[/quote]
,
is irreducible for every natural number 
.
be positive real numbers satisfying![\[
a^2 + c^2 = b(a + c).
\]](http://latex.artofproblemsolving.com/9/c/0/9c0730d15b1177ee23d3f65bc2fe8b2cc85d0c62.png)
Let![\[
m = \min \left( \frac{a^2 + ab + b^2}{ab + bc + ca} \right).
\]](http://latex.artofproblemsolving.com/e/b/6/eb6f340defa4b38e35292062b202d138b9de843b.png)
Find the value of 
.
be a function such that
for all 
.
.
be a monic polynomial with complex coefficients such that there exist a polynomial 
with complex coefficients for which ![\[P(x^2-2)=P(x)Q(x).\]](http://latex.artofproblemsolving.com/a/6/2/a62b06890b6bbcc5c9634f6de8c7ae5fe9f37828.png)
Determine all complex numbers that could be the root of 
be real numbers such that 
Prove that 
Where 
.
such that for all 
be points on the same circumference with 
.
and 
,
cuts the segment 
into equal parts.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
such that for any 
:
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
that satisfy 
and![\[
f(f(n)) = f(f(n+2)+2) = n
\]](http://latex.artofproblemsolving.com/4/d/8/4d8d7420e51a7529996473badd8a4e9976c2f01a.png)
for all integers 
.
such that, for all 
,
,
,![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
.
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
with exactly 9 elements that is subset of 
.
is an empty subset.
Y by laegolas, MathAwesome123, 62861, claserken, efang, Generic_Username, lucasxia01, rkm0959, anantmudgal09, soojoong, CQYIMO42, mhq, parmenides51, Adventure10, Mango247, Bet667
To avoid clogging the fora with a horde of geometry problems, I'll post them all here.
Day I
Day II
Enjoy the problems!
Day I
Problem 1: Let 
and 
be fixed points on a line 
, and let 
be some point lying on 
such that 
lie on in that order. Let 
be some circle passing through and , and let 
be its center. Let 
be some point lying on the circumcircle of 
. Show that the circumcircle of 
is tangent to if and only if the circumcircle of 
is tangent to .
Problem 2: Let
be an acute triangle, and let 
and 
be the feet of the altitudes from 
and 
to 
and 
, respectively. Let 
meet the circumcircle of at points 
and 
. Let be a circle passing through and such that its center lies on arc 
, and let 
be the midpoint of 
. Let 
and 
exist on such that 
and 
are tangent to .
Let
, 
, 
, and 
. Show that meets 
at the orthocenter of .
Problem 3: Let be an acute triangle, let 
be its orthocenter, and let and be the feet of the altitudes from and to the sides of the triangle, respectively. Let be the midpoint of , and let and be the feet of the perpendiculars from to 
, respectively. Let meet at a point , and let 
be a point lying on 
such that 
.
Prove that
.
and 
be fixed points on a line 
, and let 
be some point lying on 
such that 
lie on in that order. Let 
be some circle passing through and , and let 
be its center. Let 
be some point lying on the circumcircle of 
. Show that the circumcircle of 
is tangent to if and only if the circumcircle of 
is tangent to .Problem 2: Let
be an acute triangle, and let 
and 
be the feet of the altitudes from 
and 
to 
and 
, respectively. Let 
meet the circumcircle of at points 
and 
. Let be a circle passing through and such that its center lies on arc 
, and let 
be the midpoint of 
. Let 
and 
exist on such that 
and 
are tangent to .Let
, 
, 
, and 
. Show that meets 
at the orthocenter of .Problem 3: Let
be an acute triangle, let 
be its orthocenter, and let and be the feet of the altitudes from and to the sides of the triangle, respectively. Let be the midpoint of , and let and be the feet of the perpendiculars from to 
, respectively. Let meet at a point , and let 
be a point lying on 
such that 
.Prove that
.Day II
Problem 4: Let be an acute triangle, let 
be its incircle, and let 
be the midpoint of minor arc 
on the circumcircle of . Let touch 
at points 
, respectively, and let be the foot of the altitude from 
to . Denote by 
the intersection of 
and 
. Let 
and 
denote the and -excenters of triangle 
, respectively.
Prove that and lie on .
Problem 5: Let be an acute triangle, and let be the feet of the altitudes from 
, respectively. Let meet the circumcircle of at points and . Let be the intersection of 
and 
, and let 
be the intersection of 
and 
.
Prove that the circumcircles of triangles
and 
are tangent to each other.
Problem 6: Let be a point lying on segment , different from and . Let 
and 
be circles tangent to at , and suppose that and lie on the same side of . The tangent from to other than 
meets at 
. Define 
similarly.
Let be the line passing through the circumcenters of 
and 
. Prove that passes through the intersection of 
and 
.
be an acute triangle, let 
be its incircle, and let 
be the midpoint of minor arc 
on the circumcircle of . Let touch 
at points 
, respectively, and let be the foot of the altitude from 
to . Denote by 
the intersection of 
and 
. Let 
and 
denote the and -excenters of triangle 
, respectively.Prove that
and lie on .Problem 5: Let
be an acute triangle, and let be the feet of the altitudes from 
, respectively. Let meet the circumcircle of at points and . Let be the intersection of 
and 
, and let 
be the intersection of 
and 
.Prove that the circumcircles of triangles
and 
are tangent to each other.Problem 6: Let
be a point lying on segment , different from and . Let 
and 
be circles tangent to at , and suppose that and lie on the same side of . The tangent from to other than 
meets at 
. Define 
similarly.Let
be the line passing through the circumcenters of 
and 
. Prove that passes through the intersection of 
and 
.Enjoy the problems!
