[/quote]
,
.
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of 
are pairwise different; and
denotes the integer part of real number 
,
![\[ p = \frac{2ab}{a + b} \]](http://latex.artofproblemsolving.com/0/f/4/0f4297ec59fe97c92fda75af4cc835be6b9a2507.png)
, ![\[ p, a, b \in \mathbb{C} \]](http://latex.artofproblemsolving.com/b/4/4/b4499970ac49d770a810d7345fedef7773cd03f5.png)
have the same incenter 
is concyclic iff 
is concurrent
share circumcircle 
and incircle 
so that points 
and 
occur in this order along 
be the triangle formed by lines 
and 
and define triangles 
similarly. Furthermore, let 
and 
be the circumcircle and incircle of triangle 
similarly.
and the two common external tangents to 
are either concurrent or pairwise parallel.
,
and 
similarly. Prove that points 
,
such that 
for all 
.
such that the equation
.
on the altitude from 
.
intersects 
.
such that it's on the segment 
and 
.
and circle 
is 
, 
)
and circle 
is 
,
)
.
is tangent to circle 
.
and 
.
are 
and 
Find the product of all possible values of 
be a set consisting of 
positive integers, none of which is a sum of two other distinct members of 
so that 
does not divide 
for all 
.
Y by anantmudgal09, nguyendangkhoa17112003, megarnie, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with incenter 
, and let 
be a point on line 
satisfying 
. Let the excircle of triangle opposite the vertex 
be tangent to 
at 
. Define points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively.
Prove that if quadrilateral
is cyclic, then 
is tangent to the circumcircle of 
.
Ankan Bhattacharya
be a triangle with incenter 
, and let 
be a point on line 
satisfying 
. Let the excircle of triangle opposite the vertex 
be tangent to 
at 
. Define points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively.Prove that if quadrilateral
is cyclic, then 
is tangent to the circumcircle of 
.Ankan Bhattacharya
This post has been edited 1 time. Last edited by tastymath75025, Jan 21, 2019, 5:08 PM
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
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 = -14.75, xmax = 11.45, ymin = -3.79, ymax = 7.45; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wwzzff = rgb(0.4,0.6,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen qqccqq = rgb(0,0.8,0);
/* draw figures */
draw((-4.781523029254206,2.683377195388904)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr);
draw(circle((0.7015597483382036,1.4355753715441475), 2.546602274766395), linewidth(0.4) + blue);
draw((-4.781523029254206,2.683377195388904)--(2.0614153440904808,-0.7175562789934126), linewidth(0.4) + wrwrwr);
draw((2.0614153440904808,-0.7175562789934126)--(6.871149068786946,5.218374928896586), linewidth(0.4) + wrwrwr);
draw((-0.7268440288168109,-0.6727049217884695)--(-0.66,3.58), linewidth(0.4) + wrwrwr);
draw((-7.217533164119876,-0.5682970536602701)--(-2.76,-0.64), linewidth(0.4) + wrwrwr);
draw((-7.217533164119876,-0.5682970536602701)--(-0.09465334849420755,0.9812603385617286), linewidth(0.4) + wrwrwr);
draw(circle((-0.9294856026840693,-10.516443266922067), 11.768821318012856), linewidth(0.4) + linetype("2 2") + wwzzff);
draw((-7.217533164119876,-0.5682970536602701)--(-0.66,3.58), linewidth(0.4) + wrwrwr);
draw(circle((-3.943883513499975,1.5047383027747945), 3.8846624103269236), linewidth(0.4) + ffxfqq);
draw((-0.66,3.58)--(-2.76,-0.64), linewidth(0.4) + qqccqq);
draw((-2.76,-0.64)--(4.7,-0.76), linewidth(0.4) + qqccqq);
draw((4.7,-0.76)--(-0.66,3.58), linewidth(0.4) + qqccqq);
draw((-3.9387665820599382,1.505851473169865)--(3.0875794770264293,0.545579304049496), linewidth(0.4) + wrwrwr);
/* dots and labels */
dot((-0.66,3.58),dotstyle);
label("$A$", (-0.69,3.87), NE * labelscalefactor);
dot((-2.76,-0.64),dotstyle);
label("$B$", (-3.15,-1.15), NE * labelscalefactor);
dot((4.7,-0.76),dotstyle);
label("$C$", (4.93,-1.19), NE * labelscalefactor);
dot((-0.09465334849420755,0.9812603385617286),linewidth(4pt) + dotstyle);
label("$I$", (0.15,0.57), NE * labelscalefactor);
dot((6.871149068786946,5.218374928896586),linewidth(4pt) + dotstyle);
label("$I_B$", (7.13,5.43), NE * labelscalefactor);
dot((-4.781523029254206,2.683377195388904),linewidth(4pt) + dotstyle);
label("$I_C$", (-5.33,2.89), NE * labelscalefactor);
dot((-1.8356826708257379,1.2174376805311355),linewidth(4pt) + dotstyle);
label("$C_1$", (-2.35,1.71), NE * labelscalefactor);
dot((3.0875794770264293,0.545579304049496),linewidth(4pt) + dotstyle);
label("$B_1$", (3.09,-0.03), NE * labelscalefactor);
dot((2.0614153440904808,-0.7175562789934126),linewidth(4pt) + dotstyle);
label("$A_1$", (2.21,-1.05), NE * labelscalefactor);
dot((-0.7268440288168109,-0.6727049217884695),linewidth(4pt) + dotstyle);
label("$X$", (-0.71,-1.25), NE * labelscalefactor);
dot((-7.217533164119876,-0.5682970536602701),linewidth(4pt) + dotstyle);
label("$D$", (-7.99,-0.45), NE * labelscalefactor);
dot((-3.9387665820599382,1.505851473169865),linewidth(4pt) + dotstyle);
label("$M$", (-4.53,1.51), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/2/b/b/2bb203c2dcc9d9b73723cb5ce0bf06c76f0da4cf.png)
,
is the foot of the 
,
and 
are the 
and 
are collinear. Thus, 
Suppose the tangents to 
at 
,
is the 
.
,
.
(again not that difficult), the above solution is quite easy to find (Probably saying this because I personally found P1 harder than this 
).
denote the Bevan point of 
are collinear on a line perpendicular to 
.
,
Writing 
and 
,
.
.
into this equation gives us an equation in 
.
refer to the 
excenters of 
respectively. Define 
and 
.
.
is the Bevan point of 
,
.
and 
is the external angle bisector of 
.
.
is a rectangle, we would be done by Power of the Point, since then 
.
on 
,
.
we wish to show that:
and so we are done.
.
,
.
is cyclic, where 
.
satisfying 
.
was a diameter, or that 
was the angle bisector of 
which was implying the harmonic quad).
be the excenters, let 
be the foot of the altitude from 
,
concurrent in 
and 
is cyclic with diameter 
differ, then 
which implies 
,
,
is cyclic.By Pappus, 
.
,
is harmonic. It follows that 
,
is cyclic, implying that 
and 
.
will be also tangent, so 
since 
meets 
on the circle.
to be the bevan (because I couldn't find that).
denote the midpoints of small 
resp. Let 
Similarly, let 
Now applying Pascal's on cyclic hexagon 
we get 
is tangent to 
let 
be the incircle touch points on 
resp. then 
are concurrent. Thus the result is obvious for any arbitrary 
.
is the angle bisector and 
the tangent we have 
but 
is the Humpty point of 
is tangent to 
is cyclic.
,
.
be the tangency points of the incircle. Denote by 
,
and the circumcircle.
