and 
.
is given. Let 
be the midpoint of 
,
be the foot of altitude from 
to 
be the intersection point of 
and 
.
lies on the arc 
.
.
be a trapezium inscribed in a circle 
with diameter 
. The circle with center 
and radius 
meets 
)
at 
.
is perpendicular to 
.
Y by anantmudgal09, Vietjung, Supercali, kiyoras_2001, ike.chen, megarnie, tiendung2006, Adventure10, Rounak_iitr
Let 
be a triangle with incenter 
. Points 
and 
are chosen on segment 
such that the incircles of 
and 
are tangent at 
, and the incircles of 
and 
are tangent at 
. Prove that 
.
Ankan Bhattacharya
be a triangle with incenter 
. Points 
and 
are chosen on segment 
such that the incircles of 
and 
are tangent at 
, and the incircles of 
and 
are tangent at 
. Prove that 
.Ankan Bhattacharya
be arbitrary points on segment 
,
,
,
be the incircles of triangles 
.
and 
have the same exsimilicenter.![[asy]
import olympiad;
size(10cm);
pair A = dir(110), B = dir(210), C = dir(330), K = (B+C)/2, L = (K+C)/2, Ibk = incenter(A,B,K), Ick = incenter(A,C,K), Ibl = incenter(A,B,L), Icl = incenter(A,C,L), P = extension(Ibk,Icl, Ick,Ibl), I = incenter(A,B,C), U = foot(Ibl,B,C), V = foot(Icl,B,C);
D(incircle(A,B,K),green);
D(incircle(A,C,L),green);
D(incircle(A,B,L),orange);
D(incircle(A,C,K),orange);
D(P--Ibl,orange);
D(P--Ibk,green);
D(L--A--B--C--A--K--P);
D(B--I--C,dashed+blue);
D("A",A,A);
D("B",B,B);
D("C",C,dir(270));
D("K",K,dir(270));
D("L",L,dir(270));
D("U",U,dir(270));
D("V",V,dir(270));
D("I_{BK}",Ibk,dir(140));
D("I_{CK}",Ick,dir(80));
D("I_{BL}",Ibl,dir(140));
D("I_{CL}",Icl,dir(80));
D("I",I,dir(110));
D("T",P,dir(0));
[/asy]](http://latex.artofproblemsolving.com/7/3/a/73a4d550a18ec1e44fd1ff1c2211286c09aaf32c.png)
have centers 
,
and 
cocnur on 
and 
are fixed.
be the touchpoints of 
is projective.
,
,
,
,
,
,
,
.![\[
a(d^2+xy) = b^2x+c^2y.
\]](http://latex.artofproblemsolving.com/8/a/f/8afb27c604f612805d5834ef12536d2bca175b61.png)
Now, we see that ![\[
u = \frac{c+x-d}{2}, v = \frac{b+y-d}{2},
\]](http://latex.artofproblemsolving.com/b/7/9/b792980d018c10c1365d00d05189b72e380d1e46.png)
so we can solve for ![\[
d = \frac{a+b+c}{2}-(u+v).
\]](http://latex.artofproblemsolving.com/d/7/9/d799b60104d310f885b11796f66e76dadab1e2e2.png)
Then, ![\[
x = 2u - c+d = u-v + \frac{a+b-c}{2}
\]](http://latex.artofproblemsolving.com/4/d/3/4d3f5a5cfb491e161ccc2bbf4524815bd3216349.png)
and ![\[
y = 2v - b + d = v-u+\frac{a-b+c}{2}.
\]](http://latex.artofproblemsolving.com/4/e/5/4e5f1677eb35c6a855e70f6d071820094f0a671c.png)
Plugging in, we have, for appropriate constants 
:![\[
\begin{aligned}
\frac{b^2-c^2}{a}(u-v) + A &= \frac{b^2x+c^2y}{a} = d^2+xy \\
&= \left(\frac{a+b+c}{2} - (u+v)\right)^2 + \left(u-v+\frac{a+b-c}{2}\right)\left(v-u + \frac{a-b+c}{2}\right) \\
&= B - (a+b+c)(u+v) + (u+v)^2 - (u-v)^2 + (c-b)(u-v)\\
&= B - (a+b+c)(u+v) + (c-b)(u-v) + 4uv.