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
path carc(pair A, pair B, pair C, real d=0, bool dir=CW) {
pair O=circumcenter(A,B,C);
return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir);
}
pair O=(0,1), M=-O, X=dir(200), Y=dir(-20), Q=dir(-50), P=extension(M,Q,X,Y);
D(unitcircle,heavygreen);
D(CP(O,X,degrees(X),degrees(Y)),red);
D(carc(P,Y,Q,-10,CCW),purple);
D(carc(P,Q,X,10),purple);
D(X--P--M--X--Q--Y--M);
D("O",O,N);
D("P",P);
D("Q",Q,SE);
D("X",X,SW);
D("Y",Y,SE);
D("M",M);
[/asy]](http://latex.artofproblemsolving.com/5/3/5/535ac404b850cda083d682e5707a5fc0205ec5f1.png)
.
.
.
is tangent to 
iff 
iff 
is tangent to 
iff 
iff ![[asy]
unitsize(3cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair A=dir(115), B=dir(205), C=dir(-25), D=foot(A,B,C), E=foot(B,C,A), F=foot(C,A,B), H=A+B+C, M=(B+C)/2, S_1=IP(L(E,F,10,10),D(unitcircle,heavygreen)), S_2=OP(L(E,F,10,10),unitcircle), T_1=OP(D(arc(A,S_1,S_2,CW),red),D(arc(M,B,C,CW),orange)), T_2=OP(CP(A,S_1),CP(M,B)), P=extension(E,F,B,C);
D(A--B--C--cycle);
DPA(B--P--S_1^^P--T_2--M--T_1^^A--D);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("D",D);
D("E",E,NNE);
D("F",F,NW);
D("H",H,SE);
D("M",M);
D("P",P,W);
D("S_1",S_1,NE);
D("S_2",S_2,NW);
D("T_1",T_1,SW);
D("T_2",T_2,dir(0));
[/asy]](http://latex.artofproblemsolving.com/2/0/b/20bcede3a86e3ead3dc2b7c1a58f0953a97ae60e.png)
.
,
,
.
and 
,
,
lie on 
,
is the polar of 
.
.
,
.
,
.
,![[asy]
unitsize(3cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair A=dir(115), B=dir(205), C=dir(-25), E=foot(B,C,A), F=foot(C,A,B), H=A+B+C, M=(B+C)/2, X=(B+F)/2, Y=(C+E)/2, Z=(X+Y)/2, P=extension(E,F,X,Y);
D(arc(M,C,B),heavygreen);
D(A--B--C--cycle);
DPA(Y--P--E^^H--P^^A--Z--M);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("E",E,NE);
D("F",F,NW);
D("H",H,SE);
D("M",M);
D("P",P,SW);
D("X",X,SE);
D("Y",Y,NE);
D("Z",Z,SW);
[/asy]](http://latex.artofproblemsolving.com/0/2/8/028d7229adbde1f129e611e0124e8c351324767e.png)
,
;
.
.
,
be on the unit circle. Since 
and 
,
and 
Also, 
and so 
.
.
:
goes to 
,
,
goes to 
and 
goes to 
.
,![\begin{align*}p-e&=\frac{(f-e)\left(\overline{(x-e)}(y-e)-(x-e)\overline{(y-e)}\right)}{\left(\overline{(x-e)}-\overline{(y-e)}\right)(f-e) - (x-e-(y-e))\left(\overline{f}-\overline{e}\right)}\\&=\frac{(f-e)\left[\left(\overline{f}-2\overline{e}-1\right)(1-e)-(f-2e-1)(1-\overline{e})\right]}{2\left[\left(\overline{f}-\overline{e}-2\right)(f-e)-(f-e-2)\left(\overline{f}-\overline{e}\right)\right]}\\&=\frac{-(f-e)(e-1)\left(f^2-3ef-3f+e\right)}{-4(f-e)(ef+1)},\end{align*}](http://latex.artofproblemsolving.com/2/b/d/2bd6c1c4060dc0d154646da6967c4d1e2c2c7071.png)
so 
Now to prove the claim, we need to show that 
.
and ^2-4[(e-1)(f+1)+ef+1](ef+1)}{4(e+f)(ef+1)}\\&=\frac{(e-1)(f+1)\left[(e+f)^2-4(ef+1)\right]+4\left[(e+f)^2-(ef+1)^2\right]}{4(e+f)(ef+1)}\\&=\frac{(e-1)(f+1)\left[(e+f)^2-4(ef+1)\right]+4(e+f+ef+1)(e+f-ef-1)}{4(e+f)(ef+1)}\\&=\frac{(e-1)(f+1)\left[(e+f)^2-4(ef+1)\right]+4(e+1)(f+1)(e-1)(1-f)}{4(e+f)(ef+1)}\\&=\frac{(e-1)(f+1)(e^2+f^2-6ef+4(e-f)}{4(e+f)(ef+1)}.\end{align*}](http://latex.artofproblemsolving.com/0/d/4/0d460b1494b45bcef5b5921d87061bccf676b96d.png)
Thus 
The claim follows.
is the midpoint of 
is a parallelogram, 
are collinear. As 
,
.
,![[asy]
unitsize(3cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair A=dir(115), B=dir(205), C=dir(-25), E=foot(B,C,A), F=foot(C,A,B), H=A+B+C, M=(B+C)/2, N=(E+F)/2, X=(B+F)/2, Y=(C+E)/2, Z=(X+Y)/2, P=extension(E,F,X,Y);
D(arc(M,C,B),heavygreen);
D(CP(Z,M),red);
D(A--B--C--cycle);
DPA(Y--P--E^^H--P^^A--Z--M);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("E",E,NE);
D("F",F,NW);
D("H",H,SE);
D("M",M);
D("N",N,NW);
D("P",P,SW);
D("X",X,SE);
D("Y",Y,NE);
D("Z",Z,SW);
[/asy]](http://latex.artofproblemsolving.com/2/b/0/2b0413f8570dec1183631d61176d10f08a2a411a.png)
.
.
