[/quote]
,
Prove that
,
,
,
,
,
initially contains one coin. The following operations are allowed
,
,
;
,
,
and 
.
coins.
its circumradius. Let 
be the reflection of the point 
across the line 
,
be the reflection of the point 
,
be the reflection of the point 
.
.
be a real number and let 
be arbitrary real numbers in 
.
and 
are then defined recursively by
.
for all sufficiently large 
.
is a prime number
is inscribed in circle 
.
denote its incenter and 
its 
.
meets 
.
at 
and 
meets 
a second time at 
,
meets 
a second time at 
.
passes through the antipodal to 
intersects the lines 
,
at points 
,
,
of the segments 
,
,
,
.
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
,
.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
denote the set of single-variable polynomials in ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/3/2/032a637d4ae21da2c0ddd8dc127723807f11cc5f.png)
(i.e. functions taking polynomials to polynomials)![$p, q \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/7/e/7/7e7eb925258a33e456ef4ff722c1f943dc155d10.png)
,
;![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
,
has an integer root if and only if 
does.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
that verify the following equation for all 
:
is a function for which the expression 
for all 
is always a perfect square. Prove that 
for all 
.
such that 
.
Y by anantmudgal09, NJOY, lc426, Aritra12, sk2005, icematrix2, k12byda5h, jhu08, GrantStar, Mango247, Mango247, Mango247, Rounak_iitr, Siddharth03, GeoKing
Let 
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and 
, and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly.
Prove that the common chords of, and are concurrent on line .
Proposed by Anant Mudgal
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and 
, and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly.Prove that the common chords of
, and are concurrent on line .Proposed by Anant Mudgal
This post has been edited 1 time. Last edited by franzliszt, Oct 25, 2020, 12:25 AM
at 
be the intersection of the perpendicular to 
,
the power of point 
.
(simple computation.)
.
.
is the radical center of 
.
,
,
are collinear, so we are done.
when setting 
denote the intersection of the tangents and 
as an incenter configuration.
is 
,
passes 
,
(from the frame 
and from 
Now, using linearity, we can just find the concurrence point.
on this problem 
for this problem.
.
,
,
.
and 
with respect to 
,
is the radical axis of ![[asy] unitsize(2.5cm); pair A = dir(75); pair B = dir(200); pair C = dir(-20);
pair T_a = 2*B*C/(B+C); pair T_b = 2*C*A/(C+A); pair T_c = 2*A*B/(A+B); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = A+B+C; pair H_a = -B*C/A; pair H_b = -C*A/B; pair H_c = -A*B/C;
pair Q = 2*C*H_b/(H_b+C); pair P = 2*B*H_c/(H_c+B);
filldraw(T_a--T_b--T_c--cycle, invisible, magenta); draw(A--B--C--cycle, blue); draw(A--H_a, lightblue); draw(B--H_b, lightblue); draw(C--H_c, lightblue); filldraw(unitcircle, invisible, blue);
draw(H_b--Q, red); draw(H_c--P, red); draw(Q--P, dashed+red); pair O = origin; pair N = midpoint(O--H); draw(A--N, red);
filldraw(circumcircle(T_a, P, Q), invisible, deepgreen); pair R1 = IP(unitcircle, circumcircle(T_a, P, Q)); pair R2 = OP(unitcircle, circumcircle(T_a, P, Q)); draw(R1--R2, deepgreen+1);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T_a$", T_a, dir(T_a)); dot("$T_b$", T_b, dir(T_b)); dot("$T_c$", T_c, dir(T_c)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$H_a$", H_a, dir(H_a)); dot("$H_b$", H_b, dir(H_b)); dot("$H_c$", H_c, dir(H_c)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); dot("$O$", O, dir(270)); dot("$N$", N, dir(330)); dot(R1); dot(R2);
/* TSQ Source:
!unitsize(2.5cm); A = dir 75 B = dir 200 C = dir -20
T_a = 2*B*C/(B+C) T_b = 2*C*A/(C+A) T_c = 2*A*B/(A+B) D = foot A B C E = foot B C A F = foot C A B H = A+B+C H_a = -B*C/A H_b = -C*A/B H_c = -A*B/C
Q = 2*C*H_b/(H_b+C) P = 2*B*H_c/(H_c+B)
T_a--T_b--T_c--cycle 0.1 pink / magenta A--B--C--cycle blue A--H_a lightblue B--H_b lightblue C--H_c lightblue unitcircle 0.1 lightcyan / blue
H_b--Q red H_c--P red Q--P dashed red O = origin R270 N = midpoint O--H R330 A--N red
circumcircle T_a P Q 0.02 yellow / deepgreen R1 .= IP unitcircle circumcircle T_a P Q R2 .= OP unitcircle circumcircle T_a P Q R1--R2 deepgreen+1
*/ [/asy]](http://latex.artofproblemsolving.com/9/a/6/9a632c1c504e8c2ded30bafeecf1a2137444535d.png)
] Line 
.
and 
the antipodes of 
and 
,![[asy] unitsize(2.5cm); pair A = dir(75); pair B = dir(200); pair C = dir(-20);
pair T_a = 2*B*C/(B+C); pair T_b = 2*C*A/(C+A); pair T_c = 2*A*B/(A+B); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = A+B+C; pair H_a = -B*C/A; pair H_b = -C*A/B; pair H_c = -A*B/C; pair Bp = -B; pair Cp = -C;
pair Q = 2*C*H_b/(H_b+C); pair P = 2*B*H_c/(H_c+B); pair K = extension(B, H_c, C, H_b); pair L = extension(Cp, H_c, Bp, H_b);
filldraw(T_a--T_b--T_c--cycle, invisible, magenta); draw(A--B--C--cycle, blue); draw(A--H_a, lightblue); draw(B--H_b, lightblue); draw(C--H_c, lightblue); filldraw(unitcircle, invisible, blue); draw(B--Bp, cyan); draw(C--Cp, cyan);
draw(B--K--C, lightgreen); draw(Cp--L--Bp, deepgreen); draw(H_b--Q, red); draw(H_c--P, red); draw(Q--P, dashed+red); pair O = origin; pair N = midpoint(O--H); draw(A--N, red); draw(O--K, red+dashed);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T_a$", T_a, dir(T_a)); dot("$T_b$", T_b, dir(T_b)); dot("$T_c$", T_c, dir(T_c)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$H_a$", H_a, dir(H_a)); dot("$H_b$", H_b, dir(H_b)); dot("$H_c$", H_c, dir(H_c)); dot("$B'$", Bp, dir(Bp)); dot("$C'$", Cp, dir(Cp)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); dot("$K$", K, dir(K)); dot("$L$", L, dir(135)); dot("$O$", O, dir(270)); dot("$N$", N, dir(330));
/* TSQ Source:
!unitsize(3.5cm); A = dir 75 B = dir 200 C = dir -20
T_a = 2*B*C/(B+C) T_b = 2*C*A/(C+A) T_c = 2*A*B/(A+B) D = foot A B C E = foot B C A F = foot C A B H = A+B+C H_a = -B*C/A H_b = -C*A/B H_c = -A*B/C B' = -B C' = -C
Q = 2*C*H_b/(H_b+C) P = 2*B*H_c/(H_c+B) K = extension B H_c C H_b L = extension Cp H_c Bp H_b R135
T_a--T_b--T_c--cycle 0.1 pink / magenta A--B--C--cycle blue A--H_a lightblue B--H_b lightblue C--H_c lightblue unitcircle 0.1 lightcyan / blue B--Bp cyan C--Cp cyan
B--K--C lightgreen Cp--L--Bp deepgreen H_b--Q red H_c--P red Q--P dashed red O = origin R270 N = midpoint O--H R330 A--N red O--K red dashed
*/ [/asy]](http://latex.artofproblemsolving.com/f/a/e/fae1f5aa1c249cb53e962c77a3f9c5a58c6606da.png)
We observe that:
since 
is the pole of line 
(again by Brokard).