,
,
.![[asy] size(11cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair E = foot(I, C, A); pair B_1 = A+C-E; pair F = foot(I, A, B); pair C_1 = A+B-F; pair T = foot(I, B, C); pair Z = foot(A, I, foot(T, E, F));
filldraw(A--B--C--cycle, invisible, blue); draw(circumcircle(A, E, F), orange); pair D = extension(A, Z, B, C); draw(arc(dir(270), C, B), orange); draw(D--B, blue);
filldraw(A--B_1--C_1--cycle, invisible, blue); pair M = midpoint(A--D); filldraw(A--D--I--cycle, invisible, red); draw(I--M, red); filldraw(Z--E--F--cycle, invisible, deepgreen); filldraw(unitcircle, invisible, blue); draw(incircle(A, B, C), dotted+blue); draw(circumcircle(A, B_1, C_1), dashed+red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$I$", I, dir(270)); dot("$E$", E, dir(30)); dot("$B_1$", B_1, dir(B_1)); dot("$F$", F, dir(F)); dot("$C_1$", C_1, dir(C_1)); dot("$Z$", Z, dir(Z)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M));
/* TSQ Source:
!size(11cm); A = dir 130 B = dir 210 R225 C = dir 330 R315 I = incenter A B C R270 E = foot I C A R30 B_1 = A+C-E F = foot I A B C_1 = A+B-F T := foot I B C Z = foot A I foot T E F
A--B--C--cycle 0.1 lightcyan / blue circumcircle A E F orange D = extension A Z B C !draw(arc(dir(270), C, B), orange); D--B blue
A--B_1--C_1--cycle 0.1 deepgreen / blue M = midpoint A--D A--D--I--cycle 0.1 lightred / red I--M red Z--E--F--cycle 0.1 deepgreen / deepgreen unitcircle 0.1 lightcyan / blue incircle A B C dotted blue circumcircle A B_1 C_1 dashed red
*/ [/asy]](http://latex.artofproblemsolving.com/1/5/7/157411b3583da3c05337ba393eca43d89d8453a9.png)
![\[ \frac{ZF}{ZE} = \frac{BF}{CE} = \frac{AC_1}{AB_1} \]](http://latex.artofproblemsolving.com/5/9/a/59a3aaf6ffb44324eea96fb149006e355ab187af.png)
which together with 
implies that 
and 
are (directly) similar. (See IMO Shortlist 2006 G9 for a similar application of spiral similarity.)![\[ \measuredangle DAC_1 = \measuredangle ZAF = \measuredangle ZEF = \measuredangle AB_1C_1 \]](http://latex.artofproblemsolving.com/4/1/4/414cf86361e657de63dc255b1ad814931369d714.png)
we have 
,
the last step since 
is right with hypotenuse 
.![[asy] unitsize(100); pair A, B, C, I, A1, B1, C1, D, E, F, J, K, Z, M; B = dir(193); C = reflect((0, 1), (0, -1)) * B; I = intersectionpoints(circle(dir(270), abs(dir(270)-B)), -B--(-C))[1]; A = 2 * foot(origin, I, dir(270)) - dir(270); D = extension(B, C, I, rotate(90, I) * A); E = foot(I, C, A); F = foot(I, A, B); A1 = B + C - foot(I, B, C); B1 = C + A - E; C1 = A + B - F; M = (A + D)/2; Z = foot(I, A, D); K = 2 * foot(circumcenter(A, B1, C1), A, I) - A;
draw(unitcircle); filldraw(circumcircle(A, B1, C1), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavycyan+dashed); draw(incircle(A, B, C), heavycyan+dotted); filldraw(circumcircle(A, B, C), invisible, cyan); filldraw(circumcircle(B, I, C), invisible, heavycyan+dashed); draw(arc(circumcenter(D, B1, C1), B1, D), heavyred); draw(A--D, heavyred); draw(A--B--C--cycle, heavymagenta); draw(D--B, lightmagenta); draw(A--I--D, heavycyan); draw(I--K, heavyred); draw(M--B1, dashed+heavyred);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(10)); dot("$I$", I, dir(-I)); dot("$A_1 = V$", A1, dir(A1)); dot("$B_1$", B1, dir(10)); dot("$C_1$", C1, dir(130)); dot("$D$", D, dir(D)); dot("$E$", E, dir(40)); dot("$F$", F, dir(170)); dot("$M$", M, dir(130)); dot("$Z$", Z, dir(Z)); // dot("$K$", K, dir(K-circumcenter(A, B1, C1)));
clip(box((-3, -0.4), (1.2, 1.25))); [/asy]](http://latex.artofproblemsolving.com/2/8/6/2861458811aa0950edd6856310be76556cd14265.png)
.
,
(which is the circumcenter of 
)
.
then 
.
by definition, which would imply 
,
and 
after the previous claim. 
was tangent to the circumcircle of 
,
,
and 
denote the circumradius and 
is harmonic.
.
.
be the intouch point on 
be the circumcenter of 
.![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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., xmax = 6., ymin = -3., ymax = 4.; /* image dimensions */
/* draw figures */ draw(circle((2.7,0.6), 2.7658633371878665), linewidth(0.4)); draw((1.892579421398536,1.2)--(3.5074205786014643,0.), linewidth(0.4) + dotted); draw((1.445226775603362,3.064861893765504)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.109814760356735,0.)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((0.8894305730917524,2.690894150918487)--(3.9547732243966385,-1.864861893765504), linewidth(0.4) + linetype("2 2")); draw((1.445226775603362,3.064861893765504)--(0.,0.), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(5.4,0.), linewidth(0.4)); draw((5.4,0.)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.5571674061519136,1.864861893765503)--(2.7,3.3658633371878666), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((-3.5571674061519136,1.864861893765503)--(-3.109814760356735,0.), linewidth(0.4)); draw((-3.109814760356735,0.)--(1.892579421398536,1.2), linewidth(0.4)); draw((1.892579421398536,1.2)--(1.892579421398536,0.), linewidth(0.4) + dotted); draw((2.7,3.3658633371878666)--(2.7,-2.1658633371878664), linewidth(0.4)); draw((0.8894305730917524,2.690894150918487)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((2.7,3.3658633371878666)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((-3.5571674061519136,1.864861893765503)--(4.217574445111032,0.9163536869921585), linewidth(0.4)); draw((1.445226775603362,3.064861893765504)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2")); draw((3.5074205786014643,0.)--(4.217574445111032,0.9163536869921585), linewidth(0.4) + dotted); draw((0.6380284948350587,1.353053551156462)--(3.5074205786014643,0.), linewidth(0.4) + dotted); draw((0.8894305730917524,2.690894150918487)--(2.7,-2.1658633371878664), linewidth(0.4) + linetype("2 2")); /* dots and labels */ dot((0.,0.),linewidth(4.pt) + dotstyle); label("$B$", (-0.32,-0.42), NE * labelscalefactor); dot((2.7,0.6),linewidth(4.pt) + dotstyle); label("$O$", (2.78,0.76), NE * labelscalefactor); dot((2.7,-2.1658633371878664),linewidth(4.pt) + dotstyle); label("$M$", (2.78,-2.), NE * labelscalefactor); dot((1.892579421398536,1.2),linewidth(4.pt) + dotstyle); label("$I$", (1.5,1.32), NE * labelscalefactor); dot((1.445226775603362,3.064861893765504),linewidth(4.pt) + dotstyle); label("$A$", (1.52,3.22), NE * labelscalefactor); dot((-3.109814760356735,0.),linewidth(4.pt) + dotstyle); label("$D$", (-3.22,-0.46), NE * labelscalefactor); dot((1.892579421398536,0.),linewidth(4.pt) + dotstyle); label("$L$", (1.72,-0.42), NE * labelscalefactor); dot((4.217574445111032,0.9163536869921585),linewidth(4.pt) + dotstyle); label("$B_1$", (4.3,1.08), NE * labelscalefactor); dot((0.6380284948350587,1.353053551156462),linewidth(4.pt) + dotstyle); label("$C_1$", (0.2,1.02), NE * labelscalefactor); dot((-3.5571674061519136,1.864861893765503),linewidth(4.pt) + dotstyle); label("$G$", (-3.48,2.02), NE * labelscalefactor); dot((3.5074205786014643,0.),linewidth(4.pt) + dotstyle); label("$V$", (3.38,-0.4), NE * labelscalefactor); dot((2.7,3.3658633371878666),linewidth(4.pt) + dotstyle); label("$N$", (2.78,3.52), NE * labelscalefactor); dot((0.8894305730917524,2.690894150918487),linewidth(4.pt) + dotstyle); label("$Z$", (0.36,2.5), NE * labelscalefactor); dot((0.8894305730917507,-1.4908941509184854),linewidth(4.pt) + dotstyle); label("$T$", (0.46,-1.78), NE * labelscalefactor); dot((3.9547732243966385,-1.864861893765504),linewidth(4.pt) + dotstyle); label("$A'$", (4.04,-1.7), NE * labelscalefactor); dot((-0.8322939923766871,1.5324309468827513),linewidth(4.pt) + dotstyle); label("$X$", (-0.98,1.74), NE * labelscalefactor); dot((5.4,0.),linewidth(4.pt) + dotstyle); label("$C$", (5.38,-0.36), NE * labelscalefactor); dot((2.1804415825316923,0.),linewidth(4.pt) + dotstyle); label("$K$", (2.2,-0.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]](http://latex.artofproblemsolving.com/c/d/f/cdf5b2c67e5f87e85aa1885f81dc4a200e60bac9.png)
.