\end{aligned}
\]](http://latex.artofproblemsolving.com/8/e/e/8ee8f49000073ab9666c9692492b56ca1e870983.png)
Thus in particular we have, for constants 
:![\[
4uv + uP + vQ + R = 0\implies v = - \frac{Pu+R}{4u+Q}.
\]](http://latex.artofproblemsolving.com/9/3/e/93e5107e2e2f0df380a3242793b19830826add01.png)
Thus 
is a M\"obius function and so is projective if we parameterize 
;
is projective since we project through 
onto 
,
.
is also projection through 
.
is also projective.
are projective, and so it suffices to check 
for 
values of 
.
,
.
,
,
,
,
.
,
.
.
be the common exsimilicenter in the original problem. Let 
be the insimilicenter of 
be the insimilicenter of 
.
are the insimilicenters of 
and 
,
collinear, 
collinear, 
collinear, and 
collinear, so 
are collinear.
meet 
.
to 
.
,
-
being projective, which is true because 
is a rotation.
.
be a variable point along 
are inradii of 
,
.
is not depending on 
be the incenters of 
,
is constant. To prove this, denote another position of 
and let 
be the corresponding incenters. Then as rotation by 
is projective map, we get
which is constant as claimed. To finish, note that
so we are done.
lie in this order. Let 
be the inradii of 
be the incenters of 
and 
.
and similarly, 
.
By plugging in 
,
and 
,
Finally, note that 
,
and 
.
.
so 
denote the incenters of 
respectively. Let 
denote their respective incircles.
.
.
concur at a point on 
say at 
Observe that 
This is true because 
are the exsimilicenter and insimilicenter of 
,
their centers, and analogously for 
.
concur at 
goes through 
from 
Let 
be the insimilicenter of 
By Monge's Theorem on 
,
are collinear. Hence, 
,
.
,
Finally, from the definition of 
,
.
![[asy]
size(12cm);
defaultpen(fontsize(10pt));
pen pri=royalblue;
pen sec=deepcyan;
pen tri=rgb(41, 207, 255);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
real r=1.7353;
pair A, B, C, I, IB, P,Q, IC, K, L, JB, JC, T;
A=dir(120);
B=dir(210);
C=dir(330);
I=incenter(A,B,C);
IB=(7B+r*I)/(7+r);
P=IB+unit(I-B)*length(IB-foot(IB,B,C));
Q=I+unit(C-I)*length(P-I);
IC=incenter(C,extension(A,C,Q,Q+A-foot(A,C,I)),extension(B,C,Q,Q+A-foot(A,C,I)));
K=extension(B,C,A,2*foot(B,A,IB)-B);
L=extension(B,C,A,2*foot(C,A,IC)-C);
JB=incenter(A,B,L);
JC=incenter(A,C,K);
T=extension(IB,IC,B,C);
draw(B--I--C,sec);
draw(Q--T,sec);
draw(JC--T--IC,sec);
fill(A--B--C-- cycle,fil);
draw(B--A--C--T,pri);
draw(K--A--L,pri);
filldraw(incircle(A,B,L),tfil,tri);
filldraw(incircle(A,B,K),tfil,tri);
filldraw(incircle(A,C,L),tfil,tri);
filldraw(incircle(A,C,K),tfil,tri);
dot("$A$",A,N);
dot("$B$",B,S);
dot("$C$",C,SE);
dot("$T$",T,SW);
dot("$I_B$",IB,dir(285));
dot("$I_C$",IC,dir(255));
dot("$J_B$",JB,NW);
dot("$J_C$",JC,NE);
dot("$P$",P,dir(85));
dot("$Q$",Q,dir(120));
dot("$K$",K,S);
dot("$L$",L,S);
dot("$I$",I,N);
[/asy]](http://latex.artofproblemsolving.com/6/5/b/65b0d803dcebc037b39af7f527f8302abf90a43f.png)
,
,
,
denote the incenters of 
denote the inradius of 
for all 
.
of 
,
,
and the result follows. 