are collinear by Pascal's theorem on 
.
by a 
-
meet at 
.![[asy] unitsize(2.5cm); pair A = dir(75); pair B = dir(200); pair C = dir(-20);
pair T_a = 2*B*C/(B+C); pair T_b = 2*C*A/(C+A); pair T_c = 2*A*B/(A+B); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = A+B+C; pair H_a = -B*C/A; pair H_b = -C*A/B; pair H_c = -A*B/C;
pair Q = 2*C*H_b/(H_b+C); pair P = 2*B*H_c/(H_c+B);
filldraw(T_a--T_b--T_c--cycle, invisible, magenta); draw(A--B--C--cycle, blue); draw(A--H_a, lightblue); draw(B--H_b, lightblue); draw(C--H_c, lightblue); filldraw(unitcircle, invisible, blue);
draw(H_b--Q, deepgreen); draw(H_c--P, deepgreen); draw(Q--P, deepgreen); pair O = origin;
filldraw(circumcircle(T_a, P, Q), invisible, deepgreen); pair X = midpoint(B--C); pair Y = midpoint(B--H_c); pair Z = extension(E, F, B, P); pair Z2 = extension(E, F, C, Q);
draw(B--H_c, blue); draw(Z--Z2, red); draw(X--Z, orange); draw(P--O--T_a, orange); filldraw(circumcircle(B, X, Y), invisible, lightblue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T_a$", T_a, dir(T_a)); dot("$T_b$", T_b, dir(T_b)); dot("$T_c$", T_c, dir(T_c)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot(H_a); dot(H_b); dot("$H_c$", H_c, dir(H_c)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); dot("$O$", O, dir(20)); dot("$X$", X, dir(315)); dot("$Y$", Y, dir(200)); dot("$Z$", Z, dir(170)); dot(Z2);
/* TSQ Source:
!unitsize(2.5cm); A = dir 75 B = dir 200 C = dir -20
T_a = 2*B*C/(B+C) T_b = 2*C*A/(C+A) T_c = 2*A*B/(A+B) D = foot A B C E = foot B C A F = foot C A B H = A+B+C H_a .= -B*C/A H_b .= -C*A/B H_c = -A*B/C
Q = 2*C*H_b/(H_b+C) P = 2*B*H_c/(H_c+B)
T_a--T_b--T_c--cycle 0.1 pink / magenta A--B--C--cycle blue A--H_a lightblue B--H_b lightblue C--H_c lightblue unitcircle 0.1 lightcyan / blue
H_b--Q deepgreen H_c--P deepgreen Q--P deepgreen O = origin R20
circumcircle T_a P Q 0.02 yellow / deepgreen X = midpoint B--C R315 Y = midpoint B--H_c R200 Z = extension E F B P R170 Z2 .= extension E F C Q
B--H_c blue Z--Z2 red X--Z orange P--O--T_a orange circumcircle B X Y 0.1 lightcyan / lightblue
*/ [/asy]](http://latex.artofproblemsolving.com/2/8/d/28d66f2e6c98271ab65382dd69cb7c4fabb3d716.png)
Since 
,
.
and 
on the perpendicular bisector of 
:
and 
.
,
is cyclic. Then ![\[ ZP \cdot Z T_a = ZX \cdot ZY = ZB^2 \]](http://latex.artofproblemsolving.com/4/8/9/489a0b84e419999910afcfa3ce9c2e9b7a97c9b1.png)
with the last equality since the circumcircle of 
is tangent to 
coincides with the radical axis of 
at 
and 
;![[asy] unitsize(2.5cm); pair A = dir(75); pair B = dir(200); pair C = dir(-20);
pair T_a = 2*B*C/(B+C); pair T_b = 2*C*A/(C+A); pair T_c = 2*A*B/(A+B); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = A+B+C; pair H_a = -B*C/A; pair H_b = -C*A/B; pair H_c = -A*B/C;
filldraw(T_a--T_b--T_c--cycle, invisible, magenta); draw(A--B--C--cycle, blue); draw(A--H_a, lightblue); draw(B--H_b, lightblue); draw(C--H_c, lightblue); filldraw(unitcircle, invisible, blue);
pair U = 2*B*H_a/(B+H_a); pair V = 2*C*H_a/(C+H_a); pair U1 = extension(U, H, T_b, T_c); pair V1 = extension(V, H, T_b, T_c); draw(U--U1, dotted+deepgreen); draw(V--V1, dotted+deepgreen); filldraw(circumcircle(U, U1, T_c), invisible, deepgreen); filldraw(circumcircle(V, V1, T_b), invisible, deepgreen); draw(U--V, orange);
pair Z1 = IP(circumcircle(T_c, U, U1), circumcircle(T_b, V, V1)); pair Z2 = OP(circumcircle(T_c, U, U1), circumcircle(T_b, V, V1)); draw(Z1--T_a, red+1);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T_a$", T_a, dir(T_a)); dot("$T_b$", T_b, dir(T_b)); dot("$T_c$", T_c, dir(T_c)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(180)); dot("$H_a$", H_a, dir(H_a)); dot(H_b); dot(H_c); dot("$U$", U, dir(U)); dot("$V$", V, dir(V)); dot(Z1); dot(Z2);
/* TSQ Source:
!unitsize(2.5cm); A = dir 75 B = dir 200 C = dir -20
T_a = 2*B*C/(B+C) T_b = 2*C*A/(C+A) T_c = 2*A*B/(A+B) D = foot A B C E = foot B C A F = foot C A B H = A+B+C R180 H_a = -B*C/A H_b .= -C*A/B H_c .= -A*B/C
T_a--T_b--T_c--cycle 0.1 pink / magenta A--B--C--cycle blue A--H_a lightblue B--H_b lightblue C--H_c lightblue unitcircle 0.1 lightcyan / blue
U = 2*B*H_a/(B+H_a) V = 2*C*H_a/(C+H_a) U1 := extension U H T_b T_c V1 := extension V H T_b T_c U--U1 dotted deepgreen V--V1 dotted deepgreen circumcircle U U1 T_c 0.02 yellow / deepgreen circumcircle V V1 T_b 0.02 yellow / deepgreen U--V orange
Z1 .= IP circumcircle T_c U U1 circumcircle T_b V V1 Z2 .= OP circumcircle T_c U U1 circumcircle T_b V V1 Z1--T_a red+1
*/ [/asy]](http://latex.artofproblemsolving.com/5/3/b/53b1b476842c270d200626ba46818eb533db5458.png)
We need to show 
.
is apparently cyclic: the sides 
and 
are reflections across a line perpendicular to 
,
and 
are reflections across a line perpendicular to 
and 
are homothetic (their opposite sides are parallel), and their incenters are respectively 
meet 
,
with 
of 
of triangle 
,
,
lie on 
.
.
.
![[asy]
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label("$R$", (5.255869510243989,2.6078226626510923), E * labelscalefactor);
dot((-0.9913917832358846,6.238636671659605),linewidth(4pt) + dotstyle);
label("$S$", (-0.8407087072295467,6.532494890149667), NE * labelscalefactor);
dot((-5.978807234881507,3.7628225543592846),linewidth(4pt) + dotstyle);
label("$K$", (-5.832282122786005,4.055759989301052), W * labelscalefactor);
dot((-6.307449999448479,-0.591218968114576),linewidth(4pt) + dotstyle);
label("$L$", (-6.137111033659681,-0.28805199064882614), NE * labelscalefactor);
dot((1.6493553853279797,3.1409617572717043),linewidth(4pt) + dotstyle);
label("$P'$", (1.7884406490559157,3.4461021675537005), SE * labelscalefactor);
dot((-5.107959247719024,-0.21345199184111419),linewidth(4pt) + dotstyle);
label("$Q'$", (-5.108313459461022,-0.13563753521198832), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/8/1/a816584cd9e7177936026206bf6c54e0854de8eb.png)
determine triangle 
as shown. Let 
be the reflection of 
intersect 
at 
.
and 
so
and since 
we have 
which is purely imaginary because 
as shown in the above diagram, we have by the previous claim (and its symmetric variants) that
is cyclic (and symmetric variants)
we have that 
is the radical axis of 
and 
that we have 
and 
collinear. Let 
intersect 
,
intersect 
.
cyclic, which by Reim is equivalent to 
.