.
and 
.
meet 
.
and 
are collinear. Since 
and 
,
is a trapezoid with 
.
.
and 
are isogonal in angle 
.
.
.
be the reflection of 
is a rectangle, we have 
,
coincides with 
.
,
.
,
and 
are bisectors of angle 
.
,
is tangent to 
.
be the 
.
and 
concur at the circumcenter 
.
,
.
.
.
.
is cyclic.
is the 
and 
.
as required. 
and 
and thus 
.
and 
implying that 
is a harmonic quadrilateral, so the internal bisectors of 
and 
meet at a point 
on 
we conclude that 
and 
to happen. Hereby, 
and 
bisects 
.
.
so 
and 
is cyclic. Further, 
,
is inscribed. We infer that 
is a cyclic pentagon, implying that 
,
,
thus, 
is tangent to 
.
the second intersection of 
onto 
meet at some point. Recall that if 
are isogonal conjugates WRT 
to the sides of the triangle are concyclic. Conversely, taking the limit case, The perpendiculars raised from 
in 
the three center of the excircles, noticing that 
,
are collinear. Now, from quadrilateral 
,
bisect 
but 
,
is isosceles, and also 
so 
is the Bevan point of 
.
the midpoint of 
in circle 
is a parallelogram. To prove 
collinear, it suffices to prove 
,
(which i think is a nice way to characterise such triangles). To prove this, using Pappus for 
and 
it follows that 
are collinear. Using now a theorem(whose name i do not know in English) it follows that 
,
,
and 
it follows that 
.
is right and 
so 
bisects both 
and 
,
is cyclic. This means 
,
are cyclic as well. We have that 
so 
hence 
.
it follows that 
is cyclic. Doing some angle chasing, the desired tangency is equivalent to 
,
(, so, even more 
,
and it is enough to prove that 
is symmedian in 
,
.
,
,
;
;
;
;
.![[asy]
size(10cm);
defaultpen(fontsize(10pt));
pen pri=royalblue+linewidth(0.5);
pen sec=rgb(41, 207, 255)+linewidth(0.5);
pen tri=springgreen+linewidth(0.5);
pen qua=chartreuse+linewidth(0.5);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pen qfil=invisible;
pair B, C, L, I, A, A1, B1, C1, EE, F, J, D, M, K, IB, IC;
B=dir(190); C=dir(350); L=dir(270);
I=intersectionpoint(arc(L, length(B-L), 90, 180, CCW), (-B) -- (-C));
A=intersectionpoint(I -- (I+100*(I-L)), circle((0, 0), 1));
A1=B+C-foot(I, B, C);
B1=foot(A1, A, C);
C1=foot(A1, A, B);
EE=foot(I, A, C);
F=foot(I, A, B);
J=intersectionpoints(circumcircle(A, B, C), circumcircle(A, EE, F))[1];
D=extension(A, J, B, C);
M=(A+D)/2;
K=dir(90);
IB=extension(B, I, A, K);
IC=extension(C, I, A, K);
draw(A -- B -- C -- A, pri);
filldraw(circumcircle(A, B, C), fil, pri);
draw(IB -- A1 -- IC -- IB, sec);
filldraw(circumcircle(A, B1, C1), tfil, tri);
draw(B -- D -- A, pri);
filldraw(circumcircle(A, EE, F), sfil, sec);
filldraw(circumcircle(B, I, C), qfil, qua);
draw(B -- IB, tri+dashed); draw(C -- IC, tri+dashed);
draw(M -- B1, pri);
draw(D -- I -- J, sec); draw(A -- L, pri);
clip((L + (100, -1/8)) -- (L+(-100, -1/8)) -- (-100, 100) -- (100, 100) -- cycle);
dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$L$", L, S);
dot("$I$", I, NE);
dot("$A_1$", A1, SE);
dot("$B_1$", B1, NE);
dot("$C_1$", C1, SW);
dot("$E$", EE, NE);
dot("$F$", F, NW);
dot("$J$", J, dir(150));
dot("$D$", D, SW);
dot("$M$", M, dir(195));
dot("$K$", K, N);
dot("$I_B$", IB, NE);
dot("$I_C$", IC, W);
[/asy]](http://latex.artofproblemsolving.com/b/f/6/bf62377a3f80a2433e15d25f98e300e0da171364.png)
.
,
,
,
,
,
,
and also 
,
.
so 
whence 
.
,
where 
,
.
collinear, so we have 
.
,
,
,
.![\[\begin{vmatrix}0 & y & z \\ a & b & -c \\ s-a & s-b & 0\end{vmatrix}=0,\]](http://latex.artofproblemsolving.com/b/5/5/b551e72b2255b30aadb4f7f736058f7898a82093.png)
which gives 
.
,![\[s^2(a-b)(a-c)=bc(s-a)^2.\]](http://latex.artofproblemsolving.com/8/2/8/828a4820280c3d82fcf4d13ec51b3cb02858b1a8.png)
We see that![\begin{align*}
s^2(a-b)(a-c)=bc(s-a)^2 &\iff (a+b+c)^2(a-b)(a-c) = bc(s-a)^2 \\
&\iff bc\left[(a+b+c)^2-(b+c-a)^2\right]+a^2(a+b+c)^2 = a(b+c)(a+b+c)^2 \\
&\iff bc[2(b+c)][2a]=a(a+b+c)^2(b+c-a) \\
&\iff 4bc(b+c)=(a+b+c)^2(b+c-a) \\
&\iff \boxed{bc(b+c)=2s^2(s-a)}.
\end{align*}](http://latex.artofproblemsolving.com/0/5/6/056b8274d328eacc5a47c4d08ab0049d4781db34.png)
We now compute 
.
,
such that 
.
is the external angle bisector which is parameterized by 
(an easy way to see this is that 
is on it), so 
.![\[[AKI]=[AKD]\implies \frac{1}{(b-c)(a+b+c)}\begin{vmatrix}1 & 0 & 0\\ 0 & b & -c \\ a & b & c\end{vmatrix}=\frac{1}{b-c}\begin{vmatrix}1 & 0 & 0\\ 0 & b & -c \\ 0 & y & z\end{vmatrix}.\]](http://latex.artofproblemsolving.com/4/b/5/4b581bea6ba2a96b0ce6d79444ac2477cefec115.png)
This implies 
,
can be solved to give![\[\boxed{D=\left(0,\frac{b(s-c)}{b-c},-\frac{c(s-b)}{b-c}\right)=(0:b(s-c):-c(s-b))}.\]](http://latex.artofproblemsolving.com/7/0/7/707be4bdb9b2707566e428b7abe5fb28c88fae0c.png)
Let ![\[E=(s(b-c):b(s-c):-c(s-b)).\]](http://latex.artofproblemsolving.com/6/a/0/6a0dc56d0500a44fce5c8705390d64e0b14d894b.png)
The first big claim is that 
are collinear. This is equivalent to![\[\begin{vmatrix}s(b-c) & b(s-c) & -c(s-b) \\ s-a & s-b & 0 \\ s-a & 0 & s-c\end{vmatrix}=0,\]](http://latex.artofproblemsolving.com/1/c/f/1cfe3cd54f69e97921733b85cc4be73e81d9a50d.png)
which is equivalent to![\[\Gamma:=s(b-c)(s-b)(s-c)-(s-a)b(s-c)^2+(s-a)c(s-b)^2=0.\]](http://latex.artofproblemsolving.com/e/d/2/ed28ee277ce473c53abf2b1ce8184438e997f0a9.png)
We have![\begin{align*}
\Gamma=0 &\iff s(b-c)(s-b)(s-c)=(s-a)[b(s-c)^2-c(s-b)^2] \\
&\iff s(b-c)(s-b)(s-c)=(s-a)(b-c)(s^2-bc) \\
&\iff s(s-b)(s-c)=(s-a)(s^2-bc) \\
&\iff s^3-s^2(b+c)+sbc=s^3-sbc-as^2+abc \\
&\iff bc(2s-a)=s^2(b+c-a) \\
&\iff bc(b+c)=2s^2(s-a),
\end{align*}](http://latex.artofproblemsolving.com/8/c/c/8ccce0ff7e6f31fb39c074b7a374d3059230d45d.png)
which as we saw before, is directly equivalent to 
are collinear.