to 
,
and similarly 
.
also passes through 
,
whence 
be the inscribed circles of 
.
the corresponding centers and 
the touching points with 
and 
have a common external homothetic center 
and 
respectively well let 
be the other internal tangents and
be the internal homothetic center of 
and 
that of 
.
concur at 
, 
and the result follows (you need a simple lemma that states that if 
intersects 
has an inscribed circle 
.
and prove that the line that connects the homothetic centers 
intersects 
![[asy] unitsize(15); real r1, r2; path w1, w2;
pair J1 = (-0.5, 2.7); pair J2 = (1, 3); real r1 = abs(J1.y); real r2 = abs(J2.y); path w1 = circle(J1, r1); path w2 = circle(J2, r2);
pair X1 = intersectionpoints(w1, w2)[0]; pair X2 = intersectionpoints(w1, w2)[1]; pair H = extension((1.9, 0), (2, 1), X1, X2); pair P = intersectionpoints(circle(H, sqrt(abs(H-X1)*abs(H-X2))), w1)[1]; pair Q = intersectionpoints(circle(H, sqrt(abs(H-X1)*abs(H-X2))), w2)[0]; pair I = extension(J1, P, J2, Q); pair B = extension(J1, P, (0, 0), (1, 0)); pair C = extension(J2, Q, (0, 0), (1, 0)); pair A = extension(B, reflect(B, I) * C, C, reflect(C, I) * B); pair K = extension(B, C, A, reflect(A, J2) * C); pair L = extension(B, C, A, reflect(A, J1) * B); pair I_B = incenter(A, B, K); pair J_B = incenter(A, B, L); pair I_C = incenter(A, C, K); pair J_C = incenter(A, C, L);
pair R = extension(P, Q, B, C);
filldraw(A--B--C--cycle, invisible, blue); draw(incircle(A, B, K)^^incircle(A, C, L), deepcyan); draw(L--A--K, blue); draw(w1^^w2, deepcyan); draw(I--P^^I--Q, red); add(pathticks(I--P, 2, 0.5, 0, 10, red)); add(pathticks(I--Q, 2, 0.5, 0, 10, red));
draw(B--P^^C--Q, orange); draw(Q--R--B, deepgreen); draw(J_C--R, deepgreen+dashed); draw(I_C--R, deepgreen+dashed);
dot("$A$", A, dir(A-I)); dot("$B$", B, dir(230)); dot("$C$", C, dir(310)); dot("$I$", I, dir(2*I-P-Q)); dot("$K$", K, dir(270)); dot("$L$", L, dir(270)); dot("$P$", P, dir(80)); dot("$Q$", Q, dir(20)); dot("$R$", R, dir(-90));
dot("$I_B$", I_B, dir(-90)); dot("$I_C$", I_C, dir(-90)); dot("$J_B$", J_B, dir(-90)); dot("$J_C$", J_C, dir(-90)); [/asy]](http://latex.artofproblemsolving.com/f/7/2/f72d3face50ea50c75e3e94ff3a931e19f189b57.png)
,
meet at a point 
(possibly at infinity).
we have the equality 
.
.
and 
.
,
.![\[\frac{I_BP}{PI} \cdot \frac{IQ}{QJ_C} \cdot \frac{J_CR}{RI_B} = -1 \implies IP = IQ.\]](http://latex.artofproblemsolving.com/6/c/a/6ca213f974b4add570ca6c7ba4e12fc238136e1a.png)
,
,
,
be the incircles of 
at 
and 
.
be the circle with center 
be the circle through 
be the radical center, it follows that 
and 
are the common internal tangents to 
and 
.
.
,
,
and 
at 
.
are tangent at 
are tangent at 
lies on line 
.
first; then the remaining points 
be the incenters of 
.
.
and 
(easily follows by two applications of sine law). Multiplying and taking into account that 
we get 
.
and 
.
.
,
which implies 
as desired.