.
and 
,
,
is tangent to 
is the polar of 
are inverses in 
,
.![[asy]
unitsize(3cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair A=dir(115), B=dir(205), C=dir(-25), D=IP(D(incircle(A,B,C),red),D(B--C)), E=IP(incircle(A,B,C),D(C--A)), F=IP(incircle(A,B,C),D(A--B)), M_A=(0,-1), H=foot(A,B,C), L=extension(M_A,D,A,H);
D(unitcircle,heavygreen);
D(circumcircle(A,E,F),purple);
D(E--L--F--cycle);
DPA(A--H^^M_A--L);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("D",D,SSW);
D("E",E,dir(0));
D("F",F,W);
D("H",H,SW);
D("I",incenter(A,B,C));
D("L",L,SW);
D("M_A",M_A);
[/asy]](http://latex.artofproblemsolving.com/a/d/d/add6c9a37ec6d09dee6bbb3cb6dc76c4c8b9051e.png)
.
.
,
.
,
.
,
.
.
as 
Suppose 
.![\begin{align*}(b+c)(b-c)(-a+b+c)\ell&=a^2\left[(a-b+c)\left(a^2+b^2-c^2\right)-(a+b-c)\left(a^2-b^2+c^2\right)\right]\\&=a^2\left[-2a^2(b-c)+2a\left(b^2-c^2\right)\right]\\&=2a^3(b-c)(-a+b+c),\end{align*}](http://latex.artofproblemsolving.com/5/c/2/5c28c897bca78f67462de33edf3129b80277f6fd.png)
so 
.
.
for some 
.
,
to 
,
,
,
.
is the perpendicular bisector of 
that 
lie on ![[asy]
unitsize(3.2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
path carc(pair A, pair B, pair C, real d=0, bool dir=CW) {
pair O=circumcenter(A,B,C);
return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir);
}
pair A=dir(115), B=dir(205), C=dir(-25), D=foot(A,B,C), E=foot(B,C,A), F=foot(C,A,B), H=A+B+C, M=(B+C)/2, S_1=IP(L(E,F,10,10),D(unitcircle,heavygreen)), S_2=OP(L(E,F,10,10),unitcircle), P=IP(D(B--E),D(D--F)), Q=IP(D(C--F),D(D--E)), K=extension(E,F,B,C), T_1=OP(D(carc(S_1,D,S_2,5),red),D(arc(M,B,C,CW),orange)), T_2=OP(circumcircle(D,S_1,S_2),CP(M,B)), X=extension(M,H,A,K);
D(circumcircle(D,P,Q),red);
D(circumcircle(D,E,F),dashed+fuchsia);
D(carc(B,H,C,CW),chartreuse+dashed);
DPA(S_1--K--C--A--B^^D--H);
DPA(A--K^^M--X,dashed+pathpen);
D(K--T_2,dashed+red);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("D",D,SSW);
D("E",E,NNE);
D("F",F,NW);
D("H",H,N);
D("K",K,SW);
D("M",M);
D("P",P,W);
D("Q",Q,ENE);
D("X",X,NW);
D("S_1",S_1,NE);
D("S_2",S_2,WNW);
D("T_1",T_1,SW);
D("T_2",T_2,NE);
[/asy]](http://latex.artofproblemsolving.com/9/1/4/9142ac3390df8620839f23dac564462a27557974.png)
are concyclic.
and 
,
and the 
-
be the radical center of 
and the 
,
.
is the Miquel point of 
.
are concylic, and the claim follows. 
is the polar of 
in 
.
,
bisects 
.
,
on the plane, if there exists an inversion around ![[asy]
unitsize(0.5cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair O1=origin, O2=(4,0), O=(-8/3,0), X=dir(40), Y=IP(O--X,unitcircle), Xp=5/2 *(Y-O)+O, Yp=5/2 *(X-O)+O;
D(CR(O,sqrt(275/18)),heavygreen+dashed);
DPA(unitcircle^^CR(O2,5/2),red);
D(O--Yp);
D("O",O,SW);
D("X",X,SSW);
D("Y",Y,NW);
D("X'",Xp,SE);
D("Y'",Yp,NE);
[/asy]](http://latex.artofproblemsolving.com/6/e/1/6e1944df4c9601aaece7b2ceee28e2cc16475dc6.png)
,
be their respective inverses. Let 
be the radius of inversion. Then 
.![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
path carc(pair A, pair B, pair C, real d=0, bool dir=CW) {
pair O=circumcenter(A,B,C);
return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir);
}
pair A=origin, B=(3,0), P=(2,0), C=(2,1.3), D=(2,1/2), A_1=tangent(A,C,1.3,2), A_2=tangent(A,D,1/2,2), B_1=tangent(B,C,1.3), B_2=tangent(B,D,1/2), T=extension(A_1,B_1,A_2,B_2), K=extension(A_1,A_2,B_1,B_2);
D(carc(A_1,P,B_1,-5,CCW),red);
D(CP(D,A_2),red);
D(carc(A_1,A_2,B_1,-3,CCW),heavygreen);
DPA(arc(A,P,A_1)^^arc(B,B_1,P),lightmagenta);
DPA(carc(A_1,B_2,P)^^carc(P,A_2,B_1),springgreen);
DPA(A--T--A_1--K--B_1^^T--A_2);
DPA(B_1--B--B_2^^A_1--A--A_2,dashed+red);
D("A",A,SW);
D("B",B,SE);
D("K",K,SSE);
D("P",P);
D("T",T,SE);
D("A_1",A_1,W);
D("A_2",A_2,N);
D("B_1",B_1,NE);
D("B_2",B_2,NNE);
[/asy]](http://latex.artofproblemsolving.com/b/0/4/b04fd286063910386cc4c6ea35ee6e37f08f223d.png)
are concylic. Let 
:
concur at the Gergonne point of 
.