inversion, it is quite susceptible to barycentric coordinates. Let 
be on 
are isogonal. Given that 
,![\[D'=(0:b(s-c):-c(s-c)),\]](http://latex.artofproblemsolving.com/1/4/4/144e531dd7b754f5e882e89d0281b7bebd771fa8.png)
recalling that 
.
be the 
be the ![\[AB_1\cdot AB_1'=bc,\]](http://latex.artofproblemsolving.com/0/4/e/04ec6d4a3eeaa2170d1713b7ce7ded955518a402.png)
so 
.![\[BB_1'=c-\frac{bc}{s-c}=c\frac{s-b-c}{s-c}=-\frac{c(s-a)}{s-c},\]](http://latex.artofproblemsolving.com/1/a/9/1a9c1cf51375a039cd11c188ce86e3def2196068.png)
so![\[B_1'=(BB_1':AA_1':0)=(a-s:b:0),\]](http://latex.artofproblemsolving.com/9/1/5/915b019144688dc7c8ecab443add7d8d99dbe583.png)
and similarly 
.
and 
.
under the ![\[[AD'B_1']=[AD'C_1'].\]](http://latex.artofproblemsolving.com/b/c/a/bca1647793177cb1c58b5ed6e06cf0198a7c15a6.png)
This is equivalent to![\[\frac{1}{c-s}\begin{vmatrix}1 & 0 & 0\\ 0 & b(s-b) & -c(s-c) \\ a-s & b & 0\end{vmatrix}=\frac{1}{b-s}\begin{vmatrix}1 & 0 & 0\\ 0 & b(s-b) & -c(s-c) \\ a-s & 0 & c\end{vmatrix},\]](http://latex.artofproblemsolving.com/0/0/2/0026d9beca6b455bc386588b6c20b8e04cbd89ab.png)
which is immediate upon expansion as they are both equal to 
.
collinear where ![\[EA^2=EB_1\cdot EC_1,\]](http://latex.artofproblemsolving.com/d/3/2/d32b464085c054d18050d79059ee4a9dfdec7cfe.png)
so 
as 
.
tangent to 
you'll get some practice in June anyways.
and 
be the reflection of 
similarly. Let the radical axis of 
and 
meet 
are concurrent, prove that the circumcircle of 
is tangent to 
.
at 
bisects 
.
are collinear.
.
and 
are collinear. This implies the desired conclusion.
,
,
.
.
,
.
Proof of Claim 3. Let 
are collinear, so 
.
,
.
)
.
.
is cyclic.
sending 
,
.
and the result follows. 
are collinear.
.
.![\[\measuredangle AM_AT = \measuredangle TAI + \measuredangle IAC + \measuredangle CBA = \measuredangle IAA_1 + 90^\circ + \measuredangle EAI = \measuredangle AM_AE,\]](http://latex.artofproblemsolving.com/e/d/b/edb33830ef802fb7ca9c6c7dee450d33031bd0ef.png)
as desired. 
,
,
and 
,
.
isogonal in 
and so 
is tangent to ![[asy]
size(12cm);
defaultpen(fontsize(10pt));
pair A = dir(105.3); pair B = dir(193.3); pair C = dir(346.7);
pair I = incenter(A,B,C);
pair A1 = B+C-foot(I,B,C);
pair B1 = A+C-foot(I,A,C);
pair C1 = A+B-foot(I,A,B);
pair D = extension(I,I+(A-I)*dir(90), B, C);
pair Ma = intersectionpoints(circumcircle(A1,B1,C1),circumcircle(A,B,C))[0];
pair Ic = extension(A,Ma,A1,C1); pair Ib = extension(A,Ma,A1,B1);
pair E = foot(A,B,C);
pair O = (A+D)/2;
pair X = 2O-I;
pair T = intersectionpoints(Ma--3I-2Ma, circumcircle(A,B,C))[0];
filldraw(A--B--C--cycle,orange+white+white,orange);
filldraw(anglemark(I,E,A),orange,orange);
filldraw(anglemark(I,A,A1),orange,orange);
filldraw(anglemark(I,X,A),orange,orange);
draw(circumcircle(A,B,C), red);
draw(circumcircle(A1,B1,C1), red);
draw(circle((A+D)/2,abs(A-D)/2), red);
draw(X--Ib, red+white);
draw(D--B, orange);
draw(X--B1, red+white+dashed);
draw(D--I--A,red);
draw(Ma--T, red+white);
draw(Ic--A1--Ib, red+white+dashed);
draw(A--A1, red+white);
draw(A--T, red+white);
draw(A--E, red+white);
dot("$A$", A, dir(135));
dot("$B$", B, dir(225));
dot("$C$", C, dir(-45));
dot("$I$", I, dir(45));
dot("$A_1$", A1, dir(270));
dot("$B_1$", B1, dir(0));
dot("$C_1$", C1, dir(180));
dot("$D$", D, dir(180));
dot("$M_A$", Ma, dir(90));
dot("$I_B$", Ib, dir(90));
dot("$I_C$", Ic, dir(90));
dot("$E$", E, dir(-90));
dot("$O$", O, dir(90));
dot("$X$", X, dir(135));
dot("$T$", T, dir(-90));
[/asy]](http://latex.artofproblemsolving.com/7/9/1/791bc6f0a6e9b9e63521b9d49596a2c028900c58.png)
and let 
.
concur at 
so 
Now 
and 
are collinear, so by Pappus 
is collinear.
be an altitude in 
at 
Note that 
so 
This implies that 
is the Apollonian circle of 
and we conclude that 
and 
.
,
.
is cyclic.
.
.
hence 
are collinear. Hence 
intersect on the antipode of 
where 
.
and since 
and 
,
.
hence 
,
is the diameter, as desired 
.
therefore 
.
is cyclic to a circle with diameter 
,
therefore 
is cyclic. This combined with the fact that 
readily implies that 
.
,
is cyclic. Now, it's time for a second angle-chase: 
therefore 
,
.
so 
.
,
,
is tangent to the circumcircle of 
,
and 
,
.
by 
coincides with 
,![\[\measuredangle B_1OC_1=\measuredangle B_1A_1C_1=\measuredangle B_1OC_1-(\measuredangle OBC+\measuredangle BCO)=\measuredangle B_1OC_1+\measuredangle BOC \Rightarrow \measuredangle BOC=0.\]](http://latex.artofproblemsolving.com/4/a/b/4ab52b3287eb82dde1baa9f743229dbf3d8a8a63.png)
Therefore, 
and 
are perpendicular to 
.
and 
.![\[ (A,P;B_{1},C_{1}) \stackrel{A_1}{=} (A,Q,I_{B},I_{C})=-1\]](http://latex.artofproblemsolving.com/7/f/1/7f19dd306634d041b7154f41fcb871f763e5c45b.png)
Now notice that 
,
in 
,![\[(A,P';B_{1},C_{1})\stackrel{I}{=}(R,M;B_{1},C_{1})=-1\]](http://latex.artofproblemsolving.com/2/b/e/2be9883fe34fcbf9b2345229a408bcff708a6d9a.png)
which implies that 
as desired.
and 
are concyclic. Therefore by spiral similarity, 
.