.
and 
so we want 
Since 
is cyclic we have 
so by angle chasing and repeatedly applying law of sines![\begin{align*}
\frac{ZP \cdot ZQ}{YP \cdot YQ} &= \frac{[ZPQ]}{[YPQ]} = \frac{ZP \sin(\angle ZPQ)}{YQ \sin(\angle YQP)} \\
&= \frac{(ZB+BP) \sin(2B)}{(YC+CQ)\sin(2C)} \\
&= \frac{\sin(2B)}{\sin(2C)} \cdot \frac{c \frac{\sin(C)}{\sin(2C)} + c \frac{\sin(90-B)}{\sin(C)} \frac{\sin(90-B)}{\sin(2B)}}{b \frac{\sin(B)}{\sin(180-2B)}+b \frac{\sin(90-C)}{\sin(B)} \frac{\sin(90-C)}{\sin(2C)}} \\
\end{align*}](http://latex.artofproblemsolving.com/b/0/d/b0d3cf6d6ce70a8ec963844c201f6db317fccda1.png)
Scaling so 
,
Recall that we wanted to show that this was equal to 
,
Thus it suffices to show that 
which rearranges to 
which is true upon expansion and noting that 
.
be the orthic triangle of 
;
.
,
.
are collinear and we're done.
with 
.
.
,
.
and the foot from 
,
.
.
,
.
,![[asy]
unitsize(2inch);
pair A, B, C, D, E, F, H, P, Q, R, M, N, S, T, U, V, W, X, O;
O = 0;
A = dir(110); B = dir(200); C = dir(340);
D = foot(A, B, C); E = foot(B, C, A); F = foot(C, A, B);
H = A+B+C; P = 2D-H; Q = 2E-H; R = 2F-H;
M = .5B+.5C; N = .5C+.5A;
S = .5B+.5R; T = .5C+.5Q; U = .5C+.5P; V = .5A+.5R;
W = extension(A, H, S, V); X = extension(B, H, S, V);
draw(A--B--C--cycle);
draw(A--P, lightred); draw(B--Q, lightred); draw(C--R, lightred);
draw(A--R--B--P--C--Q--cycle);
draw(circumcircle(A, B, C), dashed);
draw(circumcircle(M, D, S), dotted+blue); draw(circumcircle(N, E, U), dotted+blue);
draw(A--W--X--B, magenta); draw(M--S, magenta);
dot("$A$", A, dir(130)); dot("$B$", B, dir(155)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(105)); dot("$A'$", M, dir(M)); dot("$B'$", N, dir(B)); dot("$X_1$", S, dir(20)); dot("$Y_1$", T, dir(T)); dot("$X_2$", U, dir(U)); dot("$Y_2$", V, dir(305)); dot("$H$", H, dir(300)); dot("$O$", 0, dir(270)); dot("$A_2$", W, dir(W)); dot("$B_2$", X, dir(X)); dot("$H_A$", P, dir(P)); dot("$H_B$", Q, dir(Q)); dot("$H_C$", R, dir(R));
[/asy]](http://latex.artofproblemsolving.com/b/4/5/b45a8385ad91d79e6306b2ae3357dece099a20b0.png)
be the feet from 
to the opposite side. By more 
.
is the midpoint of 
so 
,
,
,
.
to be real. This expression is![\[\frac{b+c-(b-\frac{ab}{c})}{b+c-(c-\frac{ac}{b})} \div \frac{a+b+c-\frac{bc}{a}-(b-\frac{ab}{c})}{a+b+c-\frac{bc}{c}-(c-\frac{ac}{b})}.\]](http://latex.artofproblemsolving.com/6/f/d/6fdb47a9c3cf291c2dcb5f756fbc76780fa558fb.png)
The first fraction is 
.![$a+c-\frac{bc}{a}+\frac{ab}{c} = \frac{1}{ac}[ac(a+c)-bc^2+a^2b] = \frac{1}{ac}(a+c)(ac+ab-bc)$](http://latex.artofproblemsolving.com/c/b/3/cb3a7e21fb84ec239f70826044e0edc76cc70c32.png)
.
,
.
,
is cyclic, and so are 
and 
.
is the radical axis of 
and 
as follows.
.
and 
.
,
,
,
collinear. Similarly, 
are collinear. So from 
and 
,
.
have a common chord, as do 
and 
.
.
as the tangential triangle of 
be the reflections of 
be the poles of 
and 
w.r.t. 
be defined similarly.![[asy]
size(12cm,0);
defaultpen(fontsize(10pt));
pair inv(pair P){
return P/(abs(P)^2);
}
pair A = dir(110);
pair B = dir(215);
pair C = dir(325);
pair H = orthocenter(A,B,C);
pair T_a = inv((B+C)/2);
pair T_b = inv((A+C)/2);
pair T_c = inv((A+B)/2);
pair H_a = reflect(B,C)*H;
pair H_b = reflect(A,C)*H;
pair H_c = reflect(A,B)*H;
pair B_a = inv((C+H_b)/2);
pair C_a = inv((B+H_c)/2);
pair A_b = inv((H_a+C)/2);
pair A_c = inv((H_a+B)/2);
pair B_c = inv((A+H_b)/2);
pair C_b = inv((A+H_c)/2);
draw(circumcircle(T_a,B_a,C_a),green);
draw(circumcircle(T_b,A_b,C_b),green);
draw(circumcircle(T_c,A_c,B_c),green);
draw(A_b--A_c ^^ B_c--B_a ^^ C_a--C_b);
draw(circle((0,0),1));
draw(B_a--C_a^^C_b--A_b^^A_c--B_c,blue);
draw(A--H_a^^B--H_b^^C--H_c,red+dashed);
draw(A--B--C--cycle,red+linewidth(1));
draw(T_a--T_b--T_c--cycle,linewidth(0.8));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H_a$",H_a,dir(270));
dot("$H_b$",H_b,dir(H_b));
dot("$H_c$",H_c,dir(135));
dot("$T_a$",T_a,dir(270));
dot("$T_b$",T_b,dir(45));
dot("$T_c$",T_c,dir(150));
dot("$B_a$",B_a,dir(0));
dot("$C_a$",C_a,dir(195));
dot("$A_b$",A_b,dir(300));
dot("$C_b$",C_b,dir(135));
dot("$A_c$",A_c,dir(225));
dot("$B_c$",B_c,dir(80));
dot("$H$",H,dir(40));
[/asy]](http://latex.artofproblemsolving.com/2/3/0/23043ff6eee41d7e34e94767504b4c906e1fc928.png)
lie on a line that is perpendicular to ![[asy]
import olympiad;
import cse5;
size(10cm,0);
defaultpen(fontsize(10pt));
pair inv(pair P){
return P/(abs(P)^2);
}
pair A = dir(110);
pair B = dir(215);
pair C = dir(325);
pair H = orthocenter(A,B,C);
pair H_b = reflect(A,C)*H;
pair H_c = reflect(A,B)*H;
pair T = extension(B,C,H_b,H_c);
pair B_a = inv((C+H_b)/2);
pair C_a = inv((B+H_c)/2);
pair O_a = B+C;
pair N = H/2;
draw(circle((0,0),1));
draw(B--H_b^^C--H_c,red);
draw(A--B--C--cycle,red+linewidth(1));
draw(B--T--H_b);
draw(T--B_a,blue);
draw(H_c--C_a--B ^^ H_b--B_a--C, blue);
draw(arc(O_a,1,180,0),heavygreen);
draw(arc(A,abs(H-A),180,0,CCW),heavygreen);
draw(A--O_a,dashed);
dot("$A$",A,dir(A));
dot("$B$",B,2*dir(B));
dot("$C$",C,dir(C));
dot("$H_b$",H_b,dir(H_b));
dot("$H_c$",H_c,dir(150));
dot("$B_a$",B_a,dir(0));
dot("$C_a$",C_a,dir(135));
dot("$H$",H,dir(90));
dot("$T$",T,dir(180));
dot("$O_a$",O_a,dir(270));
dot("$N$",N,dir(15));
[/asy]](http://latex.artofproblemsolving.com/c/8/2/c822b719136f1bd8d67c3fd38dcebebeac2d4c91.png)
are colinear with 
and 
.
is the radical axis of 
and 
,
of 
,
,
.
painlessly, we observe the following.
