[/quote]
,
is mundane if there exist positive integers 
and 
less than 
such that 
is a positive integer. Find, with proof, all prime numbers that are not mundane.
be a fixed positive integer. Find the greatest real constant 
that has the following property: Any 
real numbers, not necessarily distinct, that lie in the interval ![$[100,101]$](http://latex.artofproblemsolving.com/8/e/3/8e3fe24e3b899bbf2128eebe08756e1ff963aaa0.png)
,
and 
such that
.
, 
and 
.
.
-
board.
board consisting of 
unit segments. A configuration of several 
is a divisor of 
.
,
and 
are equal, and they intersect at 
and 
intersect at point 
lying inside triangle 
,
and 
intersect at point 
lying inside triangle 
.
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
be a triangle with circumcenter 
.
be the intersections of the perpendicular bisectors of 
and 
.
and 
and the point 
is given. Let 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
,
and 
are integers, 
.
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
satisfying 
,![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
Y by Rounak_iitr, ys-lg
Let triangle
with incenter 
circumscribed in 
. Let 
be midpoint of arc 
and 
, respectively. 
lies on so that 
, and 
is tangency point of 
-excircle of 
. Point 
is in so that 
and 
. Extend 
to meet at 
, and extend 
to meet line 
at L. Let line 
and 
meet at 
. Proof that 
.
with incenter 
circumscribed in 
. Let 
be midpoint of arc 
and 
, respectively. 
lies on so that 
, and 
is tangency point of 
-excircle of 
. Point 
is in so that 
and 
. Extend 
to meet at 
, and extend 
to meet line 
at L. Let line 
and 
meet at 
. Proof that 
.
![[asy]
size(8cm);
defaultpen(fontsize(10pt));
pair A = dir(125), B = dir(210), C = dir(330), M = dir(90), N = dir(270), D = dir(55), I = incenter(A,B,C), T = intersectionpoint(unitcircle,I--I+dir(M--I)*100), F = extension(A,T,I,I+dir(0)), G = F+dir(N--F)*abs(A-F)*abs(T-F)/abs(N-F), L = extension(A,G,I,F), K = extension(D,I,A,T), S = extension(M,L,N,K), T1 = 2*foot(T,M,N)-T;
draw(A--B--C--A--N^^unitcircle, heavyblue);
draw(D--A^^L--I^^B--C^^T--T1, heavyblue+linewidth(1.2));
draw(A--T--M^^D--K, orange);
draw(N--G^^A--L, purple);
draw(M--S^^N--S, red);
draw(L--S, red+dashed);
draw(circumcircle(A,G,F), cyan);
draw(circumcircle(M,F,I), magenta);
draw(S--T1, heavygreen+dashed);
dot("$A$", A, dir(120));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
dot("$D$", D, dir(60));
dot("$M$", M, dir(90));
dot("$N$", N, dir(270));
dot("$I$", I, dir(330));
dot("$F$", F, dir(45));
dot("$K$", K, dir(210));
dot("$G$", G, dir(135));
dot("$S$", S, dir(150));
dot("$L$", L, dir(210));
dot("$T$", T, dir(225));
dot("$T'$", T1, dir(315));
[/asy]](http://latex.artofproblemsolving.com/3/a/d/3ad374e36a1c49e2307f938a7f8ce304bb7c1585.png)
be the 
.
,
,
,
concur. By (degenerate) Reim's we have 
cyclic, so by radical axis it suffices to show 
cyclic.
.
which implies 
,
only depends on 
,
that 
as 
)
and 
.
are collinear
and 
.
gives 
,
and 
.
are collinear
.
.
implying
The last equality is due to the Angle Bisector Theorem.
.
are collinear.
gives 
collinear as desired. 
which solves the problem. 
,
.
since 
is an isosceles trapezoid.