,
,
.
,
,
.
,
and 
are circles centered at 
,
be the radical center of 
.
.
,
,
be the intersection of 
is self-polar with respect to 
.
and 
are tangent to 
are collinear by the self-polar 
is a cyclic quadrilateral by the Radical Lemma.
passes through 
.
is indeed 
.
and 
are also tangent to 
and 
.
is harmonic by Brokard's theorem on 
and 
.
are collinear.
to 
,
by equal tangents, from which the conclusion follows immediately.
be the intersection of the circumcircle of triangle 
,
.
;
,
is the antipode of 
.
indeed is equivalent to 
,
;
,
bisects 
.
,
not containing 
.
;
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(12.156cm);
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 = -4.18, xmax = 16.08, ymin = -4.6, ymax = 7.04; /* image dimensions */
/* draw figures */
draw(circle((2.4249716406414117,2.4397927316014285), 1.891658788228061));
draw(circle((2.4374294214904597,0.3344277681123695), 2.105401820506235));
draw((0.74,1.58)--(6.580150549681395,1.6145571038442688));
draw((6.580150549681395,1.6145571038442688)--(2.4498872023395073,-1.770937195376689));
draw((2.4249716406414117,2.4397927316014285)--(4.12,1.6));
draw((4.12,1.6)--(2.4498872023395073,-1.770937195376689));
draw((2.4249716406414117,2.4397927316014285)--(0.74,1.58));
draw((0.74,1.58)--(2.4498872023395073,-1.770937195376689));
draw(circle((5.353735192308161,0.9887524999208022), 1.3768536709079602));
draw(circle((3.651386998014646,3.065597335524897), 3.268512459136021));
/* dots and labels */
dot((0.74,1.58),dotstyle);
label("$X$", (0.4,1.52), NE * labelscalefactor);
dot((4.12,1.6),dotstyle);
label("$Y$", (4.12,1.06), NE * labelscalefactor);
dot((2.4249716406414117,2.4397927316014285),dotstyle);
label("$O$", (2.5,2.64), NE * labelscalefactor);
dot((2.4249716406414117,2.4397927316014285),linewidth(3.pt) + dotstyle);
dot((2.4498872023395073,-1.770937195376689),linewidth(3.pt) + dotstyle);
label("$R$", (2.3,-2.2), NE * labelscalefactor);
dot((6.580150549681395,1.6145571038442688),dotstyle);
label("$P$", (6.74,1.5), NE * labelscalefactor);
dot((4.499403423970918,-0.09098971964018299),linewidth(3.pt) + dotstyle);
label("$Q$", (4.52,-0.52), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/d/8/5d8a70a433d007284e9343ae449b90b7dcb5eb09.png)
.
.
it fixes both these circles. Point 
on the arc 
(not containing 
is tangent to 
.
is tangent to 
is also tangent to ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(12.156cm);
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 = -4.3, xmax = 15.96, ymin = -5.34, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((2.44,3.82)--(1.1,-0.78));
draw((2.44,3.82)--(6.06,-0.74));
draw(circle((3.56583172248972,0.9968664112747335), 3.039338797691526));
draw(circle((2.44,3.82), 4.015619503887289));
draw((4.162025606230455,1.6508185733671596)--(1.1,-0.78));
draw((1.4987147362735018,0.5887222289985883)--(6.06,-0.74));
draw((1.2176186147722652,-0.005046895013017139)--(3.58,-0.76));
draw((3.58,-0.76)--(5.312832812773074,1.0142858966291055));
draw((2.44,3.82)--(2.4770072185731933,-0.7688951030760227));
draw(circle((3.58,-0.76), 2.4800806438501146));
draw((-1.9961017692645115,-0.8049685626553591)--(6.222833878717919,2.4726441279216766));
draw((-1.9961017692645115,-0.8049685626553591)--(6.06,-0.74));
draw((-1.9961017692645115,-0.8049685626553591)--(5.312832812773074,1.0142858966291055));
draw((2.44,3.82)--(0.6224276993739685,0.2392694974353349));
draw((2.44,3.82)--(6.222833878717919,2.4726441279216766));
draw((2.44,3.82)--(3.58,-0.76));
draw((2.44,3.82)--(3.56583172248972,0.9968664112747335));
/* dots and labels */
dot((2.44,3.82),dotstyle);
label("$A$", (2.52,4.02), NE * labelscalefactor);
dot((1.1,-0.78),dotstyle);
label("$B$", (0.88,-1.16), NE * labelscalefactor);
dot((6.06,-0.74),dotstyle);
label("$C$", (6.18,-0.96), NE * labelscalefactor);
dot((1.4987147362735018,0.5887222289985883),linewidth(3.pt) + dotstyle);
label("$F$", (1.26,0.62), NE * labelscalefactor);
dot((4.162025606230455,1.6508185733671596),linewidth(3.pt) + dotstyle);
label("$E$", (4.24,1.78), NE * labelscalefactor);
dot((0.6224276993739685,0.2392694974353349),linewidth(3.pt) + dotstyle);
label("$S_1$", (0.26,0.54), NE * labelscalefactor);
dot((6.222833878717919,2.4726441279216766),linewidth(3.pt) + dotstyle);
label("$S_2$", (6.3,2.42), NE * labelscalefactor);
dot((-1.9961017692645115,-0.8049685626553591),linewidth(3.pt) + dotstyle);
label("$T$", (-1.92,-0.68), NE * labelscalefactor);
dot((3.58,-0.76),linewidth(3.pt) + dotstyle);
label("$M$", (3.58,-1.12), NE * labelscalefactor);
dot((1.2176186147722652,-0.005046895013017139),linewidth(3.pt) + dotstyle);
label("$T_1$", (1.3,-0.36), NE * labelscalefactor);
dot((5.312832812773074,1.0142858966291055),linewidth(3.pt) + dotstyle);
label("$T_2$", (5.48,0.88), NE * labelscalefactor);
dot((2.46833655502056,0.30626717745053217),linewidth(3.pt) + dotstyle);
label("$H$", (2.48,0.52), NE * labelscalefactor);
dot((2.4770072185731933,-0.7688951030760227),linewidth(3.pt) + dotstyle);
label("$D$", (2.4,-1.14), NE * labelscalefactor);
dot((3.56583172248972,0.9968664112747335),linewidth(3.pt) + dotstyle);
label("$O$", (3.64,0.84), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/d/6/d/d6ddeb3928acb90214005fbcff499eb10383f17e.png)
are antiparallel in angle 
.