.
;
.
Analogously, 
,
By noting the analogous calculations, we get
Hence the radical center is
which can be easily verified lie on the 
-
respectively. Let 
,
.
intersect 
respectively.
be the intersection between 
and the tangent to 
.
.
and 
,![\[\frac{U'-H}{N-C} = -\frac{2BC+(A+B+C)(A-C)}{\frac12(A+B-C)(A-C)} = \frac{2(A+C)}{C-A}\]](http://latex.artofproblemsolving.com/3/3/6/336b4dce27d3a1b7683b63bba9d29601d35561e4.png)
and it is easy to check that this is purely imaginary. 
intersect 
at 
;
intersect 
at 
;
at 
.
,
,
are all tangent to 
in that order, then 
if and only if the big quadrilateral is cyclic.
,
is cyclic. Therefore 
,
.
.
at 
.
by simple angle chasing, so 
satisfies 
.![\[-1 = A(A, U; B, H_a) = (F, U; B, L')\]](http://latex.artofproblemsolving.com/d/8/e/d8e6f1256296d6d0e15ca7a3fad878164abbfe55.png)
so 
,
and 
.
.
is the radical axis of 
,
.
.
concur on the Euler line. This is also easy in complex numbers, but here we present a synthetic proof: notice that 
,
and 
be the feet on 
,
,
.
,
,
are the radical axis of 
,
and 
respectively. If this is true, then they intersect at the homothety center sending 
,
)
be the intersection of 
.
be the line through 
,
are concyclic, denote it as 
be the reflexion of 
,
be the midpoint of 
and 
.
,
.
implying that 
.
are concyclic. Denote it as 
.
,
,
are concurrent, call the intersection point 
.
.
and 
.![\[ PO^2 = 2(PN^2 + NH^2) - PH^2 = 2(PH^2 +2r_N^2) -PH^2 = PH^2 +r_O^2 \]](http://latex.artofproblemsolving.com/8/f/a/8fae928c5a212612ab7d73bb76e4b4195f1ebab8.png)
where the last equality uses the fact 
.
,
lies on 
,
also lies on ![$[Y,Z; H,P] = \stackrel{F} =[FF,FZ; FM_C, FE ]= [F, FZ \cap \odot N; M_C,E] =-1$](http://latex.artofproblemsolving.com/0/5/c/05c2dcc849e4551c884879e42a9ea630acb9d048.png)
are harmonic. The homothety at center 
,
,
.
are concyclic.
,
are concyclic, denote it by 
,
,
.
,
are concyclic.
be the midpoint of 
,
are collinear and 
.
.
we have 
are collinear.
,
implying that 
are concyclic. As 
,
.
,
.
,
,
,
.
,
,
,
are concyclic. Hence 
to 
.
.
are concyclic.
must lie on the same circle, otherwise by radical theorem 
are collinear, requiring 
,
,
lies on the radical axis of 
.
such that 
,
are tangents of 
)
,
with touch points 
,
which completes the proof.
be the orthic triangle. Let 
be the reflections of the orthocenter 
,
are the triangles determined in the problem.
.
to be the pole of 
,
similarly as the pole of 
.
.
passes through 
.
and 
.
,
is the reflection of 
meet 
and 
at points 
respectively. I claim that 
is the radical axis of 
,
.
is cyclic with 
tangent to its circumcirle. Moreover, note that 
,
is cyclic. Now 
,
is the radical axis of 
is cyclic by some angle chasing, so 
,
of 
is chosen. The tangents from 
to 
meet at 
are the circumcentres of 
,
![[asy]
size(11cm);
defaultpen(fontsize(9pt));
import olympiad;
pair A = dir(50), B = dir(270), C = dir(175), O = (0,0);
pair HA = foot(B,A,C);
pair X = 2*foot(O,B,HA)-B;
pair H = orthocenter(A,B,C);
pair N = (O+H)/2;
pair W = extension(H, foot(H,C,N), O, (A+X)/2);
pair O2 = circumcenter(A,H,B);
pair K = 2*O2 - H;
pair L = foot(O,H,W);
draw(circumcircle(A,H,B), orange);
draw(circumcircle(A,W,X), purple);
draw(unitcircle, royalblue);
draw(A--B--X--A, royalblue);
draw(K--L, fuchsia);
draw(W--H, fuchsia);
draw(C--O2, dashed+fuchsia);
draw(X--W--A, springgreen);
dot("$A$", A);
dot("$B$", B);
dot("$X$", X);
dot("$C$", C);
dot("$O$", O);
dot("$H_A$", HA);
dot("$N$", N);
dot("$H$", H);
dot("$W$", W);
dot("$L$", L);
dot("$K$", K);
dot("$O_2$", O2);
[/asy]](http://latex.artofproblemsolving.com/3/d/3/3d305357d2d683f402b7c5540a73a6fcb10efed9.png)
and 
So, we can observe that 
are collinear.
our claim follows: 
intersect 
[
is parallel to 
]
t
at 
similarly. Set 
.
similarly.
is tangent to 
at 
.
.
.
.
lie on a circle.
.
Similarly, 
.
Similarly, 
.
of 
.
.
.
.
.
and 
.
, say 
is given by 
.
, then we want 
.
lies on the radical axis of 
is the radical axis of 
.
,
which is 
then 
is a linear function of 
for any real 
and points 
.
,
with respect to 
and from 
after which it's simple to see that for some point 
symmetric in ![[asy]
// Generated By AsyPadv1.2
import olympiad;
import markers;
import math;
import graph;
// change the unit size to fit your needs
unitsize(1cm);
// colored pens
pen c000000 = rgb("000000");
// dependency level 0
/* You can change the coordinates of these points of dependency level 0.
The drawing will retain the same relationships and qualities.
Please be aware that as a result of this some of the image may be clipped off. */
pair A = (3.33, 18.31); dot(A, c000000); label("$A$", A, dir(104.7967622450561));
pair B = (2.65, 15.11); dot(B, c000000); label("$B$", B, dir(234.4623222080241));
pair C = (6.02, 15.10); dot(C, c000000); label("$C$", C, dir(315.0));
// dependency level 1
// Do not change anything below, unless you are experienced in Asymptote.