(it is well known that 
)
,
,
and 
.
are concyclic ; 
and 
.![[asy]
size(250);
pair A=dir(115),B=dir(-160),C=dir(-20), D = dir(65), N = dir(-90),M=dir(90),I = incenter(A,B,C), T = intersectionpoint(unitcircle,I--I+dir(M--I)*100),F= extension(A,T,I,foot(I,M,N)),G = IP(F--3*F-2*N,unitcircle),L= extension(A,G,F,I),K = extension(D,I,A,F),X=extension(M,L,N,K),Y=IP(F--3*F-2*M,unitcircle),Z=IP(I--100*I-99*G,unitcircle),U=extension(M,T,N,G);
draw(unitcircle,cyan+linewidth(0.8));
dot("$A$",A,dir(A));
dot("$B$",B,dir(140));
dot("$C$",C,dir(C));
dot("$N$",N,dir(N));
dot("$M$",M,dir(M));
dot("$F$",F,dir(F));
dot("$T$",T,dir(T));
dot("$I$",I,dir(I));
dot("$G$",G,dir(G));
dot("$D$",D,dir(D));
dot("$K$",K,dir(210));
dot("$L$",L,dir(L));
dot("$X$",X,dir(-90));
dot("$Y$",Y,dir(Y));
dot("$Z$",Z,dir(Z));
dot("$U$",U,dir(20));
draw(A--D^^L--I^^B--C^^T--Z,green);
draw(B--A--C^^A--T--M,purple);
draw(M--Y--D,fuchsia);
draw(Z--G--N,fuchsia);
draw(circumcircle(I,F,Y)^^circumcircle(A,G,F),dotted+brown);
draw(A--L,fuchsia);
draw(L--M^^X--N,purple);
draw(T--L,linewidth(0.7));
draw(L--U,dashed+linewidth(0.7));
[/asy]](http://latex.artofproblemsolving.com/5/4/d/54d4992c9352716ba99228834496d44229e8d964.png)
.
,
is further equivalent to points 
being collinear. By Ceva-Menelaus confi in 
,
.
,
with incenter 
,
,
as the respective midpoints of arcs 
,
and 
,
,
.
meet on ![[asy]
//setup;
size(12cm);
pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0);
blu1=RGB(226,235,255); blu2=RGB(196,216,255); lightpurple=RGB(237,186,255); // blu1 lighter
//defn
pair M,N,I,A,X,P,Z; M=(7, 17.35); N=(7,-2.82); I=(2.59, 4.5); A=foot(M,N,I); X=foot(N,M,I); P=foot(I,M,N); Z=orthocenter(M,N,I);
pair reflect(pair P,pair A,pair B){return 2*foot(P,A,B)-P;}
pair F,G,L,D,G1,Z1,K; F=extension(A,X,Z,P); G=foot(M,N,F); L=extension(A,G,Z,I); D=reflect(A,M,N); G1=reflect(G,M,N); Z1=2*P-Z; K=extension(D,I,A,X);
//draw
filldraw(Z--M--N--cycle,blu1,blu); draw(circumcircle(A,M,N),blu);
draw(Z--Z1--M--X--A--N,blu);
draw(A--Z1^^Z--D--X^^D--extension(D,I,N,X)^^N--G, purple);
draw(A--L,magenta); draw(M--L^^N--foot(M,N,K),red);
//label
void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.18),a,linewidth(.3)); label(s,P,v);}
pt("$M$",M,dir(90),blu); pt("$N$",N,dir(-90),blu); pt("$A$",A,dir(130),blu);
pt("$I$",I,dir(90),blu); pt("$X$",X,-dir(70),blu); pt("$P$",P,dir(130),blu);
pt("$Z$",Z,dir(130),blu); pt("$F$",F,dir(50),blu); pt("$G$",G,dir(130),purple);
pt("$L$",L,-dir(50),magenta); pt("$D$",D,dir(50),purple); pt("$G'$",G1,dir(50),purple);
pt("$Z'$",Z1,dir(-90),blu); pt("$K$",K,-dir(40),blu);
pt("",foot(M,N,K),(0,0),red); pt("",extension(D,I,N,X),(0,0),purple);
label("$\Omega$",(13, 16.7),blu);
[/asy]](http://latex.artofproblemsolving.com/f/e/5/fe542458210af78f824f0d90b3a0306b59a764c9.png)
,
;
as the orthocenter of 
,
meanwhile 
,
collinear;
while 
,
also collinear;
be the respective reflections of 
in 
collinear. Now we do yet more harmonic chasing:![\[(A,X;\overline{NK}\cap\Omega,D) \overset N=(I,\overline{NX}\cap\overline{DI};K,D) \overset X=(IZ;FP)=-1\text{ and}\]](http://latex.artofproblemsolving.com/a/9/1/a91187c6850bd6b93f6f478e039ca05b181e5468.png)
![\[(A,X;\overline{ML}\cap\Omega,D)\overset M=(ZI;LZ')\overset A=(MN;GG')=-1,\]](http://latex.artofproblemsolving.com/f/9/2/f928ecbee9da0746b3bd4ea891c62dfab087a907.png)
so the points coincide as desired!
and 
are concurrent on 
meet 
and 
.
are collinear and 
.