,![\[NB_1\cdot NC_1=NA^2=ND^2\]](http://latex.artofproblemsolving.com/7/4/4/744495b45faf67e6a7253ea0bf7915dca8fb6374.png)
and it follows that 
.
bisects ![[asy]
size(10cm);
defaultpen(fontsize(10pt)+linewidth(0.4));
dotfactor *= 1.2;
pair A = dir(116.8), B = dir(192.4), C = dir(347.6), I = incenter(A,B,C), Ia = 2*circumcenter(B,I,C)-I, Ib = 2*circumcenter(C,I,A)-I, Ic = 2*circumcenter(A,I,B)-I, A1 = foot(Ia,B,C), B1 = foot(Ib,C,A), C1 = foot(Ic,A,B), L = dir(90), F = foot(A,B,C), D = extension(B,C,I,I+dir(A-L)), M = (A+D)/2, T = F+dir(L--F)*abs(B-F)*abs(C-F)/abs(L-F), S = A1+dir(A--A1)*abs(B-A1)*abs(C-A1)/abs(A-A1), I1 = 2M-I;
draw(A--B--C--A, linewidth(0.8)+heavyred);
draw(unitcircle, heavyred);
draw(A--I--D--B^^F--A--D, heavyred);
draw(circumcircle(A1,B1,C1), purple);
draw(Ia--Ib--Ic--Ia, lightgreen);
draw(B--Ib--A1--Ic--C, lightgreen);
draw(circumcircle(A,D,F), heavyblue);
draw(B1--M, orange+linewidth(0.8));
draw(M--extension(A,L,B1,C1)--A, orange+dashed);
draw(L--T--S--A, magenta);
draw(arc(circumcenter(D,B1,C1),circumradius(D,B1,C1),50,135), cyan);
dot("$A$", A, dir(80));
dot("$B$", B, dir(225));
dot("$C$", C, dir(330));
dot("$A_1$", A1, dir(270));
dot("$B_1$", B1, dir(10));
dot("$C_1$", C1, dir(130));
dot("$D$", D, dir(175));
dot("$F$", F, dir(300));
dot("$L$", L, dir(90));
dot("$I_A$", Ia, dir(270));
dot("$I_B$", Ib, dir(45));
dot("$I_C$", Ic, dir(105));
dot("$M$", M, dir(115));
dot("$I$", I, dir(310));
dot("$T$", T, dir(225));
dot("$S$", S, dir(315));
dot("$I'$", I1, dir(180));
[/asy]](http://latex.artofproblemsolving.com/5/5/0/5500d6b3918e111b5c69a2a90253a1f5f7613149.png)
.
,
and 
as desired. 
and 
.
.
as desired. 
be the Bevan point (center of 
)
,
lies on 
,
,
is cyclic with diameter 
is the 
are angle bisectors. Therefore 
where 
and the result follows. 
,
so config cases imply 
as in IMO 2013/3.
.
.
must lie on both 
,
must lie on both 
so the three circles cannot have a common chord and so 
and 
,
intersect 
.
.
.
,
collinear. Let 
collinear. But this is clear: wrt 
)
be 
is 
,![\[\measuredangle B_1I''M = \measuredangle C_1B_1M+\measuredangle B_1MA=\measuredangle MC_1B_1+\measuredangle B_1C_1A=\measuredangle MC_1A, \]](http://latex.artofproblemsolving.com/1/a/e/1ae656a945358559925d0db4aaf967c3f4e96e0e.png)
![\[\measuredangle II'A=\measuredangle IEA = \measuredangle MEA=\measuredangle MC_1A.\]](http://latex.artofproblemsolving.com/b/c/b/bcb98528ed38ffc96e365c5f47a16dce45fb69bf.png)
Thus 
,
excentral and 
to be foot from 
is cyclic with diameter 
which means 
,
and 
cyclic with diameter 
and 
,
in 
collinear, which would show by angle bisection that![\[\frac{AB_1}{AC_1} = \frac{IB_1}{IC_1} = \frac{A'B_1}{A'C_1}\]](http://latex.artofproblemsolving.com/3/1/e/31e14b7343b03a04a163cdfaa521383e057cef86.png)
and thus 
is harmonic. Let 
are isogonal. It follows that![\[\angle MA'A = \angle MAA_1 = \angle MAX = \angle MTX\]](http://latex.artofproblemsolving.com/0/f/2/0f2617f4f9a3da23498ade9fa149726d49fd0520.png)
but since 
isogonal it follows that 
hence 
means that 
as desired. It follows that 
,
are both tangent to 
.
so indeed 
respectively. Let 
and 
,
is cyclic quadrilateral. Since 
is cyclic, we get that 
hence 
is cyclic quadrilateral. Also, 
is the power of 
is tangent to 
and 
,
.
.
and 
,
passes through the midpoint of ![[asy]import geometry;
size(10cm);defaultpen(fontsize(10pt));
pair O,A,B,C,I,A1,B1,C1,Ia,Ib,Ic,D,M,K;
O=(0,0);A=dir(110);B=dir(193.3);C=dir(347.2);I=incenter(A,B,C);A1=2midpoint(B--C)-foot(I,B,C);
B1=2midpoint(A--C)-foot(I,A,C);C1=2midpoint(A--B)-foot(I,A,B);Ia=intersectionpoint(line(A,I),perpendicular(A1,line(A1,B)));
Ib=intersectionpoint(line(B,I),perpendicular(B1,line(B1,C)));
Ic=intersectionpoint(line(C,I),perpendicular(C1,line(C1,A)));
path w=circumcircle(A,B,C);path o=circumcircle(A1,B1,C1);D=intersectionpoint(perpendicular(I,line(A,I)),line(B,C));
M=intersectionpoints(w,A--Ia)[1];K=extension(B,C,A,I);
draw(A--B--C--cycle,red+1);draw(w,heavyblue+1);draw(o,cyan+1);draw(Ia--Ib--Ic--cycle,orange);
draw(Ia--A1,heavygreen);draw(Ib--A1,heavygreen);draw(Ic--A1,heavygreen);
draw(B--Ib,magenta);draw(A--Ia,magenta);draw(C--Ic,magenta);
draw(D--B,red+0.3);draw(A--D,red+0.3);draw(I--D,red+0.3);
draw(B1--C1,dashed+red+0.3);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$A_1$",A1,dir(A1));
dot("$B_1$",B1,dir(B1));
dot("$C_1$",C1,dir(C1));
dot("$I$",I,dir(I));
dot("$I_A$",Ia,dir(Ia));
dot("$I_B$",Ib,dir(Ib));
dot("$I_C$",Ic,dir(Ic));
dot("$D$",D,dir(D));
dot("$M$",M,dir(M));
dot("$K$",K,dir(K));
[/asy]](http://latex.artofproblemsolving.com/a/5/4/a54d73272dda2f15e34b398fe7db942af30b8306.png)
.![\[CB_1 \left(\frac{CA}{CA_1} + \frac{BA}{BA_1}\right) = BC\]](http://latex.artofproblemsolving.com/6/6/8/668148d1c593455801400edc969473a7ea6e6aa5.png)
![\[(s-a)\cdot \frac{s(b+c) - 2bc}{(s+b)(s-c)} = a \Rightarrow s^2 (b+c-a) - s(2bc) + abc = 0\]](http://latex.artofproblemsolving.com/e/b/7/eb79cb034ca545568c2075a86508190b962c713d.png)
![\[s \left(\frac{(b+c)^2 - a^2}{2} - 2bc\right) + abc = 0 \Rightarrow s(a-b+c)(a+b-c) = 2abc\]](http://latex.artofproblemsolving.com/4/e/a/4eadc17381010920e96eb2665dea4b05840cc15d.png)
![\[\frac{K^2}{s-a} = \frac{abc}{2} = K\cdot r_a \Rightarrow r_a = \frac{a}{\sin A} = 2R\]](http://latex.artofproblemsolving.com/f/e/1/fe14955629cd5b08738abd6694effb47bae9c03a.png)
Therefore, 
.
,
so 
.
collinear. Next,![\[\angle ANX = \angle AA_1 B = 180 - \angle BAA_1 -\angle B = 180 - \angle B - \angle TAC= 180 - \angle TAAN = \angle ANT\]](http://latex.artofproblemsolving.com/6/e/7/6e795d01eb6ce947c76343d8519d5d9d7daf1b72.png)
Therefore, 
are collinear, so 
are collinear. Observe that 
,
is cyclic. We can also angle chase to get![\[\angle ADB = \angle AIN = 90 - \angle ANI = 90 - \angle AA_1 X = \angle XAA_1\]](http://latex.artofproblemsolving.com/f/7/0/f70fbad36246b94d1ac3847b9078c611f98ab539.png)
This means 
,
are collinear. However, 
(since 
is external bisector of 
,
,
collinear, this means 
are also collinear.