path segA_B = A--B; draw(segA_B, c000000);
path segB_C = B--C; draw(segB_C, c000000);
path segC_A = C--A; draw(segC_A, c000000);
path ccA_B_C = circumcircle(A, B, C);
pair ccA_B_Ccenter = circumcenter(A, B, C); real ccA_B_Crad = circumradius(A, B, C); draw(ccA_B_C, c000000);
path pbB_C = ((B+C)/2-20.0*unit(((C-B).y, -(C-B).x)))--((B+C)/2+20.0*unit(((C-B).y, -(C-B).x))); draw(pbB_C, c000000);
path pbA_C = ((A+C)/2-20.0*unit(((C-A).y, -(C-A).x)))--((A+C)/2+20.0*unit(((C-A).y, -(C-A).x)));
// dependency level 2
path perA_segB_C = (A-20.0*(dir(segB_C).y, -dir(segB_C).x))--(A+20.0*(dir(segB_C).y, -dir(segB_C).x));
path perB_segC_A = (B-20.0*(dir(segC_A).y, -dir(segC_A).x))--(B+20.0*(dir(segC_A).y, -dir(segC_A).x));
pair O = intersectionpoint(pbB_C, pbA_C); dot(O, c000000); label("$O$", O, dir(315.0));
pair B_ccA_B_C_tangent1 = (B.x+(B.y-ccA_B_Ccenter.y)/sqrt((B.x-ccA_B_Ccenter.x)^2+(B.y-ccA_B_Ccenter.y)^2), B.y-(B.x-ccA_B_Ccenter.x)/sqrt((B.x-ccA_B_Ccenter.x)^2+(B.y-ccA_B_Ccenter.y)^2));
path tl1B_ccA_B_C = (B-20.0*unit(B_ccA_B_C_tangent1-B))--(B_ccA_B_C_tangent1+20.0*unit(B_ccA_B_C_tangent1-B)); draw(tl1B_ccA_B_C, c000000);
pair C_ccA_B_C_tangent1 = (C.x+(C.y-ccA_B_Ccenter.y)/sqrt((C.x-ccA_B_Ccenter.x)^2+(C.y-ccA_B_Ccenter.y)^2), C.y-(C.x-ccA_B_Ccenter.x)/sqrt((C.x-ccA_B_Ccenter.x)^2+(C.y-ccA_B_Ccenter.y)^2));
path tl1C_ccA_B_C = (C-20.0*unit(C_ccA_B_C_tangent1-C))--(C_ccA_B_C_tangent1+20.0*unit(C_ccA_B_C_tangent1-C)); draw(tl1C_ccA_B_C, c000000);
// dependency level 3
pair H = intersectionpoint(perA_segB_C, perB_segC_A); dot(H, c000000); label("$H$", H, dir(315.0));
// dependency level 4
pair N = (H + O)/2; dot(N, c000000); label("$N$", N, dir(315.0));
// dependency level 5
path lineA_N = (A-20.0*unit(N-A))--(N+20.0*unit(N-A));
// dependency level 6
pair Op = intersectionpoint(pbB_C, lineA_N); dot(Op, c000000); label("$O'$", Op, dir(315.0));
// dependency level 7
path segA_Op = A--Op; draw(segA_Op, c000000);
// dependency level 8
path perH_segA_Op = (H-20.0*(dir(segA_Op).y, -dir(segA_Op).x))--(H+20.0*(dir(segA_Op).y, -dir(segA_Op).x)); draw(perH_segA_Op, c000000);
// dependency level 9
pair X = intersectionpoint(tl1B_ccA_B_C, perH_segA_Op); dot(X, c000000); label("$X$", X, dir(257.9715975988014));
pair Y = intersectionpoint(tl1C_ccA_B_C, perH_segA_Op); dot(Y, c000000); label("$Y$", Y, dir(315.0));
// clip the drawing view
clip((0.0, 13.0)--(0.0, 20.0)--(10.0, 20.0)--(10.0, 13.0)--cycle);
[/asy]](http://latex.artofproblemsolving.com/d/1/b/d1bdfc0bb2d94e9ff0f09174fc741386a8172647.png)
.
by angles, and similarly, 
,
as desired.
,![[asy]
// Generated By AsyPadv1.2
import olympiad;
import markers;
import math;
import graph;
// change the unit size to fit your needs
unitsize(1cm);
// colored pens
pen c000000 = rgb("000000");
// dependency level 0
/* You can change the coordinates of these points of dependency level 0.
The drawing will retain the same relationships and qualities.
Please be aware that as a result of this some of the image may be clipped off. */
pair A = (3.87, 19.06);
pair B = (3.19, 15.86); dot(B, c000000); label("$B$", B, dir(234.4623222080241));
pair C = (6.56, 15.85); dot(C, c000000); label("$C$", C, dir(315.0));
// dependency level 1
// Do not change anything below, unless you are experienced in Asymptote.
path segA_B = A--B;
path segB_C = B--C; draw(segB_C, c000000);
path segC_A = C--A;
path ccA_B_C = circumcircle(A, B, C);
pair ccA_B_Ccenter = circumcenter(A, B, C); real ccA_B_Crad = circumradius(A, B, C);
path pbB_C = ((B+C)/2-20.0*unit(((C-B).y, -(C-B).x)))--((B+C)/2+20.0*unit(((C-B).y, -(C-B).x)));
path pbA_C = ((A+C)/2-20.0*unit(((C-A).y, -(C-A).x)))--((A+C)/2+20.0*unit(((C-A).y, -(C-A).x)));
// dependency level 2
path perA_segB_C = (A-20.0*(dir(segB_C).y, -dir(segB_C).x))--(A+20.0*(dir(segB_C).y, -dir(segB_C).x));
path perB_segC_A = (B-20.0*(dir(segC_A).y, -dir(segC_A).x))--(B+20.0*(dir(segC_A).y, -dir(segC_A).x));
pair O = intersectionpoint(pbB_C, pbA_C); dot(O, c000000); label("$O$", O, dir(315.0));
pair B_ccA_B_C_tangent1 = (B.x+(B.y-ccA_B_Ccenter.y)/sqrt((B.x-ccA_B_Ccenter.x)^2+(B.y-ccA_B_Ccenter.y)^2), B.y-(B.x-ccA_B_Ccenter.x)/sqrt((B.x-ccA_B_Ccenter.x)^2+(B.y-ccA_B_Ccenter.y)^2));
path tl1B_ccA_B_C = (B-20.0*unit(B_ccA_B_C_tangent1-B))--(B_ccA_B_C_tangent1+20.0*unit(B_ccA_B_C_tangent1-B));
pair C_ccA_B_C_tangent1 = (C.x+(C.y-ccA_B_Ccenter.y)/sqrt((C.x-ccA_B_Ccenter.x)^2+(C.y-ccA_B_Ccenter.y)^2), C.y-(C.x-ccA_B_Ccenter.x)/sqrt((C.x-ccA_B_Ccenter.x)^2+(C.y-ccA_B_Ccenter.y)^2));
path tl1C_ccA_B_C = (C-20.0*unit(C_ccA_B_C_tangent1-C))--(C_ccA_B_C_tangent1+20.0*unit(C_ccA_B_C_tangent1-C));
// dependency level 3
pair H = intersectionpoint(perA_segB_C, perB_segC_A);
pair T = intersectionpoint(pbB_C, tl1B_ccA_B_C); dot(T, c000000); label("$T$", T, dir(267.42663924422766));
// dependency level 4
pair N = (H + O)/2;
path ccT_C_B = circumcircle(T, C, B);
pair ccT_C_Bcenter = circumcenter(T, C, B); real ccT_C_Brad = circumradius(T, C, B); draw(ccT_C_B, c000000);
// dependency level 5
path lineA_N = (A-20.0*unit(N-A))--(N+20.0*unit(N-A));
// dependency level 6
pair Op = intersectionpoint(pbB_C, lineA_N);
// dependency level 7
path segA_Op = A--Op;
// dependency level 8
path perH_segA_Op = (H-20.0*(dir(segA_Op).y, -dir(segA_Op).x))--(H+20.0*(dir(segA_Op).y, -dir(segA_Op).x)); draw(perH_segA_Op, c000000);
// dependency level 9
pair X = intersectionpoint(tl1B_ccA_B_C, perH_segA_Op); dot(X, c000000); label("$X$", X, dir(156.923826164661));
pair Y = intersectionpoint(tl1C_ccA_B_C, perH_segA_Op); dot(Y, c000000); label("$Y$", Y, dir(331.75364744524285));
// dependency level 10
path ccT_X_Y = circumcircle(T, X, Y);
pair ccT_X_Ycenter = circumcenter(T, X, Y); real ccT_X_Yrad = circumradius(T, X, Y); draw(ccT_X_Y, c000000);
path segT_X = T--X; draw(segT_X, c000000);
path segT_Y = T--Y; draw(segT_Y, c000000);
// dependency level 11
pair P = intersectionpoints(pbB_C, ccT_X_Y)[0]; dot(P, c000000); label("$P$", P, dir(315.0));
// dependency level 12
path segT_P = T--P; draw(segT_P, c000000);
// clip the drawing view
clip((0.0, 13.0)--(0.0, 20.0)--(10.0, 20.0)--(10.0, 13.0)--cycle);
[/asy]](http://latex.artofproblemsolving.com/6/0/c/60cf79b00236dae2f5a3a7f550dd9724025bfa30.png)
meet 
again at 
and 
Using the similar triangles from earlier, this is 
This gives 
as desired.
symmetric in 
is symmetric in 
in both cancel out). Hence, for this 
.