and so 
(by the shooting lemma), and 
(by power of point 
)
It follows that 
lines 
are concurrent. Let 
,
lies on 
lie on the polar of 
.
it follows that 
and 
lie on the polar of 
,
so ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(13.958cm);
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 = -4.3, xmax = 15.64, ymin = -5.02, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((2.94,4.04)--(1.68,-0.24));
draw((2.94,4.04)--(6.76,-0.28));
draw((1.8771590475233597,0.4297148598412539)--(4.22,-0.26));
draw((4.22,-0.26)--(5.63550837773795,0.9916763895738355));
draw((3.756333712630654,0.7106956247075447)--(4.22,-0.26));
draw((2.94,4.04)--(3.756333712630654,0.7106956247075447));
draw((0.1232176454868604,0.1674594102411332)--(4.5110167554759,2.263352779147671));
draw((0.1232176454868604,0.1674594102411332)--(2.914897397305532,0.851969457802577));
draw((0.1232176454868604,0.1674594102411332)--(5.63550837773795,0.9916763895738355));
draw((1.68,-0.24)--(4.5110167554759,2.263352779147671));
draw((2.0743180950467197,1.0994297196825078)--(6.76,-0.28));
draw((1.68,-0.24)--(6.76,-0.28));
/* dots and labels */
dot((2.94,4.04),dotstyle);
label("$A$", (3.02,4.24), NE * labelscalefactor);
dot((1.68,-0.24),dotstyle);
label("$B$", (1.64,-0.62), NE * labelscalefactor);
dot((6.76,-0.28),dotstyle);
label("$C$", (6.84,-0.08), NE * labelscalefactor);
dot((2.0743180950467197,1.0994297196825078),linewidth(3.pt) + dotstyle);
label("$F$", (1.84,1.32), NE * labelscalefactor);
dot((4.5110167554759,2.263352779147671),linewidth(3.pt) + dotstyle);
label("$E$", (4.6,2.38), NE * labelscalefactor);
dot((2.914897397305532,0.851969457802577),linewidth(3.pt) + dotstyle);
label("$H$", (2.82,1.08), NE * labelscalefactor);
dot((4.22,-0.26),linewidth(3.pt) + dotstyle);
label("$M$", (4.08,-0.66), NE * labelscalefactor);
dot((1.8771590475233597,0.4297148598412539),linewidth(3.pt) + dotstyle);
label("$X$", (1.92,0.12), NE * labelscalefactor);
dot((5.63550837773795,0.9916763895738355),linewidth(3.pt) + dotstyle);
label("$Y$", (5.72,1.12), NE * labelscalefactor);
dot((0.1232176454868604,0.1674594102411332),linewidth(3.pt) + dotstyle);
label("$P$", (0.2,0.28), NE * labelscalefactor);
dot((3.756333712630654,0.7106956247075447),linewidth(3.pt) + dotstyle);
label("$Z$", (3.84,0.84), NE * labelscalefactor);
dot((3.7563337126306546,0.7106956247075447),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/e/c/fec9780732cb896b60c234cdccb8d8b021e05491.png)
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(14.182cm);
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 = -4.38, xmax = 15.88, ymin = -5.16, ymax = 6.48; /* image dimensions */
/* draw figures */
draw((3.16,3.8)--(1.92,-0.62));
draw((1.92,-0.62)--(6.56,-0.7));
draw((6.56,-0.7)--(3.16,3.8));
draw(circle((3.1295768633540373,2.0354580745341613), 1.7648041743972807));
draw(circle((4.270423136645964,1.1045419254658377), 2.9152244808882344));
draw(circle((3.326968536652661,1.8743889623608976), 1.9328363511972426));
draw((0.6984874677017939,-0.12153028343233199)--(4.8353096510531275,1.5826784030179188));
draw((0.6984874677017939,-0.12153028343233199)--(5.133719280703113,1.1877244814223489));
draw((0.6984874677017939,-0.12153028343233199)--(3.72031361693487,0.3724596193518222));
draw((3.16,3.8)--(3.72031361693487,0.3724596193518222));
/* dots and labels */
dot((3.16,3.8),dotstyle);
label("$A$", (3.24,4.), NE * labelscalefactor);
dot((1.92,-0.62),dotstyle);
label("$B$", (1.7,-1.08), NE * labelscalefactor);
dot((6.56,-0.7),dotstyle);
label("$C$", (6.44,-1.16), NE * labelscalefactor);
dot((2.237737496441112,0.5125804308626747),linewidth(3.pt) + dotstyle);
label("$F$", (2.36,0.66), NE * labelscalefactor);
dot((4.8353096510531275,1.5826784030179188),linewidth(3.pt) + dotstyle);
label("$E$", (4.44,1.62), NE * labelscalefactor);
dot((3.099153726708075,0.2709161490683233),linewidth(3.pt) + dotstyle);
label("$H$", (3.06,-0.12), NE * labelscalefactor);
dot((1.407004792701848,1.651688305819772),linewidth(3.pt) + dotstyle);
label("$Q$", (1.06,1.74), NE * labelscalefactor);
dot((2.1827618721519153,0.31661893138021663),linewidth(3.pt) + dotstyle);
label("$X'$", (2.16,-0.14), NE * labelscalefactor);
dot((5.133719280703113,1.1877244814223489),linewidth(3.pt) + dotstyle);
label("$Y'$", (5.22,1.3), NE * labelscalefactor);
dot((3.6582405764275143,0.7521717064012827),linewidth(3.pt) + dotstyle);
label("$Z'$", (3.74,0.88), NE * labelscalefactor);
dot((3.72031361693487,0.3724596193518222),linewidth(3.pt) + dotstyle);
label("$T$", (3.8,0.5), NE * labelscalefactor);
dot((0.6984874677017939,-0.12153028343233199),linewidth(3.pt) + dotstyle);
label("$P'$", (0.78,0.), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/6/b/d/6bd6c49d74aa2818e1626ee72203c50a998f0f89.png)
.