are collinear, then we're done, since 
,
with 
,
,
so 
and 
,![\[\angle OIA = 180 - \angle OAI - \angle AOI = 180 - \frac{1}{2}A - \angle AB_1C_1 - 2\angle AXN\]](http://latex.artofproblemsolving.com/0/9/0/090bda47f3f38cfb3fd099c26b95cf5bdd1c6baf.png)
![\[ = 180 - \frac{1}{2}A - \angle AB_1C_1 - \angle AC_1B_1 + \angle AB_1C_1 = 180 - \frac{1}{2}A - \angle AC_1B_1 = \angle C_1IA\]](http://latex.artofproblemsolving.com/5/6/c/56c94ea7f41b8afa2d3da3db996af1b8bede6c97.png)
Therefore, 
are coillinear, so 
are collinear (since 
)
![[asy]
size(12cm); defaultpen(fontsize(6pt)); dotfactor*=0.75;
pair A=dir(111.32); pair B=dir(192.94); pair C=dir(-13.14); pair I=incenter(A,B,C); pair P=foot(I,B,C); pair E=foot(I,A,C); pair F=foot(I,A,B); pair Q=extension(A,I,B,C); pair A_1=B+C-P; pair B_1=A+C-E; pair C_1=A+B-F; pair K=intersectionpoints(circumcircle(A,F,E),circumcircle(A,B,C))[1]; pair M=intersectionpoints((I--(10*Q-9*I)),unitcircle)[0]; pair D=extension(A,K,B,C); pair N=(A+D)/2; pair I_A=2*M-I; pair I_B=2*intersectionpoints((I--(10*(extension(B,I,A,C))-9*I)),unitcircle)[0]-I; pair I_C=2*intersectionpoints((I--(10*(extension(C,I,A,B))-9*I)),unitcircle)[0]-I;
draw(10*A-9*B--10*B-9*A); draw(10*C-9*B--10*B-9*C); draw(10*A-9*C--10*C-9*A); draw(10*I_A-9*I_B--10*I_B-9*I_A); draw(10*I_C-9*I_B--10*I_B-9*I_C); draw(10*I_A-9*I_C--10*I_C-9*I_A); draw(unitcircle); draw(circumcircle(A,A_1,B_1),red); draw(circumcircle(A,F,E),darkgreen); draw(circumcircle(P,F,E)); draw(circumcircle(I,P,Q),linewidth(0.4)+gray); draw(circumcircle(foot(I_B,A,C),foot(I_B,B,C),foot(I_B,A,B)),linewidth(0.4)+orange); draw(circumcircle(foot(I_A,A,C),foot(I_A,B,C),foot(I_A,A,B)),linewidth(0.4)+orange); draw(circumcircle(foot(I_C,A,C),foot(I_C,B,C),foot(I_C,A,B)),linewidth(0.4)+orange); draw(circumcircle(I_A,I_B,I_C),blue); draw(circumcircle(B,I,C),blue); draw(circumcircle(D,C_1,B_1),red); draw(10*A-9*D--10*D-9*A,red); draw(10*B_1-9*C_1--10*C_1-9*B_1); draw(10*I-9*D--10*D-9*I,red); draw(B--K--C^^F--K--E,darkgreen); draw(A--A_1); draw(10*A-9*I_A--10*I_A-9*A^^10*B-9*I_B--10*I_B-9*B^^10*C-9*I_C--10*I_C-9*C^^10*A_1-9*I_A--10*I_A-9*A_1^^10*B_1-9*I_B--10*I_B-9*B_1^^10*C_1-9*I_C--10*I_C-9*C_1,linewidth(0.4)+gray);
dot("$A$",A,dir(105)*3); dot("$B$",B,dir(210)*3); dot("$C$",C,dir(-20)*4); dot("$I$",I,dir(-90)*1.5); dot("$P$",P,dir(-100)*3); dot("$E$",E,dir(30)*1.5); dot("$F$",F,dir(F)*4.5); dot("$Q$",Q,dir(-80)*3.5); dot("$A_1$",A_1,dir(-90)*3); dot("$B_1$",B_1,dir(40)*2.5); dot("$C_1$",C_1,dir(-150)*4); dot("$K$",K,dir(90)*3); dot("$M$",M,dir(-90)*1.5); dot("$D$",D,dir(-80)*2); dot("$N$",N,dir(100)*1.5); dot("$I_A$",I_A,dir(I_A)*4); dot("$I_B$",I_B,dir(I_B)*3); dot("$I_C$",I_C,dir(I_C)*3);
clip((2.7,2.2)--(2.7,-2.7)--(-3,-2.7)--(-3,2.2)--cycle);
[/asy]](http://latex.artofproblemsolving.com/3/5/c/35c6df0ef9b04d43d3d7af13f6f558f1fd8a7bfc.png)
,
,
concur at ![[asy]
size(6cm); defaultpen(fontsize(8pt)); dotfactor*=0.75;
pair A=dir(130); pair B=dir(200); pair C=dir(340); pair P=(-0.3,0.1); pair D=foot(P,B,C); pair E=foot(P,A,C); pair F=foot(P,A,B); pair Q=extension(A,foot(A,E,F),B,foot(B,D,F)); pair X=extension(A,Q,F,E); pair Y=extension(A,Q,F,P);
draw(A--B--C--cycle); draw(D--E--F--cycle); draw(A--Q ^^ B--Q ^^ C--Q,linewidth(0.4)); draw(F--P ^^ E--P ^^ D--P,linewidth(0.4)); draw(A--P^^C--P^^B--P,linewidth(0.4)); draw(circumcircle(A,E,F));
dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(90)); dot("$F$",F,dir(180)); dot("$E$",E,dir(2)); dot("$D$",D,dir(-90)); dot("$Y$",Y,dir(30)); dot("$X$",X,dir(X)); dot("$Q$",Q,dir(-90));
[/asy]](http://latex.artofproblemsolving.com/1/5/5/155cafc5d2f45ff4551c110d81b8b8f8c6836b6a.png)
lie on the same side on the angle bisector of 
and 
.
,![\[\measuredangle QAF=\measuredangle EFP=\measuredangle EAP\]](http://latex.artofproblemsolving.com/d/5/5/d55407b914b78e11a551194c0e65a37e0afed206.png)
so 
are isogonal. By symmetry, 
are also isogonal and 
are too.
.
.
also lies on 
.
are collinear by radical axis on 
and 
.
bisects 
,
.![\[\frac{KF}{KE}=\frac{KB}{KC}=\frac{BP}{CP}=\frac{AC_1}{AB_1}\]](http://latex.artofproblemsolving.com/b/c/6/bc65ecb87e12582206da361356b041519a7995da.png)
by angle bisector theorem, and from 
the conclusion follows.
.
.
.![\[\measuredangle NAI=\measuredangle KAM=\measuredangle KBM=\measuredangle KBC+\measuredangle CBM=\measuredangle AC_1I+\measuredangle IAC_1=\measuredangle AIN.\]](http://latex.artofproblemsolving.com/c/7/4/c74cdb44849d12fda97b72e4f10159fb084d9cd3.png)
,
,
is tangent to 
;
which implies the conclusion.
be the 
we get 
hence 
hence 
and since by more orthic duality we have 
and 
we get 
hence 
passing through 
to get the desired result.
cyclic.
(we must take in count Claim 2)
Hence 
is cyclic becuase its the apolonian circle of 
we have 
and 
we get by the previous ratios that 
hence we get 
and now since 
we get 
(Claim 3 goes brrr) and becuase 
we can get 
tangent to 
thus we are done 
and 
is cyclic.
,
,
,
.![\[\measuredangle B_1AC_1=\measuredangle CNB=\measuredangle B_1NC_1,\]](http://latex.artofproblemsolving.com/0/0/5/0059de03aee05be98dd1f5a0da1fa7366442339c.png)
so 
is cyclic. Let 
,
,
are concurrent. Also notice that 
.
,
is cyclic and the claim is proven.
to prove that 
.
is tangent to 
.
.
.
.
.
.![\[MB_1 \cdot MC_1=MA^2=MD^2,\]](http://latex.artofproblemsolving.com/7/8/e/78ef80603b7259733a1d12ec705f0fb491411f85.png)
so 
.
.
is the altitude from 
using LoC yields the claim is equivalent to
as desired.
so 
is collinear.
.
and 
is spiral similar to 
through 
by ratio lemma in 
is concyclic and thus an Appolonius circle of 
.
are harmonic conjugates wrt 
via cross ratio, so if 
then 
by ratio lemma in 
,
is tangent to 
as desired.
and 
and note 
.
and let 
and 
harmonic. Now we need to show that 
and Menelaus on 
to show 
collinear.
is obviously an Apollonian circle. From here we can let 
then we can easily how that 
is the 
is harmonic.
hence by power of point we get that.