,
so 
,
be the tangential triangle, 
the orthic triangle, and 
the medial triangle. Denote by 
,
,
the reflections of 
over 
,
,
,
,
,
the antipodes of 
,
,
on 
.
intersect 
at 
and 
at 
.![[asy] size(5cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue+white; pen sec=fuchsia; pen tri=red; pen qua=purple+pink; pen qui=heavycyan; pen fil=invisible;
pair O,A,B,C,D,EE,F,H,TA,TB,TC,NN,HB,HC,X,Y,Z,K,Z,Bp,Cp; O=origin; A=dir(120); B=dir(195); C=dir(345); D=foot(A,B,C); EE=foot(B,C,A); F=foot(C,A,B); H=A+B+C; TA=2/(1/B+1/C); TB=2/(1/C+1/A); TC=2/(1/A+1/B); NN=H/2; HB=-A*C/B; HC=-A*B/C; X=extension(TA,C,H,H+rotate(90)*(A-NN)); Y=extension(TA,B,H,H+rotate(90)*(A-NN)); K=2A-H; Z=extension(B,HC,C,HB); Bp=-B; Cp=-C;
draw(H--O,pri2+dashed); draw(A--NN,tri+dashed); draw(HC--Y--B,sec); draw(H--A,pri2); draw(B--HB,pri2); draw(C--HC,pri2); draw(Y--H,qui); filldraw(unitcircle,fil,pri); filldraw(A--B--C--cycle,fil,pri);
dot("\(A\)",A,NW); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(O\)",O,S); dot("\(H\)",H,S); dot("\(H_B\)",HB,NE); dot("\(H_C\)",HC,NW); dot("\(X\)",Y,W); dot("\(N\)",NN,S); [/asy]](http://latex.artofproblemsolving.com/5/e/5/5e504c95c2e9563e4db71adb8bcd5e541ba8d564.png)
in 
in 
be the pole of 
via 
.
,
,
,![\[X=\frac2{\frac1c-\frac b{ac}}=\frac{2ac}{a-b}.\]](http://latex.artofproblemsolving.com/4/3/1/43191f71b9241c34f813f417a452ebdc8210b4c2.png)
![\[\frac{x-h}{n-a}=\frac{\frac{2ac}{a-b}-(a+b+c)}{\frac{b+c-a}2}=\frac{\frac{(a+b)(c+b-a)}{a-b}}{\frac{b+c-a}2}=\frac{a+b}{2(a-b)}\in\mathbb I,\]](http://latex.artofproblemsolving.com/f/4/9/f4940e6c8f2779c6023c21d768149317eb0338b4.png)
as desired. 
and 
of ![[asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri1=lightblue; pen pri2=lightblue+white; pen sec=fuchsia; pen tri=lightred; pen qua=purple+pink; pen qui=heavycyan; pen sex=olive; pen fil=invisible; pen qfil=invisible; pen sefil=invisible;
pair O,A,B,C,D,EE,F,H,TA,TB,TC,NN,HB,HC,X,Y,Z,K,Z,Bp,Cp,L,Xs,Ys,M,LB,LC,V; O=origin; A=dir(110); B=dir(205); C=dir(335); D=foot(A,B,C); EE=foot(B,C,A); F=foot(C,A,B); H=A+B+C; TA=2/(1/B+1/C); TB=2/(1/C+1/A); TC=2/(1/A+1/B); NN=H/2; HB=-A*C/B; HC=-A*B/C; X=extension(TA,B,H,H+rotate(90)*(A-NN)); Y=extension(TA,C,H,H+rotate(90)*(A-NN)); K=2A-H; Z=extension(B,HC,C,HB); Bp=-B; Cp=-C; L=(A+K)/2; Xs=(B+HC)/2; Ys=(C+HB)/2; M=(B+C)/2; LB=(B+H)/2; LC=(C+H)/2; V=A/2;
draw(B--HC,sec); draw(C--HB,sec); filldraw(circumcircle(D,LB,LC),qfil,qua+dashed); filldraw(circumcircle(D,M,L),qfil,qua); draw(Ys--M--Xs,tri); draw(A--D,pri2); draw(B--HB,pri2); draw(C--HC,pri2); draw(A--O--H,pri1); filldraw(unitcircle,fil,pri); filldraw(A--B--C--cycle,fil,pri); filldraw(D--LB--Xs--cycle,sefil,sex); filldraw(D--LC--Ys--cycle,sefil,sex); filldraw(D--NN--V--cycle,sefil,sex);
dot("\(A\)",A,NW); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(D\)",D,S); dot("\(M_A\)",M,S); dot("\(H_B\)",HB,dir(75)); dot("\(H_C\)",HC,dir(120)); dot("\(X^*\)",Xs,W); dot("\(Y^*\)",Ys,dir(15)); dot("\(H\)",H,dir(120)); dot("\(L_B\)",LB,N); dot("\(L_C\)",LC,dir(120)); dot("\(V\)",V,NE); dot("\(N\)",NN,dir(300)); dot("\(O\)",O,S); [/asy]](http://latex.artofproblemsolving.com/e/7/c/e7c43cff0c4dad76609458cc2de0c831c142994c.png)
is the circle 
containing 
,
and centered at the midpoint 
of 
.
,
be the midpoints of 
,
,
,
,
and ![\[\frac{NK}{DN}=\frac{AH}{AO}=2\cos A=\frac{HH_C}{BH}=\frac{L_BM_B}{DL_B},\]](http://latex.artofproblemsolving.com/0/8/c/08c587226a035e9ae0f2fac788dce7996c42a9b1.png)
so 
.
.
.
is the radical center of 
,
,
,
is 
.
and 
by Claim 1. Since 
is tangential and 
,![[asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri1=lightblue; pen pri2=lightblue+white; pen sec=fuchsia; pen tri=lightred; pen qua=purple+pink; pen qui=heavycyan; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible;
pair O,A,B,C,D,EE,F,H,TA,TB,TC,NN,HA,HB,HC,X,Y,Z,K,Z,Bp,Cp,L,Xs,Ys,M,P,UB,UC,VB,VC; O=origin; A=dir(70+200); B=dir(210+200); C=dir(330+200); D=foot(A,B,C); EE=foot(B,C,A); F=foot(C,A,B); H=A+B+C; TA=2/(1/B+1/C); TB=2/(1/C+1/A); TC=2/(1/A+1/B); NN=H/2; HA=-B*C/A; HB=-A*C/B; HC=-A*B/C; X=extension(TA,B,H,H+rotate(90)*(A-NN)); Y=extension(TA,C,H,H+rotate(90)*(A-NN)); K=2A-H; Z=extension(B,HC,C,HB); Bp=-B; Cp=-C; L=(A+K)/2; Xs=(B+HC)/2; Ys=(C+HB)/2; M=(B+C)/2; P=extension(EE,F,B,TA); UB=2/(1/C+1/HA); UC=2/(1/B+1/HA); VB=2/(1/A+1/HC); VC=2/(1/A+1/HB);
filldraw(circumcircle(TB,UB,VB),qfil,qua); filldraw(circumcircle(TC,UC,VC),qfil,qua); draw(TA--(intersectionpoints(circumcircle(TB,UB,VB),circumcircle(TC,UC,VC))[1]+0.2*(D-TA)),tri+dashed); draw(UB--UC,sec); draw(B--HA--C,pri1); draw(A--HA,pri1); filldraw(TA--TB--TC--cycle,sfil,sec); filldraw(unitcircle,fil,pri); filldraw(A--B--C--cycle,fil,pri);
dot("\(A\)",A,A); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(T_A\)",TA,N); dot("\(T_B\)",TB,SW); dot("\(T_C\)",TC,SE); dot("\(H_A\)",HA,dir(120)); dot("\(D\)",D,dir(-20)); dot("\(U_B\)",UB,NW); dot("\(U_C\)",UC,1.5N); [/asy]](http://latex.artofproblemsolving.com/5/8/3/583d008dfd8c490280b13d0968332ac540473d40.png)
is the radical center of 
,
,
,
,
,
.