.
is cyclic.
Note that 
.
is cyclic.
.
we have that 
are concurrent so 
are collinear. Now we only need to prove 
are collinear. Let 
meet 
.
Also Note that 
so 
so 
so 
are collinear.
are intersection points of 
and 
respectively. So the Miquel point of quadrilateral 
.
, points 
are collinear and if 
Thus quadrilaterals 
and 
are cyclic and by concurrency of radical axes of circles , lines 
are concurrent at point 
and we're done.
m
are collinear. Now , by angle chasing one can see that :
Thus quadrilateral 
and circle 
is the Miquel point of this quadrilateral. Now since quadrilateral 
is cyclic , we can get :
So points 
are collinear. Now if line 
, gives us that points 
are collinear and also again by Pascal on 
, points 
are collinear too. So note that by concurrency of 
and since quadrilateral 
is cyclic too and by Lemma , point 
and 
). 
Also lol, I literally had to take a 2 hour anime (``Samurai Champloo") break and then finally after coming back was I able to get rid of the clog in my head and finally solve the problem :rofl: 
After all this, I got rid of the clog in my mind and then went to try the problem again and it took me quite a bit more of time, buy I managed to solve it. 
,
,
,
,
,
,
.
are collinear. Let 
.
with lines 
gives that 
and as 
,
which gives that 
are collinear. Let 
.
gives that 
which gives 
and so 
are also collinear. Now the from the Diameter of Incircle Lemma, we have that 
are collinear (can be derived from simple length computation and then showing that 
)
and so 
is the angle-bisector of 
.
Now again ( 
with lines 
gives that 
.
and 
is cyclic and similarly 
is also cyclic. Also, 
is also cyclic. Finally, 
is also cyclic. Now Radax on 
gives that 
.
gives 
.
gives 
and then another Radax on 
finally also gives 
.
simply means that 
is the external-angle-bisector of 
which is same as the external-angle-bisector of 
from the fact that ![\[-1=(X,I;F,Z)\overset{T'}{=}(P,G;Q,A).\]](http://latex.artofproblemsolving.com/a/3/8/a38e8a5ab6d47612e4bacda680af10bd9f2bea3e.png)
Now finally we also have that 
(due to the complete quadrilateral of 
)![\[-1=(X,I;F,Z)\overset{N}{=}(T,I;NG\cap MT,M),\]](http://latex.artofproblemsolving.com/0/e/4/0e4e26b18dfaae514de05e6d30e7171206e3a921.png)
which finally gives that 
.![\[-1=(T,I;J,M)\overset{F}{=}(K,I;Y,P)=(P,Y;K,I)\overset{N}{=}(P,G;NK\cap\odot(ABC),A),\]](http://latex.artofproblemsolving.com/b/1/0/b1053761a780d6b9611753cf2d5b6fba628c4a52.png)
which on combining with the result we derived at the end of the previous paragraph finishes the proof.
The quote mentioned below follows from the fact that the 6 points lie on a conic and as 5 points determine a conic, which are already on a circle, the conic is itself the circle and thus the 6th point lies on the circle. Thank you soo much for this lesson!! 
is cyclic, whence 
= 
= 
,
.
reveals that 
,
,
are collinear, so if 
.
,
shows that 
,
,
are collinear so we have that 
and 
concur on ![\[ -1 = (A, D; M, N) \overset{I}{=} (N, P; T, A). \]](http://latex.artofproblemsolving.com/a/7/b/a7bf4aa06fd6347d5adf5e7ed053275a2816650f.png)
Furthermore if 
is harmonic then the problem is completed (uniqueness of harmonic conjugate, perspectivity thru AP cap GT, which lies on IF). Now observe:![\[ (X, A; G, P) \overset{N}{=} (K, A; F, \overline{NP} \cap \overline{AT}) = \overset{P}{=} (D, A; \overline{PF}\cap\omega, N), \]](http://latex.artofproblemsolving.com/3/c/1/3c186a2ebe7667f46f0d8cfa7d9238a8ff29007f.png)
and in particular if ![\[ \measuredangle MPT = \measuredangle MAT = \measuredangle ATM + \measuredangle TMA = \measuredangle MTD + \measuredangle TMA = \measuredangle MAD + \measuredangle TMA = \measuredangle FIM = \measuredangle FIT = \measuredangle FPT, \]](http://latex.artofproblemsolving.com/3/1/b/31bb4841e1ee781c44d6a60d18aa4121c557edcc.png)
so 
.
is isogonal to 
.
