgreen Q = foot T A midpoint B--C R45 T--Q deepgreen D = foot A B C S = extension Q T M N R260 A--D blue B--E blue dotted C--F blue dotted X = extension E F A D
*/ [/asy]](http://latex.artofproblemsolving.com/2/5/a/25aa2502b5bf3db505cccdf2c64583f9f652ab54.png)
passes through the orthocenter 
,
.![\[ -1 = (PT;MN)_\gamma \overset{Q}{=} (AS; MN) \overset{T}{=} (AH';XD). \]](http://latex.artofproblemsolving.com/d/c/8/dc8cb86260237a4e833f9c67609207d06f54eb5d.png)
On the other hand, 
is a complete quadrilateral, so 
.
.
.
bisects 
be the midpoint of 
such that 
.
.
and similarly for 
,
(and similar result for 
)
this expression reduces to
.
,
.
such that 
,
,
and etc.Also assume WLOG 
.![[asy]
size(6cm);
pair A,B,C,E,F,N,R,M,G,O,I,H,S,T,K,X,L,Y,Z,P;
A=dir(130);
B=dir(210);
C=dir(330);
H=orthocenter(A,B,C);
R=midpoint(B--C);
O=circumcenter(A,B,C);
I=incenter(A,B,C);
S=midpoint(A--H);
E=foot(B,C,A);
F=foot(C,A,B);
X=foot(S,A,I);
Y=foot(O,A,I);
G=extension(S,R,E,F);
K=2*X-A;
L=2*Y-A;
M=extension(A,K,B,C);
N=extension(A,K,E,F);
T=extension(B,C,E,F);
Z=circumcenter(T,M,N);
P=2*Z-T;
draw(A--B--C--A);
draw(A--L,gray);
draw(A--H,gray);
draw(F--E,gray);
draw(S--R,gray);
draw(circumcircle(A,F,E),gray);
draw(unitcircle);
draw(B--E,gray);
draw(A--R,dashed);
draw(C--F,gray);
draw(R--L,gray);
draw(P--N--M--P,gray);
dot("$A$",A,dir(120));
dot("$B$",B,dir(180));
dot("$C$",C,dir(0));
dot("$E$",E,NE);
dot("$F$",F,dir(200));
dot("$H$",H,dir(270));
dot("$R$",R,dir(50));
dot("$S$",S,dir(0));
dot("$K$",K,2*dir(240));
dot("$L$",L,dir(270));
dot("$N$",N,dir(80));
dot("$M$",M,dir(250));
dot("$P$",P,dir(0));
[/asy]](http://latex.artofproblemsolving.com/d/5/2/d52c94695b941fc0ed02b643e2ba5bb4b5a00b3d.png)
repectively.Also let 
in 
and 
in 
respectively.Now a simple angle chase shows that 
by the incenter-excenter lemma.Now angle chase to get 
where the last equality holds due to 
and whence the claim holds.
and 
.
are homothetic with center 
are collinear and we are done 
,
and 
be the midpoints of 
,
is cyclic. Now since 
it follows that 
which means that 
at 
to get the following diagram
are the feet of the altitudes of 
respectively. 