and 
are homothetic, so the four lines 
,
,
concur at the center of homothety.
be the circumcenters of 
and 
respectively, and let 
and 
be the corresponding excenters of 
,
,
and 
,
,
and 
,
bisects 
.![\[\frac{CD}{DB}=\frac{CZ}{ZB}=\frac{T_bZ}{ZT_c}=\frac{T_bA}{AT_c}\]](http://latex.artofproblemsolving.com/1/8/d/18d85eb8ba12a1a31ed40ee9af70b4557faf98e6.png)
which shows that 
are collinear. Now, as 
,
.
is parallel to 
denote the lengths of segments 
and 
.![\[T_aP=\frac{bc}{a+b}\qquad\text{and}\qquad T_aQ=\frac{bc}{a+c}.\]](http://latex.artofproblemsolving.com/0/8/2/08237f7df8097cbcedf6139f17d3007520ae347d.png)
Therefore, their ratio is 
.
,![\[CX=\frac{(s-a)(s-b)}{s}\qquad\text{and}\qquad\frac{(s-a)(s-c)}{s}.\]](http://latex.artofproblemsolving.com/7/d/c/7dca9eeb70b625eefa55f512d6195132afd00f91.png)
Now computing 
and 
shows that![\[T_aX=(s-a)\left(\frac{s-b}{s}+1\right)=\frac{(s-a)(a+c)}{s}\qquad\text{and}\qquad T_aY=\frac{(s-a)(a+b)}{s}\]](http://latex.artofproblemsolving.com/7/9/4/7943558e9ffc0525f6eca155fcc3db0b91d35127.png)
which means that 
so 
as desired. In particular, since 
,![\[\frac{RO_A}{RO_T}=\frac{AO}{AI_A}=\frac{RO}{RS}\]](http://latex.artofproblemsolving.com/e/0/b/e0b40c8ed9b692fb01bacce4719a81061796ee4f.png)
where 
is the midpoint of arc 
.
is perpendicular to 
.
,![\[KA_1\cdot KA_2=KB\cdot KC=KD\cdot KM\]](http://latex.artofproblemsolving.com/6/8/a/68a18098d860981542ef088e74d597121326bd8d.png)
where 
are concyclic. Notice that the perpendicular bisector of 
coincides with 
.
to be 
be the reflection of 
be the reflection of 
was symmetric in 
and 
,
are cyclic.![\[\frac{T_aX'}{T_aY'}=\frac{b}{c}.\]](http://latex.artofproblemsolving.com/a/4/6/a4695da3e8385155a4d1237d541fd41356618320.png)
From the quantities in claim 1, we have![\[T_aX'=T_aC-CX'=(s-a)-\frac{(s-a)(s-b)}{s}=(s-a)\left(1-\frac{s-b}{s}\right)=\frac{b(s-a)}{s}.\]](http://latex.artofproblemsolving.com/8/1/1/8118230735f45855764241f6daaac6c5dccfcd1b.png)
Similarly, we have 
so we're done. 
,
and 
and 
,
and 
,
to 
at 
,
be the reflection of 
be the hexagon, such that 
at 
,![[asy]import geometry;import olympiad;
size(12cm); defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;
pair O,A,B,C,H,h,N,D,E,F,G,K,g,j,I,J,P,Q,R,Y,X,T,U,V;
O=(0,0);A=dir(120);B=dir(215);C=dir(325);H=orthocenter(A,B,C);h=2foot(A,B,C)-H;g=2foot(B,A,C)-H;j=2foot(C,A,B)-H;N=midpoint(O--H);D=intersectionpoint(perpendicular(B,line(B,O)),perpendicular(h,line(h,O)));
E=intersectionpoint(perpendicular(C,line(C,O)),perpendicular(h,line(h,O)));
G=intersectionpoint(perpendicular(C,line(C,O)),perpendicular(g,line(g,O)));
K=intersectionpoint(perpendicular(A,line(A,O)),perpendicular(g,line(g,O)));
I=intersectionpoint(perpendicular(A,line(A,O)),perpendicular(j,line(j,O)));
J=intersectionpoint(perpendicular(B,line(B,O)),perpendicular(j,line(j,O)));
P=extension(D,B,C,E);
Q=extension(C,G,A,K);
R=extension(A,I,J,B);
Y=extension(B,C,R,E);
X=extension(P,Y,O,H);
T=extension(B,C,A,h);
V=extension(X,R,A,B);
U=extension(A,C,X,Q);
draw(A--B--C--cycle,deep);draw(circumcircle(A,B,C),deep);draw(P--Q--R--cycle,deep);draw(J--I--K--G--E--D--cycle,org);draw(R--E,light);draw(D--Q,light);draw(A--h,light);draw(R--X--P,dark);draw(X--Q,dark);
draw(T--U--V--cycle,dark);draw(X--O,dark+dashed);
dot("$O$",O,dir(0));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H_A$",h,dir(h));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$H_B$",g,dir(g));
dot("$H_C$",j,dir(j));
dot("$F$",G,dir(G));
dot("$H$",H,dir(0));
dot("$I$",I,dir(I));
dot("$J$",J,dir(J));dot("$G$",K,dir(K));dot("$R$",R,dir(R));dot("$Q$",Q,dir(Q));dot("$P$",P,dir(P));dot("$X$",X,dir(X));dot("$T$",T,dir(T));dot("$U$",U,dir(U));dot("$V$",V,dir(V));
[/asy]](http://latex.artofproblemsolving.com/3/6/6/366281e1ae0d839de42493a7f38911fa95037349.png)
be the orthocentre, circumcentre and centre of nine-point circle of triangle 
to 
.![[asy]import geometry;import olympiad;
size(5cm); defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;
pair O,A,B,C,H,h,N,D,P,L;
O=(0,0);A=dir(105);B=dir(195);C=dir(345);H=orthocenter(A,B,C);h=2foot(A,B,C)-H;N=midpoint(O--H);D=intersectionpoint(perpendicular(B,line(B,O)),perpendicular(h,line(h,O)));P=extension(C,N,H,D);L=2P-H;
draw(A--B--C--cycle,deep);draw(circumcircle(A,B,C),deep);draw(B--D--h,magenta);draw(H--D,deep);draw(C--P,light);draw(C--H,light);draw(C--h,light);draw(circumcircle(O,B,D),magenta+dashed);draw(O--H,deep);draw(circumcircle(H,L,h),light);
draw(L--O,magenta);draw(C--L,light);draw(O--h,magenta);draw(O--B,magenta);
clip((1.05,1.05)--(-1.05,1.05)--(-1.05,-1.05)--(1.05,-1.05)--cycle);
dot("$O$",O,dir(90));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H$",H,dir(H));
dot("$H_A$",h,dir(h));
dot("$N$",N,dir(N));
dot("$D$",D,dir(D));
dot("$P$",P,dir(P));
dot("$L$",L,dir(L));
[/asy]](http://latex.artofproblemsolving.com/5/0/e/50e4a4b361c3db2d3235d0cd91a7ba0927ad235a.png)
.
hence 
.
.
.
,
are collinear, we conclude that 
.
hence 
is cyclic, similarly 
and 
is cyclic. Now, by radical axis, 
.
and 
are concurrent, call that concurrency point 
and 
are cyclic and we easily get that 
is cyclic. Now by radical axis, 
is cyclic. Similarly 
is cyclic.
and nine-point circle, we get that 
.
be the reflection of 
be the reflection of 
are collinear. Thus, by angle chase 
.