,
,
,
respectively. Then, the pairs 
,
be the radius of circles 
When 
and 
,
or 
be the incenters of triangles 
respectively. We have 
(for example, by projecting from 
concur at some point 
,
and line 
,
,
be two lines and let 
.
and 
.
concur iff 
.
and let 
.
by projection through 
iff $(A, B; C, P) = (X, Y; Z, P)
.![[asy]
import olympiad;
import math;
import geometry;
size(15cm);
defaultpen(fontsize(10pt));
pair B = (0,0), C = (9,0), A = (2,6), K = (1.04,0) , L = (5.77,0), P=(1.04,0.76), Q=(5.37,1.35), Ibk = incenter(A,B,K), Ick = incenter(A,C,K), Ibl = incenter(A,B,L), Icl = incenter(A,C,L), I = incenter(A,B,C), U = extension(Q,P,B,C);
D(incircle(A,B,K),green+blue);
D(incircle(A,C,L),green+blue);
D(incircle(A,B,L),red+blue);
D(incircle(A,C,K),red+blue);
D(A--B--C--A,red);
D(A--K,blue);
D(A--L,blue);
D(I--B,cyan);
D(I--C,cyan);
D(U--Q,yellow);
D(U--B,red);
D(U--Ick,yellow);
D(U--Icl,yellow);
D("A",A,N);
D("B",B,S);
D("C",C,dir(270));
D("K",K,dir(270));
D("L",L,dir(270));
D("I_{1}",Ibk,dir(140));
D("I_{4}",Ick,S);
D("I_{2}",Ibl,S);
D("I_{3}",Icl,NE);
D("I",I,dir(110));
D("P",P,N);
D("Q",Q,SE);
D("U",U,N);
[/asy]](http://latex.artofproblemsolving.com/0/8/2/082004a110eb6190c73c81c70d569db3bcf541b4.png)
Since 
Now, from Menelaus, we get:
concur on a point on 
yields:
and 
.
, hence 
.
and 
,
.
with transversal 
.
be the incenters of 
,
denote the incircle of any 
sends the lines 
to the lines 
,
,
concur at a point 
to 
.
.
,
and 
.
we see that 
![[asy]size(12cm);
defaultpen(fontsize(10pt));
pen pri=royalblue;
pen sec=deepcyan;
pen tri=rgb(41, 207, 255);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
real r=1.7353;
pair A, B, C, I, IB, P,Q, IC, K, L, JB, JC, T;
A=dir(120);
B=dir(210);
C=dir(330);
I=incenter(A,B,C);
IB=(7B+r*I)/(7+r);
P=IB+unit(I-B)*length(IB-foot(IB,B,C));
Q=I+unit(C-I)*length(P-I);
IC=incenter(C,extension(A,C,Q,Q+A-foot(A,C,I)),extension(B,C,Q,Q+A-foot(A,C,I)));
K=extension(B,C,A,2*foot(B,A,IB)-B);
L=extension(B,C,A,2*foot(C,A,IC)-C);
JB=incenter(A,B,L);
JC=incenter(A,C,K);
T=extension(IB,IC,B,C);
draw(B--I--C,sec);
draw(Q--T,sec);
draw(JC--T--IC,sec);
fill(A--B--C-- cycle,fil);
draw(B--A--C--T,pri);
draw(K--A--L,pri);
filldraw(incircle(A,B,L),tfil,tri);
filldraw(incircle(A,B,K),tfil,tri);
filldraw(incircle(A,C,L),tfil,tri);
filldraw(incircle(A,C,K),tfil,tri);
dot("$A$",A,N);
dot("$B$",B,S);
dot("$C$",C,SE);
dot("$T$",T,SW);
dot("$I_B$",IB,dir(285));
dot("$I_C$",IC,dir(255));
dot("$J_B$",JB,NW);
dot("$J_C$",JC,NE);
dot("$P$",P,dir(85));
dot("$Q$",Q,dir(120));
dot("$K$",K,S);
dot("$L$",L,S);
dot("$I$",I,N);[/asy]](http://latex.artofproblemsolving.com/5/e/1/5e107be50d72706a9eb6e983b882e3ddd35a0272.png)
,
,
through 
we get 
thus by the Prism Lemma 
are concurren
are the Homothety centers of incircle of 
and similarly 
.![\[-1=\dfrac{IQ}{QI_C}\cdot\dfrac{IC_T}{TI_B}\cdot\dfrac{I_BP}{PI}=\dfrac{IQ}{PI}\left(\dfrac{TI_C}{QI_C}\cdot \dfrac{I_BP}{TI_B}\right)=\dfrac{IQ}{PI}\Longleftrightarrow IP=IQ\]](http://latex.artofproblemsolving.com/d/5/d/d5d0bf4ff93e30317a61b4a1388f7c3327e902c0.png)
and we are done.