.
so 
and 
be chosen dividing lines 
in the same ratio, and define 
,
as the midpoint of 
.
.
and 
.
so 
and 
are concyclic. It follows that as 
moves uniformly over line 
are concyclic, 
and 
,
the perpendicular bisector of 
,
.
the line 
we see that 
.
it follows that this holds for all positions of 
.
and 
so 
as desired.
means that 
move on two conics such that the cross-ratio of any four positions of ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(14.182cm);
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 = -4.4, xmax = 15.86, ymin = -5.26, ymax = 6.28; /* image dimensions */
/* draw figures */
draw((3.08,4.48)--(2.08,0.66));
draw((2.08,0.66)--(7.1,0.62));
draw((7.1,0.62)--(3.08,4.48));
draw(circle((4.601162548656355,2.040899856372682), 2.8745686650482893));
draw(circle((3.4341702685138764,3.2239341182799035), 1.3050433243078468));
draw(circle((3.7883405370277576,1.9678682365598092), 1.321438562656396));
draw((3.08,4.48)--(3.0495001983969527,0.6522748988175542));
draw((2.4669439221385807,1.9783972932521536)--(5.109737151916936,1.9573391798674655));
draw((2.5099785280448303,2.302517977131252)--(3.060029255089296,1.9736715137067287));
draw((3.060029255089296,1.9736715137067287)--(4.703577508507319,2.9210424918312805));
draw((3.060029255089296,1.9736715137067287)--(4.578258346148836,-0.8335775583211602));
draw((4.578258346148836,-0.8335775583211602)--(3.08,4.48));
draw((3.788340537027759,1.9678682365598097)--(3.777811480335415,0.6464716216706341));
/* dots and labels */
dot((3.08,4.48),dotstyle);
label("$A$", (3.16,4.68), NE * labelscalefactor);
dot((2.08,0.66),dotstyle);
label("$B$", (1.76,0.44), NE * labelscalefactor);
dot((7.1,0.62),dotstyle);
label("$C$", (7.16,0.26), NE * labelscalefactor);
dot((4.578258346148836,-0.8335775583211602),linewidth(3.pt) + dotstyle);
label("$M$", (4.66,-0.72), NE * labelscalefactor);
dot((3.0495001983969527,0.6522748988175542),linewidth(3.pt) + dotstyle);
label("$H$", (2.9,0.22), NE * labelscalefactor);
dot((3.788340537027759,1.9678682365598097),linewidth(3.pt) + dotstyle);
label("$I$", (3.86,2.08), NE * labelscalefactor);
dot((3.777811480335415,0.6464716216706341),linewidth(3.pt) + dotstyle);
label("$D$", (3.66,0.34), NE * labelscalefactor);
dot((3.060029255089296,1.9736715137067287),linewidth(3.pt) + dotstyle);
label("$L$", (3.14,2.1), NE * labelscalefactor);
dot((2.5099785280448303,2.302517977131252),linewidth(3.pt) + dotstyle);
label("$F$", (2.26,2.2), NE * labelscalefactor);
dot((4.703577508507319,2.9210424918312805),linewidth(3.pt) + dotstyle);
label("$E$", (4.78,3.04), NE * labelscalefactor);
dot((5.109737151916936,1.9573391798674655),linewidth(3.pt) + dotstyle);
label("$I_F$", (5.18,2.08), NE * labelscalefactor);
dot((2.4669439221385807,1.9783972932521536),linewidth(3.pt) + dotstyle);
label("$I_E$", (2.12,1.84), NE * labelscalefactor);
dot((4.161794692353406,0.6434119944832398),linewidth(3.pt) + dotstyle);
label("$J$", (4.32,0.36), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/d/c/0dc9ff1c62debf7d554bd6c5eda1ee4aaafa6db4.png)
be the intersection of lines 
.