.
, we use the linearity of PoP.
.
and 
and 
.
lies on 
and ![[asy]
size(12cm); defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;
pair O,A,B,C,H,Ha,Hb,Hc,T,D,E,F,G,I,J,U,V,K,k,P,Q,R,L,W,Z,M,m,t,i;
O=(0,0);A=dir(105);B=dir(195);C=dir(345);path w=circumcircle(A,B,C);H=orthocenter(A,B,C);Ha=2*foot(A,B,C)-H;Hb=2*foot(B,A,C)-H;Hc=2*foot(C,A,B)-H;T=foot(A,B,C);D=midpoint(B--Ha);E=midpoint(C--Ha);F=midpoint(C--Hb);G=midpoint(Hb--A);I=midpoint(Hc--A);J=midpoint(Hc--B);U=foot(B,A,C);V=foot(C,B,A);K=midpoint(C--H);
path w1=circumcircle(I,U,E);path w2=circumcircle(V,G,D);k=2*C-K;P=midpoint(B--C);Q=midpoint(A--C);R=midpoint(A--B);L=midpoint(A--H);W=2foot(V,B,C)-V;Z=2foot(T,B,A)-T;M=midpoint(B--H);m=2B-M;t=intersectionpoints(circumcircle(B,O,C),circumcircle(A,O,Ha))[1];i=intersectionpoints(circumcircle(U,Q,I),circumcircle(G,V,R))[0];
draw(A--B--C--cycle,deep);draw(w,deep);draw(B--Ha--C--Hb--A--Hc--cycle,org);draw(A--T,org);draw(B--V,org);draw(C--U,org);draw(w1,med);draw(w2,med);draw(C--V,med);draw(B--U,med);draw(circumcircle(T,U,V),med+dashed);draw(A--Z,org);draw(C--W,org);draw(circumcircle(B,O,C),org);
draw(circumcircle(i,P,t),deep+dashed);draw(circumcircle(A,Ha,t),org);
clip((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle);
dot("$O$",O,dir(90));dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$H$",H,dir(H));dot("$H_A$",Ha,dir(Ha));dot("$H_B$",Hb,dir(Hb));dot("$H_C$",Hc,dir(Hc));dot("$T$",T,dir(T));dot("$D'$",D,dir(D));dot("$E'$",E,dir(E));dot("$F'$",F,dir(F));dot("$G'$",G,dir(G));dot("$J'$",J,dir(J));dot("$I'$",I,dir(I));dot("$U$",U,dir(U));dot("$V$",V,dir(V));dot("$K$",K,dir(K));dot("$K'$",k,dir(k));dot("$P'$",P,dir(P));dot("$Q'$",Q,dir(Q));dot("$R'$",R,dir(R));dot("$L$",L,dir(L));dot("$W$",W,dir(W));dot("$Z$",Z,dir(Z));dot("$M$",M,dir(M));dot("$M'$",m,dir(m));dot("$T'$",t,dir(45));
[/asy]](http://latex.artofproblemsolving.com/6/2/c/62cc6d0cd4f48a503c9ef821a9b02b7e3ab705a3.png)
as 
.
and 
,
and 
and 
.
are concurrent.
and 
so that 
is cyclic and 
.
analogously so that 
is a convex hexagon inscribed in 
and similar.
at 
,
,
concur at the foot 
.
on the complex unit circle. Then we have 
,
,
.
.
.
.
,
.
to 
and 
are cyclic with diameters 
,
.
.
,
(and analogous results for the other five vertices of the hexagon) it follows that 
etc. are indeed the triangles defined in the problem statement (with circumcircles 
lies on the radical axis of 
.
and 
are the 
and 
exradii of 
.
a
a
and 
the reflection of ![\[BH'=BH=BH_a=BH_c\]](http://latex.artofproblemsolving.com/9/f/8/9f836cd641d7279987d1eff4e2a9bf1870428cee.png)
with diameter 
.
which implies that points 
it follows that points 
.
.
,
must be harmonic. From this it follows that the tangents to 
denote the line through 
and 
similarly. Let 
and 
denote the intersections of 
, 
and 
, 
similarly.
(and similar) is cyclic.![\[\measuredangle Z_bY_cT_c = \measuredangle H_aY_cC = 2 \measuredangle H_aCY_c = 2\measuredangle H_aAC = 2\measuredangle BCA = 2\measuredangle T_cBA = \measuredangle BT_cA = \measuredangle Z_bT_cT_b\]](http://latex.artofproblemsolving.com/a/5/2/a5292446dd06e56fa3fa1180ea2ed32866c37cd6.png)
which implies the claim.![\[\text{Pow}_{\omega_C}(T_a)=T_aZ_b \cdot T_aT_c = T_aY_c \cdot T_aT_b = \text{Pow}_{\omega_B}\]](http://latex.artofproblemsolving.com/3/f/9/3f9351710514b924d294e6c28be97467ec9d6ed5.png)
Thus, 
and 
.
lines 
, 
, 
and 
concur. The former two clearly intersect at 
, we have that 
and thus 
.![\[\measuredangle Z_aEK = \measuredangle Z_aT_aX_c = \measuredangle Z_aT_cE\]](http://latex.artofproblemsolving.com/a/d/7/ad7086595c37635646420abf7bc5a6ccd22818cf.png)
so 
is tangent to 
.![\[KT_c \cdot KZ_a = KE^2 = KA^2\]](http://latex.artofproblemsolving.com/4/9/b/49bfa7ed8f125e6b72cbe6a2d1d343247bb60137.png)
which implies the claim.
is the common chord of circles 
and 
are the common chords of 
and 
concur are the point 
- the homothetic center of the orthic and tangential triangles. Further since the incenter of 
Let 
be the tangential triangle of 
.
,
,
,
,
similarly.
,
w.r.t. ![\[h_a=b+c-bc\overline{h}=b+c-\frac{ab+bc+ca}a=-\frac{bc}a\]](http://latex.artofproblemsolving.com/7/5/5/755906ce3bc44d317280a8e2a112cb6d43d3a408.png)
So let 
be pole of 
.![\[c_b'=\frac{2 \left(-\frac{bc}a \right)b}{b-\frac{bc}a}=\frac{2bc}{c-a}\]](http://latex.artofproblemsolving.com/3/9/1/391c61c4c5b6e8e1b2ed47c83b32070a55c5af71.png)
So we just need to prove 
where 
.![\[\frac{a+b+c-\frac{2bc}{c-a}}{c-\frac{a+b+c}2}=\frac{2(a+b+c)(c-a)-4bc}{(c-a)(c-a-b)}=\frac{2(c+a)(c-a-b)}{(c-a)(c-a-b)} =-\frac{2(\overline{c}+\overline{a})}{\overline{c}-\overline{a}}\]](http://latex.artofproblemsolving.com/b/d/4/bd4ec2131345483f6de44a8349fed65ca83ca0f2.png)
Hence the quantity is imaginary and we are done.. 
(and analogous).
and 
.
is cyclic (and analogous).
And done. 
lies on radical axis of 
(and analogous).
is cyclic and 
.![\[\angle EFX_A=180^{\circ}-\angle EFB_A=\angle X_BB_AX_A=\angle FA_BX_A\]](http://latex.artofproblemsolving.com/0/4/d/04d20d7c0b5abcb47a43da59cec37b91722d4ea4.png)
And so done. 
.
,
,
,
,
similarly.
(and analogous)
so homothety at 
is the radical axis of 
has same power to both circles (and do the same for ![\[\text{Pow}_{\omega_A}\left(K_{A_C} \right)=K_{A_C}A_B \cdot K_{A_C}X_A=K_{A_C}F^2=K_{A_C}B^2=\text{Pow}_{\Gamma}\left(K_{A_C} \right)\]](http://latex.artofproblemsolving.com/0/8/5/08536e6000be06c6a4a704e7e435a810b0b863d5.png)
And done. 
is the radical axis of 
.