is the midpoint of the minor arc 
is the midpoint of the minor arc 
is the circle tangent to 
is perpendicular to the common tangent of 
and so 
is perpendicular to the common tangent of 
,![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
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pen ccqqqq = rgb(0.8,0,0);
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draw((4.607272727272718,-3.6481818181818184)--(-3.0371269115540978,2.4580737591692805), linewidth(0.5));
draw((-3.8523178858599465,-2.152425976034979)--(-0.711663246950657,0.6005210221578313), linewidth(0.5));
draw((-3.8523178858599465,-2.152425976034979)--(4.607272727272718,-3.6481818181818184), linewidth(0.5));
draw((-0.711663246950657,0.6005210221578313)--(-4.12,-3.666363636363637), linewidth(0.5));
draw((-3.0371269115540978,2.4580737591692805)--(0.24363636363635877,-3.6572727272727277), linewidth(0.5) + dotted);
draw(circle((-0.7320655081523083,-1.8385633255371434), 1.8207381627180734), linewidth(0.5) + ccqqqq);
draw(circle((0.23868179362544875,-1.2790791220359572), 4.969631206682314), linewidth(0.5));
draw(circle((-3.032172341543187,0.07988015393250965), 2.3781987662415993), linewidth(0.5));
draw((-3.0371269115540978,2.4580737591692805)--(-0.7282723118783992,-3.659297537013383), linewidth(0.5));
draw((-0.7282723118783992,-3.659297537013383)--(-0.7320655081523083,-1.8385633255371434), linewidth(0.5));
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dot((-4.12,-3.666363636363637),dotstyle);
label("$B$", (-4.048223248823088,-3.4978134199624327), NW * labelscalefactor);
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label("$C$", (4.676794204961867,-3.4815049761235826), E * labelscalefactor);
dot((-0.711663246950657,0.6005210221578313),linewidth(4pt) + dotstyle);
label("$E$", (-0.6397584865033575,0.7260735342997986), NE * labelscalefactor);
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dot((-5.582945987043152,-3.66941144050331),linewidth(4pt) + dotstyle);
label("$K$", (-5.515983194319622,-3.546738751478984), N * labelscalefactor);
dot((0.2490351695044648,-6.248699543963433),linewidth(4pt) + dotstyle);
label("$M^*$", (0.3061312561499647,-6.123472878017333), NE * labelscalefactor);
dot((-1.4645461139418077,-1.7085206351215922),linewidth(4pt) + dotstyle);
label("$N^*$", (-1.4062553469293255,-1.5734170469780957), E * labelscalefactor);
dot((-2.2819905664053017,-0.7759524769385739),linewidth(4pt) + dotstyle);
label("$M_{EF}$", (-2.825089960909309,-0.6112188604859272), S * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/a/b/0ab9fe90c02074beb035168a80dcbf21efeefd52.png)
.
are collinear and 
.
is cyclic. We prove that 
,
.
Thus the aforementioned spiral similarity also takes 
is cyclic. Note that 
is obviously cyclic, so the two cyclic quadrilaterals have the same circumcircle. Now 
,
.
,
moves through some fixed point (call it 
,
.
,
,
.
.
are tangent to 
,
is harmonic and so, 
bisects 
.
Also, 
-
But also, 
-
and 
be points on 
.
and 
similarly. Let 
and 
.
.
.
and 
meet 
and 
,
and so 
as desired.
.
which shows that 
is a kite. In particular, 
,
.
and applying the claim shows that 
is cyclic.
we know 
Notice 
so 
Hence, 
and 
is cyclic with diameter 
so 
linearly and a point 
linearly, then the point 
is 
on both lines, and 
and 
then 
and 
and 
are scaled equally. Let 
.
and 
where 
and 
which implies all such 
must lie on a line. Undoing the linear transformation means that 
and 
of 
.
and 
.
,
and 
gives 
which lies on that line. 
.
.
intersect 
and 
.
.
are concyclic.
and since 
is cyclic this implies 
is cyclic. Similarly 
is cyclic, done.
is cyclic.
is cyclic and 
.
,![\[(U,V;P,Q)\stackrel{A}{=}(B,C;AP\cap BC,\infty_{BC})\]](http://latex.artofproblemsolving.com/7/2/4/724fff1c86ec210bdf889cbbe9a339bbc40b6e53.png)
implying that 
,
be the 
is the 
-
is 
be the foot of perpendicular from 
.
is angle bisector. Therefore 
.
of 
where 
and 
.
with 
which finishes as if 
.
and the A-symmedian at 
,
and 
.
since 
and 
bisects 
.
.
and 
are similar. When going from 
as this is the scale factor between 
.
and we are done.
such that 
is perpendicular to 
are similar. We know 
and 
)
and because of similarity of 2 triangles we get 
is perpendicular to 
and 