Hence we are done.
where ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */
pen dotstyle = blue; /* point style */
real xmin = -21, xmax = 18, ymin = -11, ymax = 11; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0);
draw((-5.00130207163223,8.65950215591446)--(-9.152407121756008,-4.029074816583814)--(9.101968524336803,-4.141759165135286)--cycle, linewidth(0.5) + ccqqqq);
/* draw figures */
draw(circle((0,0), 10), linewidth(0.5) + qqwuqq);
draw((-5.00130207163223,8.65950215591446)--(-9.152407121756008,-4.029074816583814), linewidth(0.5) + ccqqqq);
draw((-9.152407121756008,-4.029074816583814)--(9.101968524336803,-4.141759165135286), linewidth(0.5) + ccqqqq);
draw((9.101968524336803,-4.141759165135286)--(-5.00130207163223,8.65950215591446), linewidth(0.5) + ccqqqq);
draw((-19.759779416499267,-3.9635954564539646)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq);
draw((-19.759779416499267,-3.9635954564539646)--(-9.152407121756008,-4.029074816583814), linewidth(0.5) + ffvvqq);
draw((-19.759779416499267,-3.9635954564539646)--(-5.00130207163223,8.65950215591446), linewidth(0.5) + ffvvqq);
draw(circle((-3.9231657357373417,4.586826023931432), 4.212964376160934), linewidth(0.5) + qqwuqq);
draw(circle((-1.0781363358948877,4.072676131983027), 6.035743729142097), linewidth(0.5) + qqwuqq);
draw(circle((-0.061728871928146724,-9.999809475503547), 10.876125385867832), linewidth(0.5) + qqwuqq);
draw(circle((-3.1536888609567755,-2.271612354266112), 6.2508635466551326), linewidth(0.5) + qqwuqq);
draw(circle((5.97349896208963,-2.3279545285418446), 3.616242395309652), linewidth(0.5) + qqwuqq);
draw((-11.056367069694842,3.4805345203159774)--(3.8660813226853814,0.6107526625138946), linewidth(0.5) + ffvvqq);
draw((-6.953736677062792,2.6915450756578996)--(2.8228754214215614,-4.10299829324825), linewidth(0.5) + ffvvqq);
draw((2.8228754214215614,-4.10299829324825)--(3.8660813226853814,0.6107526625138946), linewidth(0.5) + ffvvqq);
draw((-5.00130207163223,8.65950215591446)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq);
draw((-7.77946021700683,6.283311128060428)--(-7.199972516325446,1.9388822636727459), linewidth(0.5) + ffvvqq);
draw((-7.199972516325446,1.9388822636727459)--(0.23458513001919146,3.906990328265279), linewidth(0.5) + ffvvqq);
draw((0.23458513001919146,3.906990328265279)--(-7.77946021700683,6.283311128060428), linewidth(0.5) + ffvvqq);
draw((-7.77946021700683,6.283311128060428)--(-2.845029399842456,0.5141498919484042), linewidth(0.5) + ffvvqq);
/* dots and labels */
dot((0,0),dotstyle);
label("$O$", (0.12817675840733747,0.2926007430266706), NE * labelscalefactor);
dot((-5.00130207163223,8.65950215591446),dotstyle);
label("$A$", (-4.89166511339295,8.939495990405241), NE * labelscalefactor);
dot((-9.152407121756008,-4.029074816583814),dotstyle);
label("$B$", (-9.041014175054459,-3.7406826106163535), NE * labelscalefactor);
dot((9.101968524336803,-4.141759165135286),dotstyle);
label("$C$", (9.210318410855255,-3.856748318634858), NE * labelscalefactor);
dot((-2.845029399842456,0.5141498919484042),dotstyle);
label("$I$", (-2.7154330880460043,0.81489642910994), NE * labelscalefactor);
dot((2.8228754214215614,-4.10299829324825),dotstyle);
label("$A_1$", (2.9427701778560533,-3.798715464625606), NE * labelscalefactor);
dot((3.8660813226853814,0.6107526625138946),dotstyle);
label("$B_1$", (3.987361550022587,0.9019457101238182), NE * labelscalefactor);
dot((-6.953736677062792,2.6915450756578996),dotstyle);
label("$C_1$", (-6.835765722702887,2.9911284544568955), NE * labelscalefactor);
dot((-19.759779416499267,-3.9635954564539646),dotstyle);
label("$D$", (-19.573977177733674,-3.6826497566071015), NE * labelscalefactor);
dot((-7.199972516325446,1.9388822636727459),dotstyle);
label("$C_2$", (-7.09691356574452,2.2367013523366177), NE * labelscalefactor);
dot((0.23458513001919146,3.906990328265279),dotstyle);
label("$B_2$", (0.360308174444345,4.209818388651191), NE * labelscalefactor);
dot((-2.873314018840767,-4.067835688470851),dotstyle);
label("$A_2$", (-2.74444951505063,-3.76969903762098), NE * labelscalefactor);
dot((-7.77946021700683,6.283311128060428),dotstyle);
label("$S$", (-7.677242105837039,6.560148976025903), NE * labelscalefactor);
dot((-0.06172887192814658,-9.999809475503545),dotstyle);
label("$M_a$", (0.04112747739345968,-9.718066573569326), NE * labelscalefactor);
dot((-11.056367069694842,3.4805345203159774),dotstyle);
label("$T$", (-10.927081930355143,3.7745719835817995), NE * labelscalefactor);
dot((2.845029399842456,-0.5141498919484042),dotstyle);
label("$V$", (2.971786604860679,-0.2296949430565987), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/c/8/1/c81e6806ddf9708b88cfb6baeb4de8f0ce44d620.png)
.
,
,
,
,
cyclic and 
then as 
is the diameter of 
.
by definition of the Bevan point, and hence 
,
,
due to the excentre-orthocentre duality so 
,
is the 
and 
being defined as per the diagram below:![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */
pen dotstyle = blue; /* point style */
real xmin = -21.65110060692449, xmax = 20.977842173753757, ymin = -11.055305559931867, ymax = 12.390612969441175; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0);
draw((-5.476119010842156,8.367324577132951)--(-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894)--cycle, linewidth(0.5) + ccqqqq);
/* draw figures */
draw(circle((0,0), 10), linewidth(0.5) + qqwuqq);
draw((-5.476119010842156,8.367324577132951)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ccqqqq);
draw((-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ccqqqq);
draw((9.786194275204696,-2.056794012036894)--(-5.476119010842156,8.367324577132951), linewidth(0.5) + ccqqqq);
draw((-3.595934733432582,2.0708437955383086)--(-3.5924066956343808,-2.0681976692155213), linewidth(0.5) + ffvvqq);
draw((-5.476119010842156,8.367324577132951)--(-3.595934733432582,2.0708437955383086), linewidth(0.5) + ffvvqq);
draw((-9.244471768878217,3.813101324960256)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ffvvqq);
draw((-9.78267371561353,-2.0734741314624983)--(5.640013234593206,8.257738837818346), linewidth(0.5) + ffvvqq);
draw((5.640013234593206,8.257738837818346)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq);
draw((-9.244471768878217,3.813101324960256)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq);
draw((-9.244471768878217,3.813101324960256)--(-5.476119010842156,8.367324577132951), linewidth(0.5) + ffvvqq);
draw((-5.476119010842156,8.367324577132951)--(5.486618265075023,-8.411225757322281), linewidth(0.5) + ffvvqq);
draw((-7.836531405897288,2.6447476274567396)--(3.595927255225547,-2.0620704742838716), linewidth(0.5) + ffvvqq);
draw((3.595927255225547,-2.0620704742838716)--(5.57158428578019,0.8217731824280499), linewidth(0.5) + ffvvqq);
/* dots and labels */
dot((0,0),dotstyle);
label("$O$", (0.12962484507830163,0.33461508928060524), NE * labelscalefactor);
dot((-5.476119010842156,8.367324577132951),dotstyle);
label("$A$", (-5.3322084486960994,8.693884337679233), NE * labelscalefactor);
dot((-9.78267371561353,-2.0734741314624983),dotstyle);
label("$B$", (-9.661710449858735,-1.7302243266584978), NE * labelscalefactor);
dot((9.786194275204696,-2.056794012036894),dotstyle);
label("$C$", (9.920960140015337,-1.7302243266584978), NE * labelscalefactor);
dot((-3.595934733432582,2.0708437955383086),dotstyle);
label("$I$", (-3.4671922020414256,2.3994545052197083), NE * labelscalefactor);
dot((3.595927255225547,-2.0620704742838716),dotstyle);
label("$A_1=V$", (3.726441892198029,-1.7302243266584978), NE * labelscalefactor);
dot((5.57158428578019,0.8217731824280499),dotstyle);
label("$B_1$", (5.691369723494917,1.1672116279657274), NE * labelscalefactor);
dot((-7.836531405897288,2.6447476274567396),dotstyle);
label("$C_1$", (-7.696782618561846,2.9656201515255916), NE * labelscalefactor);
dot((-3.5924066956343808,-2.0681976692155213),dotstyle);
label("$A_2$", (-3.4671922020414256,-1.7302243266584978), NE * labelscalefactor);
dot((0.008523800852425989,-9.999996367240291),dotstyle);
label("$M_a$", (0.12962484507830163,-9.656543374940862), NE * labelscalefactor);
dot((5.640013234593206,8.257738837818346),dotstyle);
label("$M_b$", (5.757977446589726,8.593972753037018), NE * labelscalefactor);
dot((-9.244471768878217,3.813101324960256),dotstyle);
label("$M_c$", (-9.095544803552851,4.131255305684762), NE * labelscalefactor);
dot((5.486618265075023,-8.411225757322281),dotstyle);
label("$U$", (5.624762000400107,-8.091261882212832), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/d/7/ad7ee14a0e82b308b23bfe7a17fddd8d8932a635.png)
works where 
is the antipode of 
.