(both are perpendicular to 
It follows that 
so 
is the external bisector of 
(since 
)
on it. Finally, as 
we have 
so the points ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(13.958cm);
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 = -4.3, xmax = 15.64, ymin = -4.92, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((3.44,4.62)--(1.6,0.16));
draw((1.6,0.16)--(8.02,0.1));
draw((3.44,4.62)--(8.02,0.1));
draw(circle((4.822244812758319,1.4401949651400658), 3.467241090855649));
draw((3.44,4.62)--(3.3981606986899564,0.1431947598253275));
draw((4.737674778055778,3.3393253282069604)--(1.6,0.16));
draw((2.5126169814238826,2.3721042049731063)--(8.02,0.1));
draw((1.3831556524927384,1.8811332032887342)--(6.845772881121266,4.255704677193139));
draw(circle((4.13112240637916,3.0300974825700338), 2.978496297719618));
draw(circle((3.612482112063784,0.9873385470390418), 0.8709261746681332));
draw(circle((4.797755187241681,-1.1801949651400663), 3.467241090855648), dotted);
draw((5.6754716629277056,5.5769436831340595)--(3.3981606986899564,0.1431947598253275));
draw((3.3981606986899564,0.1431947598253275)--(1.6168048932090857,4.626859092358958));
draw((1.6168048932090857,4.626859092358958)--(5.6754716629277056,5.5769436831340595));
draw((2.4255114394674,1.348484607930371)--(5.801156928827508,2.1386822135305126));
/* dots and labels */
dot((3.44,4.62),dotstyle);
label("$A$", (3.52,4.82), NE * labelscalefactor);
dot((1.6,0.16),dotstyle);
label("$B$", (1.36,-0.22), NE * labelscalefactor);
dot((8.02,0.1),dotstyle);
label("$C$", (8.1,0.3), NE * labelscalefactor);
dot((2.5126169814238826,2.3721042049731063),linewidth(3.pt) + dotstyle);
label("$F$", (2.18,2.42), NE * labelscalefactor);
dot((4.737674778055778,3.3393253282069604),linewidth(3.pt) + dotstyle);
label("$E$", (4.68,3.5), NE * labelscalefactor);
dot((3.3981606986899564,0.1431947598253275),linewidth(3.pt) + dotstyle);
label("$D$", (3.32,-0.22), NE * labelscalefactor);
dot((3.415510374483363,1.9996100697198675),linewidth(3.pt) + dotstyle);
label("$H$", (3.44,2.2), NE * labelscalefactor);
dot((2.8772839735524687,1.4542390705465278),linewidth(3.pt) + dotstyle);
label("$P$", (3.,1.16), NE * labelscalefactor);
dot((4.064057019639518,1.732048225081455),linewidth(3.pt) + dotstyle);
label("$Q$", (4.14,1.86), NE * labelscalefactor);
dot((1.3831556524927384,1.8811332032887342),linewidth(3.pt) + dotstyle);
label("$S_1$", (0.92,1.84), NE * labelscalefactor);
dot((6.845772881121266,4.255704677193139),linewidth(3.pt) + dotstyle);
label("$S_2$", (6.92,4.38), NE * labelscalefactor);
dot((2.4255114394674,1.348484607930371),linewidth(3.pt) + dotstyle);
label("$S_2'$", (2.24,1.52), NE * labelscalefactor);
dot((5.801156928827508,2.1386822135305126),linewidth(3.pt) + dotstyle);
label("$S_1'$", (5.68,2.38), NE * labelscalefactor);
dot((1.6168048932090857,4.626859092358958),linewidth(3.pt) + dotstyle);
label("$Q'$", (1.18,4.44), NE * labelscalefactor);
dot((5.6754716629277056,5.5769436831340595),linewidth(3.pt) + dotstyle);
label("$P'$", (5.76,5.7), NE * labelscalefactor);
dot((3.799877893183851,1.1017069732798614),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/2/f/52f81adfd421974c6b38750e95c163d93957fde0.png)
with incenter 
opposite to 
,
of the internal bisectors of angle 
respectively, line 
.
and 
are tangent to each other. 
followed by reflection in the bisector 
.
and 
.
are the intersections of 
with 
.
and 
.
as 
are concyclic. Similarly reasoning applies to ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(13.958cm);
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 = -4.2, xmax = 15.74, ymin = -4.94, ymax = 6.28; /* image dimensions */
/* draw figures */
draw(circle((3.923956456999118,1.920274095384037), 0.7064434520308507));
draw(circle((3.9327598918653233,3.077925780289998), 1.8641286095892664));
draw((2.3784868992284762,4.107105520398066)--(0.48,1.24));
draw((0.48,1.24)--(3.650392508744856,2.571599756733878));
draw((4.404237760174842,2.438339941331338)--(5.74,1.2));
draw((5.74,1.2)--(5.796007065047093,3.020606981285099));
draw((2.3784868992284762,4.107105520398066)--(11.664332533973065,1.1549480415666769));
draw((11.664332533973065,1.1549480415666769)--(3.650392508744856,2.571599756733878));
draw(shift((2.682771805789768,2.412549269787128))*xscale(1.72165913898356)*yscale(1.72165913898356)*arc((0,0),1,-80.63054505370833,130.9563314866489));
draw(shift((4.867480845014839,2.1066455572606064))*xscale(1.302876212804682)*yscale(1.302876212804682)*arc((0,0),1,10.070050466466421,249.96264022263608));
draw((4.200287763622658,3.225725455227983)--(3.918584418776112,1.213851069058737));
draw((4.200287763622658,3.225725455227983)--(11.664332533973065,1.1549480415666769));
draw((0.48,1.24)--(11.664332533973065,1.1549480415666769));
draw((2.682771805789767,2.4125492697871267)--(11.664332533973065,1.1549480415666769), dotted);
/* dots and labels */
dot((0.48,1.24),dotstyle);
label("$A$", (0.4,0.82), NE * labelscalefactor);
dot((5.74,1.2),dotstyle);
label("$B$", (5.78,0.8), NE * labelscalefactor);
dot((3.918584418776112,1.213851069058737),dotstyle);
label("$P$", (3.88,0.8), NE * labelscalefactor);
dot((3.650392508744856,2.571599756733878),linewidth(3.pt) + dotstyle);
label("$A_1$", (3.22,2.72), NE * labelscalefactor);
dot((4.404237760174842,2.438339941331338),linewidth(3.pt) + dotstyle);