so it implies that 
are conccurent at external homothetic center of 
and 
named 
and 
intersect at I so ![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(22.15969431137711cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ real xmin = -29.88720851620066, xmax = 14.43218010655356, ymin = -15.439174074706292, ymax = 22.773450222129547; /* image dimensions */
pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen qqzzcc = rgb(0.,0.6,0.8); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen zzffqq = rgb(0.6,1.,0.);
pair K = (-6.781350498846555,-2.538620079446605), L = (4.389280938246323,-2.569998257640686);
filldraw((-3.79365102990033,12.53635641196014)--(-9.586493200442964,-2.530740465116278)--(10.73125864894795,-2.5878128017718707)--cycle, invisible, linewidth(2.) + dbwrru);
filldraw(circle((-3.105070388750036,1.90791499658331), 4.456844122406074), invisible, linewidth(2.) + linetype("0 3 4 3") + wvvxds);
filldraw(circle((-0.8118343963566139,2.349887988824497), 4.905257019704117), invisible, linewidth(2.) + linetype("0 3 4 3") + wvvxds);
filldraw(circle((-7.793395958005923,-1.302778472860014), 1.2329939178540792), invisible, linewidth(2.) + linetype("2 2") + sexdts);
filldraw(circle((5.671787825599821,-0.42356159229146023), 2.1500307304818906), invisible, linewidth(2.) + linetype("2 2") + sexdts);
/* draw figures *//* special point *//* special point *//* special point */
draw((-3.79365102990033,12.53635641196014)--(-9.586493200442964,-2.530740465116278), linewidth(2.) + dbwrru);
draw((-9.586493200442964,-2.530740465116278)--(10.73125864894795,-2.5878128017718707), linewidth(2.) + dbwrru);
draw((10.73125864894795,-2.5878128017718707)--(-3.79365102990033,12.53635641196014), linewidth(2.) + dbwrru);
draw((-3.79365102990033,12.53635641196014)--K, linewidth(2.));
draw((-3.79365102990033,12.53635641196014)--L, linewidth(2.)); /* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */
draw((5.67178782559982,-0.4235615922914604)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + linetype("2 2") + qqzzcc);
draw((3.6954595639694174,0.42305446253944107)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + linetype("2 2") + qqzzcc);
draw((-9.586493200442964,-2.530740465116278)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + qqzzcc);
draw((-25.73875160010579,-2.4542508876070626)--(-0.8118343963566137,2.349887988824497), linewidth(2.) + qqzzcc);
draw((-1.8261325606399366,2.7837665712829005)--(3.6954595639694174,0.42305446253944107), linewidth(2.) + dtsfsf);
draw((1.2198937175515965,2.2705502786417124)--(0.6494332857778794,0.9362707551806287), linewidth(2.) + dtsfsf);
draw((-1.8261325606399366,2.7837665712829005)--(-6.779651039942243,-0.6009271762414112), linewidth(2.) + dtsfsf);
draw((-3.8935572632802167,0.49235628305876683)--(-4.7122263373019635,1.6904831119827222), linewidth(2.) + dtsfsf);
draw((-6.779651039942243,-0.6009271762414112)--(-9.586493200442964,-2.530740465116278), linewidth(2.) + zzffqq);
draw((3.6954595639694174,0.42305446253944107)--(10.73125864894795,-2.5878128017718707), linewidth(2.) + zzffqq);
/* dots and labels */
label("$A$", (-3.5253347924341663,13.15983113479268), NE * labelscalefactor,uququq);
label("$B$", (-10.297192262759504,-3.2861084359974266), NE * labelscalefactor,uququq);