;
I claim that with the condition 
,
.
,
.
,
,
,
,
,![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */
pen dotstyle = blue; /* point style */
real xmin = -23.259339754245026, xmax = 19.541271812197504, ymin = -10.447898799466639, ymax = 13.092437562076874; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0);
draw((-5.529490044046149,8.332150973955915)--(-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894)--cycle, linewidth(0.5) + ccqqqq);
/* draw figures */
draw(circle((0,0), 10), linewidth(0.5) + qqwuqq);
draw((-5.529490044046149,8.332150973955915)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ccqqqq);
draw((-9.78267371561353,-2.0734741314624983)--(9.786194275204696,-2.056794012036894), linewidth(0.5) + ccqqqq);
draw((9.786194275204696,-2.056794012036894)--(-5.529490044046149,8.332150973955915), linewidth(0.5) + ccqqqq);
draw((-17.336151248875133,-2.0799125676251085)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq);
draw((-17.336151248875133,-2.0799125676251085)--(-9.78267371561353,-2.0734741314624983), linewidth(0.5) + ffvvqq);
draw((-17.336151248875133,-2.0799125676251085)--(-5.529490044046149,8.332150973955915), linewidth(0.5) + ffvvqq);
draw(circle((-4.58199211135871,5.1957067190349475), 3.2764363104863987), linewidth(0.5) + qqwuqq);
draw(circle((-0.9474979326874364,3.1364442549209683), 6.927482949728449), linewidth(0.5) + qqwuqq);
draw(circle((0.008523800852427224,-9.999996367240291), 12.597511800380618), linewidth(0.5) + qqwuqq);
draw(circle((-3.074089768471129,-2.0663682977880526), 6.708587710444645), linewidth(0.5) + qqwuqq);
draw(circle((6.710344226938108,-2.058028238075522), 3.0758502958916925), linewidth(0.5) + qqwuqq);
draw((-11.432001519352662,3.12684157533726)--(5.574190950974938,0.8002948325823711), linewidth(0.5) + ffvvqq);
draw((-7.857008451664908,2.6377615833571366)--(3.634496544141483,-2.0620375985781543), linewidth(0.5) + ffvvqq);
draw((3.634496544141483,-2.0620375985781543)--(5.574190950974938,0.8002948325823711), linewidth(0.5) + ffvvqq);
draw((-5.529490044046149,8.332150973955915)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq);
draw((-7.575344228330761,6.527952192096369)--(-7.455155307994771,3.62091525913628), linewidth(0.5) + ffvvqq);
draw((-7.455155307994771,3.62091525913628)--(-1.3174867198163902,5.47506212933665), linewidth(0.5) + ffvvqq);
draw((-1.3174867198163902,5.47506212933665)--(-7.575344228330761,6.527952192096369), linewidth(0.5) + ffvvqq);
draw((-7.575344228330761,6.527952192096369)--(-3.6344941786712726,2.0592624641139796), linewidth(0.5) + ffvvqq);
draw((-5.5257089064128975,4.203787559149821)--(-3.6309759845503167,-2.0682305449212386), linewidth(0.5) + ffvvqq);
draw((-3.6344941786712726,2.0592624641139796)--(0.008523800852428653,-9.999996367240293), linewidth(0.5) + ffvvqq);
draw((-7.575344228330761,6.527952192096369)--(0.008523800852428653,-9.999996367240293), linewidth(0.5) + ffvvqq);
draw((-3.6344941786712726,2.0592624641139796)--(-7.455155307994771,3.62091525913628), linewidth(0.5) + ffvvqq);
draw((-3.6344941786712726,2.0592624641139796)--(-1.3174867198163902,5.47506212933665), linewidth(0.5) + ffvvqq);
draw((-3.6344941786712726,2.0592624641139796)--(3.634496544141483,-2.0620375985781543), linewidth(0.5) + ffvvqq);
/* dots and labels */
dot((0,0),dotstyle);
label("$O$", (0.1472446961532296,0.3191300477166152), NE * labelscalefactor);
dot((-5.529490044046149,8.332150973955915),dotstyle);
label("$A$", (-5.403459616369786,8.678624494287465), NE * labelscalefactor);
dot((-9.78267371561353,-2.0734741314624983),dotstyle);
label("$B$", (-9.650082795227755,-1.7540245750329557), NE * labelscalefactor);
dot((9.786194275204696,-2.056794012036894),dotstyle);
label("$C$", (9.911134209747932,-1.7205865972466723), NE * labelscalefactor);
dot((-3.6344941786712726,2.0592624641139796),dotstyle);
label("$I$", (-3.497494882551642,2.392284670466186), NE * labelscalefactor);
dot((3.634496544141483,-2.0620375985781543),dotstyle);
label("$A_1=V$", (3.758546297071818,-1.7205865972466723), NE * labelscalefactor);
dot((5.574190950974938,0.8002948325823711),dotstyle);
label("$B_1$", (5.697949008676245,1.1216415145874168), NE * labelscalefactor);
dot((-7.857008451664908,2.6377615833571366),dotstyle);
label("$C_1$", (-7.710680083623329,2.960730292833004), NE * labelscalefactor);
dot((-17.336151248875133,-2.0799125676251085),dotstyle);
label("$D$", (-17.207065774927766,-1.7540245750329557), NE * labelscalefactor);
dot((-7.455155307994771,3.62091525913628),dotstyle);
label("$C_2$", (-7.30942435018793,3.9638696264215056), NE * labelscalefactor);
dot((-1.3174867198163902,5.47506212933665),dotstyle);
label("$B_2$", (-1.1902744152980995,5.802958404667093), NE * labelscalefactor);
dot((-3.6309759845503167,-2.0682305449212386),dotstyle);
label("$A_2$", (-3.497494882551642,-1.7205865972466723), NE * labelscalefactor);
dot((-7.575344228330761,6.527952192096369),dotstyle);
label("$S$", (-7.443176261333063,6.872973693828162), NE * labelscalefactor);
dot((0.008523800852428653,-9.999996367240293),dotstyle);
label("$M_a$", (0.1472446961532296,-9.678825310382122), NE * labelscalefactor);
dot((-11.432001519352662,3.12684157533726),dotstyle);
label("$T$", (-11.288543706755634,3.462299959627255), NE * labelscalefactor);
dot((3.6344941786712726,-2.0592624641139796),dotstyle);
dot((-5.5257089064128975,4.203787559149821),dotstyle);
label("$P$", (-5.403459616369786,4.532315248788324), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/c/4/9c44ca53b2ebe71a4446a62b321771d4a1b3811e.png)
,
be the incenter and circumcenter of 
be the circumcenter of 
.
are concyclic.
and 
.
lie on the circle with diameter 
.
.
s
and 
,
.
then 
however this would yield 
which is not true. Thus, 
.
is the altitude, then 
are collinear.
hence 
.
,
.
are concyclic which can be seen under 
thus, 
are collinear and since 
gives it.
.
at 
.
thus, 
which implies also 
.
and 
.
is a rectangle. 
and since 
,
.![\[\measuredangle LDC_1=\measuredangle LDI-\measuredangle C_1DI=\measuredangle DIL-\measuredangle IDB_1=\measuredangle DB_1C_1\]](http://latex.artofproblemsolving.com/3/1/5/315d5239e131356778cc9e9b155932ce21e29280.png)
As desired