label("$B_1$", (4.48,2.56), NE * labelscalefactor);
dot((2.3784868992284762,4.107105520398066),linewidth(3.pt) + dotstyle);
label("$A_2$", (2.12,4.32), NE * labelscalefactor);
dot((5.796007065047093,3.020606981285099),linewidth(3.pt) + dotstyle);
label("$B_2$", (5.88,3.14), NE * labelscalefactor);
dot((11.664332533973065,1.1549480415666769),linewidth(3.pt) + dotstyle);
label("$T$", (11.74,1.28), NE * labelscalefactor);
dot((3.918584418776112,1.213851069058737),linewidth(3.pt) + dotstyle);
dot((4.200287763622658,3.225725455227983),linewidth(3.pt) + dotstyle);
label("$Q$", (3.82,3.14), NE * labelscalefactor);
dot((2.682771805789767,2.4125492697871267),linewidth(3.pt) + dotstyle);
label("$O_1$", (2.76,2.54), NE * labelscalefactor);
dot((4.867480845014841,2.106645557260607),linewidth(3.pt) + dotstyle);
label("$O_2$", (4.94,2.22), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/2/4/f24eb98bf6d072d40505a75f23a575f5d5233c5d.png)
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(13.958cm);
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 = -3.86, xmax = 16.08, ymin = -4.94, ymax = 6.28; /* image dimensions */
/* draw figures */
draw((2.666013595902149,5.13488233125554)--(4.779210376771846,3.6825998756313716));
draw((4.786013595902149,5.124882331255539)--(2.659210376771846,3.6925998756313714));
draw((3.722611986336997,4.408741103443456)--(4.041447547001892,1.8868799644009344));
draw((4.009585795588095,3.163937273561756)--(3.7544737377507924,3.1316837942826337));
draw((3.722611986336997,4.408741103443456)--(3.38,1.89));
draw((3.6787042103246614,3.132041197713889)--(3.423907776012334,3.166699905729566));
draw((0.34111249455605996,5.14584884588453)--(8.521213506904486,5.107263463750812));
draw((0.3810063496275591,3.703346121042429)--(8.721103802069432,3.6640060387195903));
draw((2.669997880097665,5.979550580704977)--(2.6415210901571657,-0.0575288866809045));
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dot((5.5,1.88),dotstyle);
label("$A'$", (5.58,2.08), NE * labelscalefactor);
dot((4.041447547001892,1.8868799644009344),dotstyle);
label("$P$", (4.12,2.08), NE * labelscalefactor);
dot((2.659210376771846,3.6925998756313714),dotstyle);
label("$B_2'$", (2.2,3.86), NE * labelscalefactor);
dot((2.666013595902149,5.13488233125554),dotstyle);
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dot((4.786013595902149,5.124882331255539),linewidth(3.pt) + dotstyle);
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dot((4.779210376771846,3.6825998756313716),linewidth(3.pt) + dotstyle);
label("$A_2'$", (4.86,3.8), NE * labelscalefactor);
dot((3.38,1.89),linewidth(3.pt) + dotstyle);
label("$M$", (3.46,2.), NE * labelscalefactor);
dot((3.722611986336997,4.408741103443456),linewidth(3.pt) + dotstyle);
label("$Q'$", (3.74,4.58), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/2/2/c/22ca18d12a146873c2a5bec2d72b1bfde4f80195.png)
.
and note that 
are concurrent (at the Gergonne point of triangle 
)
passes through 
passes through 
be the second common point of 
and 
.
and 
is the perpendicular bisector of 
.
followed by reflection in 
(but we shall refer to them with 
versions only).
is a line parallel to 
and 
lie on it. Also, 
and similarly 
.
.
.
so it lies on the parallel midline of the perpendicular bisectors of 
.
lies on the perpendicular bisector of 
so 
as desired.
meet on 
(perpendicular bisector of 
are coaxial. Since 
and 
(i.e., the common perpendicular bisector of 
,
,
.
,
,
,
,
.
,
.
,
wrt 
.
,
,
are colinear.
lies on 
lies on 
,
,
are colinear. Meanwhile, by Brokard on 
,
,
,
.
are colinear.
.
.
,
,
.
.
.
,
is cyclic. The rest is simple angle chasing.
and 
.
,
.
,
is cyclic.
is cyclic.
.
,
and 
.
such that 
is further away from 
.
as 
and the center of 
as 
.
and 
.
concur and 
concur.
.
and 
.
concur in 
,
as well.
.
be the point on 
.
passes through 
are all tangents to 
are all tangents to 
with center 
,
.
.
.
.
is harmonic, and projecting at 
,
and 
intersect the circumcircle at 
and 
are concyclic. Let 
and 
be the reflections of 
so 
so 
is concyclic. Therefore, 
and similarly, 
.
and 
,
.
at 
at 
.
is cyclic with diameter 
and 
is cyclic with diameter 
.
and 
,
.
is cyclic. We have, therefore, that
Therefore it suffices to show that 
.
.
and 
so 
.
is similar to 
so 
which finishes the proof.
is cyclic with diameter 
.
to 
to 
.
as well. Then, 
is a parallelogram, 
is the perpendicular bisector of 
.![\[\angle SAH=\angle SMB=\angle SNF=\angle SZP\]](http://latex.artofproblemsolving.com/4/4/d/44dfdb05566262241536418c3ee378d63bf3444d.png)
and 
so there is a spiral similarity at 
we easily see that 
is cyclic so 
as desired.
inversion on 
with center 
e
.
meets 
lie on 
such that 
and 
.
.
.
thus, 
.![\[\frac{AF}{AP}=\frac{AB}{AP}.\frac{AF}{AB}=\frac{AE}{AC}.\frac{AF}{AB}=\frac{AE}{AC}.\frac{AC}{AQ}=\frac{AE}{AQ}\]](http://latex.artofproblemsolving.com/d/6/f/d6f6ae5719a7539c2f678f230120893df864dbfe.png)
Hence 
as desired
such that 
are collinear and 
.
.
.
are the feet of the altitude from 
,
implies 
.
,
are concyclic. Let 
.![\[\measuredangle KW'A=\measuredangle KFA=\measuredangle KPB=180-\measuredangle MWK\]](http://latex.artofproblemsolving.com/7/b/9/7b986a119093fba2add03d2465a1482498d2dbfe.png)
Thus, 
are collinear so 
.
.
and 
,
is a parallelogram. By the lemma, 
hence 
.
and 
as desired