label("$C$", (10.864862332007176,-3.407034462253236), NE * labelscalefactor,uququq);
dot(K,wrwrwr);
label("$K$", (-6.97172654072474,-3.527960488509046), NE * labelscalefactor,wrwrwr);
dot(L,wrwrwr);
label("$L$", (4.153467874809743,-3.6488865147648553), NE * labelscalefactor,wrwrwr);
label("$P$", (-6.548485448829407,-0.021105727090566892), NE * labelscalefactor,red);
label("$Q$", (3.911615822298124,1.0067654960838148), NE * labelscalefactor,red);
label("$I_{1}$", (-7.757745711387503,-2.0768481734393305), NE * labelscalefactor);
label("$I_{2}$", (5.423191150495744,-1.4117550290323775), NE * labelscalefactor);
label("$I_{3}$", (-3.283482739922547,2.6392668505372447), NE * labelscalefactor);
label("$I_{4}$", (-0.5626471491668312,2.820655889920959), NE * labelscalefactor);
label("$I$", (-1.5905183723412128,3.2438969818162926), NE * labelscalefactor,wrwrwr);
label("$T$", (-25.47340855786361,-1.9559221471835206), NE * labelscalefactor,wrwrwr);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/b/e/0/be0f13e6038bc80a0c81e2111a5982ed2a0cda08.png)
be 
,![\[\angle BAK_1 = \angle IAK_2 = \tfrac12 \angle BAK \text{ and } \angle K_1AL_2 = \angle K_2AL_1 = \tfrac12 \angle KAL \text{ and } \angle L_2AI = \angle L_1AC = \tfrac12 \angle CAL.\]](http://latex.artofproblemsolving.com/e/b/2/eb2f9059a96e629e196ddf0cc8c03a491bd8d5d5.png)
The angles formed by the sets of lines are the same, hence 
.
concur at 
.
through 
,
.
,
.
be the incircles with centers 
,
and the insimilicenter 
are collinear. 
,
as well, hence 
.
,
,
)
)
by simply observing angles, and hence 
concur at some point 
by exsimilicenters/insimilicenters, and hence 
be 
respectively. Furthermore, let 
denote the radii of these 
circles, and let 
be the inradius of 
and 
.
Analogously, we conclude 
.
about point 
under this rotation be 
respectively. It's easy to see 
via previous angle chasing. Thus, 
by basic projections and cross ratio manipulations. Now, the Prism Lemma implies 
are concurrent at some point 
and 
are harmonic bundles.
where lengths are directed, implying the the first result by the definition of the cross ratio. The second result follows analogously. 
.
But 
,
so 
(where the direction of lengths is irrelevant). To finish, we observe that homotheties imply 
as desired. 
and 
.
and 
and 
respectively. We begin with a crucial claim:
Since 
are insimilicenter, and exsimilicenters of incenters of 
So by Prism Lemma again implies that 
![[asy]
import graph; size(17cm);
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 = -30.53, xmax = 7.65, ymin = -2.46, ymax = 14.78; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); pen qqwuqq = rgb(0.,0.39215686274509803,0.);
draw((-6.37,9.94)--(-11.,2.)--(3.,2.)--cycle, linewidth(0.5) + zzttqq);
/* draw figures */
draw((-6.37,9.94)--(-11.,2.), linewidth(0.5) + zzttqq);
draw((-11.,2.)--(3.,2.), linewidth(0.5) + zzttqq);
draw((3.,2.)--(-6.37,9.94), linewidth(0.5) + zzttqq);
draw((-11.,2.)--(-5.545193307851359,5.133647281153756), linewidth(0.5) + xdxdff);
draw((-5.545193307851359,5.133647281153756)--(3.,2.), linewidth(0.5) + xdxdff);
draw(circle((-9.42594737269626,2.9042530770234087), 0.9042530770234085), linewidth(0.5) + red);
draw((-6.37,9.94)--(-8.753257409829024,2.), linewidth(0.5) + ubqqys);
draw(circle((-6.329313718559153,4.683190109747784), 2.6669871773326093), linewidth(0.5) + red);
draw((-6.37,9.94)--(-1.612508515592935,2.), linewidth(0.5) + ubqqys);
draw(circle((-0.8190089387275968,3.4004863958489855), 1.4004863958489857), linewidth(0.5) + red);
draw(circle((-4.872503344984124,4.886962039842536), 2.8869620398425346), linewidth(0.5) + red);
draw((-4.872503344984124,4.886962039842536)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + qqwuqq);
draw((-0.8190089387275968,3.4004863958489855)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + qqwuqq);
draw((-11.,2.)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + zzttqq);
/* dots and labels */
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dot((-11.,2.),linewidth(3.pt) + dotstyle);
label("$B$", (-11.5,1.35), NE * labelscalefactor);
dot((3.,2.),linewidth(3.pt) + dotstyle);
label("$C$", (3.09,1.4), NE * labelscalefactor);
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dot((-8.753257409829024,2.),linewidth(3.pt) + dotstyle);
label("$K$", (-8.81,1.4), NE * labelscalefactor);
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label("$P$", (-8.8,3.66), NE * labelscalefactor);
dot((-6.329313718559153,4.683190109747784),linewidth(3.pt) + dotstyle);
label("$J_B$", (-7.2,4.8), NE * labelscalefactor);
dot((-1.612508515592935,2.),linewidth(3.pt) + dotstyle);
label("$L$", (-1.79,1.4), NE * labelscalefactor);
dot((-4.872503344984124,4.886962039842536),linewidth(3.pt) + dotstyle);
label("$I_C$", (-4.83,4.9), NE * labelscalefactor);
dot((-0.8190089387275968,3.4004863958489855),linewidth(3.pt) + dotstyle);
label("$J_C$", (-0.9,3.56), NE * labelscalefactor);
dot((-25.794084103041374,1.9605476148542396),linewidth(3.pt) + dotstyle);
label("$X$", (-26.03,1.3), NE * labelscalefactor);
dot((-2.147226471973991,3.8875631783032922),linewidth(3.pt) + dotstyle);
label("$Q$", (-2.7,4.12), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/e/5/0e548a16e897ffc70ab02736b32a15945a925ea1.png)
we get that
,
which is symmetric and thus also equals 
.
and similarly 
,
Similarly, 
But 
,
,
,
(to avoid fractions), and we let 
,
and 
so we see that they must satisfy the equation 
To show that 
where 
Note that from here we're basically done since we combine this with the other equation to get some line which should match the 
so if we combine this with the first expression to remove the 
term we end up with the equation 
Note that 
so the original equation is actually 
so it would be enough that 
or that 
or that 
which is just the Law of Sines. We are done. 
respectivily and note that
Now using projections its easy to know that 
are concurrent at 
and 
we have.
Which means that 
and of the incircles of 
so we have that the inverse of the incircle of 
and 
,
,
,
and 
,
be the insimilicenters of 
are colinear.
,
and 
.
are also colinear. 
,
,
and 
.
are colinear, the perpendiculars through each of these points to 
,
and 
,
and 
as desired. 
.
.![\[\measuredangle BAI_1= \frac 12 \measuredangle BAK =\frac 12\left( \measuredangle BAC-\measuredangle KAC\right) =\measuredangle IAC-\measuredangle I_3AC=\measuredangle IAI_3\]](http://latex.artofproblemsolving.com/2/b/0/2b041c5032e9d8475cfc11209b6498b24d869cc4.png)
and similarly 
.
and 
are rotations by half of 
and the cross ratios are equal. The concurrence of 
and prism lemma again implies the result. 
Notice that the insimilcenter of 
But also, by Monge on 
we have that this insimilicenter line on line 
.
implying that 
![[asy]size(12cm);
defaultpen(fontsize(10pt));
pen pri=red;
pen sec=darkgreen;
pen tri=rgb(41, 20, 255);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
real r=1.7353;
pair A, B, C, I, IB, P,Q, IC, K, L, JB, JC, T;
A=dir(120);
B=dir(210);
C=dir(330);
I=incenter(A,B,C);
IB=(7B+r*I)/(7+r);
P=IB+unit(I-B)*length(IB-foot(IB,B,C));
Q=I+unit(C-I)*length(P-I);
IC=incenter(C,extension(A,C,Q,Q+A-foot(A,C,I)),extension(B,C,Q,Q+A-foot(A,C,I)));
K=extension(B,C,A,2*foot(B,A,IB)-B);
L=extension(B,C,A,2*foot(C,A,IC)-C);
JB=incenter(A,B,L);
JC=incenter(A,C,K);
T=extension(IB,IC,B,C);
draw(B--I--C,sec);
draw(Q--T,sec);
draw(JC--T--IC,sec);
fill(A--B--C-- cycle,fil);
draw(B--A--C--T,pri);
draw(K--A--L,pri);
filldraw(incircle(A,B,L),tfil,tri);
filldraw(incircle(A,B,K),tfil,tri);
filldraw(incircle(A,C,L),tfil,tri);
filldraw(incircle(A,C,K),tfil,tri);
dot("$A$",A,N);
dot("$B$",B,S);
dot("$C$",C,SE);
dot("$T$",T,SW);
dot("$I_B$",IB,dir(285));
dot("$I_C$",IC,dir(255));
dot("$J_B$",JB,NW);
dot("$J_C$",JC,NE);
dot("$P$",P,dir(85));
dot("$Q$",Q,dir(120));
dot("$K$",K,S);
dot("$L$",L,S);
dot("$I$",I,N);[/asy]](http://latex.artofproblemsolving.com/e/9/a/e9a9c3168b4f1dfaef2ecddc4fa1c8393971726b.png)
concur.
,
therfore 
;
,
.
,
,
.
.
,
therefore 
and we can conclude.
so 
is projective and hence we have 
.
at 
intersect 
at 
.
so 
meaning that 
,
and 
be the inradii of the incircles of 
hence 
.
so proceeding like before we get that 
they're inradii respectively. By homothety again we get 
.
we get 
,
,
,
respectively.
and 
.
which implies that 
.
,
concur at some point 
we get that 
be the inradius of some triangle with vertices ![\[\frac{KC_1}{KB_1} \cdot \frac{B_1P}{PI} \cdot \frac{IQ}{QC_1} = 1.\]](http://latex.artofproblemsolving.com/0/1/4/014dbeb8e2390e0e654683a64d0caf7eb90a12c4.png)
However, 
as desired.
be the incircles of 
be the centers of these incircles, respectively.
Therefore,
Because 
,![\[(BI;B_1B_2) = (CI;C_1C_2).\]](http://latex.artofproblemsolving.com/b/e/5/be52aa4667c41707c4b182c800c889a7c015799d.png)
By the prism lemma, ![\[(BP;B_1B_2)=(CQ;C_1C_2)=-1,\]](http://latex.artofproblemsolving.com/a/f/1/af1ed469af7b695ff2fafc9bdb79b11c528b63b6.png)
by the prism lemma, 
and 
.
at 
at 
is isosceles.
,
.
be incenters of 
respectively.
and 
are collinear due to angle bisector and homothety.
,
is sent to pencil 
so by prism lemma; 
concur at some point 
so 
and transversal 
,
be the incenters of triangles 
.![\[(AB, AX_K; AX_L, AI) = (AI, AY_K; AY_L, AC) \implies (BX_K; X_LI) = (CY_L; Y_KI),\]](http://latex.artofproblemsolving.com/1/d/3/1d3304944d4f7e5a8ec5a8e57a002b50a3508483.png)
,
concur at point, say ![\[-1 = (BP; X_KX_L) = (CQ; Y_LY_K) \implies T \in PQ\]](http://latex.artofproblemsolving.com/4/0/d/40dc22a325e7e3dd76ae79149c6e464f38bfc151.png)
with transversal 
,![\[1 = \frac{IP}{PX_K} \cdot \frac{X_KT}{TY_L} \cdot \frac{Y_LQ}{QI} = \frac{IP}{r_{XK}} \cdot \frac{r_{XK}}{r_{YL}} \cdot \frac{r_{YL}}{QI} \implies IP = IQ. \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/4/4/c44dedc943a3a0fccbc0d4e524afa9331c680b0e.png)