,
are cyclic due to right angles.
.![\[ \measuredangle TPN = \measuredangle TIN = \measuredangle MIA = \measuredangle MPA = \measuredangle DPM = \measuredangle DPN \]](http://latex.artofproblemsolving.com/c/6/4/c640e3c38fec961bdbf6db1d4d2b991c2605d1a5.png)
. 
,
,
,
concur at some point 
is cyclic since ![\[ \measuredangle GIP = \measuredangle GIF = \measuredangle GAF = \measuredangle GAT = \measuredangle GDT = \measuredangle GDP \]](http://latex.artofproblemsolving.com/c/6/2/c62efc6915f13b172c7b8edb898d0c1bbac06caa.png)
then radical axis on 
,
,
gives the result. are cyclic due to right angles. 
.
,
,
concur at 
and 
gives the result. 
is a harmonic quadrilateral.
lies on 
.![\[ (NM;AD) \overset{I}= (A,T;N,\Gamma \cap \overline{KID}) = -1 \]](http://latex.artofproblemsolving.com/e/3/f/e3f86c7addc511baf0241708d71a894d8f8ee10d.png)
and that ![\[ (NM;AD) \overset{T}= (RI;FP) \overset{M}= (A,T;\Gamma \cap \overline{MF},N) = -1 \]](http://latex.artofproblemsolving.com/b/1/e/b1e6a0d9601daa0f95fda9737bc070556407fc36.png)
lies on ![\[
(A,T;D, \Gamma \cap \overline{MQL}) \overset{Q}= (TA;GM) \overset{N}= (RI;FP) = -1
\]](http://latex.artofproblemsolving.com/3/3/9/3397950ae9563895d19cb428d081a8db8adfeba0.png)
and that ![\[ (A,T;D, \Gamma \cap \overline{NK}) \overset{K}= (AT;SN) = -1 \]](http://latex.artofproblemsolving.com/5/1/8/5184815d93956a0869374482489f9da16a2d5daa.png)
![[asy]
unitsize(3cm);
pair A = dir(120); pair B = dir(205); pair C = dir(335); pair I=incenter(A, B, C); pair M=dir(90); pair N=dir(270); pair D=dir(60); pair T=IP(Line(I, M, 10), unitcircle, 0); pair TTT =foot(T, M, N); pair P=foot(I, M, N); pair TT=2*TTT-T; pair F= extension(A,T,I,P); pair G=IP(Line(N, F, 10), unitcircle, 1); pair K=extension(A, F, D, I); pair L=extension(A, G, I, F); pair X=extension(A, M, G, D); pair Q=IP(Line(M, F, 10), unitcircle, 1); pair Y=extension(A, T, G, D); pair Z=extension(L, M, N, K);
draw(unitcircle, heavygreen); draw(A--B--C--cycle, heavygreen); draw(A--T, heavygreen); draw(N--G, heavygreen); draw(A--L, heavygreen); draw(A--K--D, heavygreen); draw(M--N, heavygreen); draw(N--Z, heavygreen); draw(M--L, heavygreen); draw(M--X--D, red); draw(P--X--N, red); draw(A--TT, red); draw(D--T, red); draw(M--Q, red); draw(D--Q, red); draw(A--N, heavygreen); draw(M--T, heavygreen);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, N); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$T'$", TT, dir(TT)); dot("$F$", F, NW); dot("$G$", G, dir(G)); dot("$K$", K, dir(K)); dot("$L$", L, NW); dot("$X$", X, dir(X)); dot("$P$", P, E); dot("$Q$", Q, dir(Q)); dot("$Y$", Y, SE); dot("$Z$", Z, dir(Z));
[/asy]](http://latex.artofproblemsolving.com/c/8/8/c88011b4482c16dfe2a543823e8c02bc6ea2dc9b.png)
,
.
,
we see that 
.
are concurrent
we find that 
lies on 
and 
are reflections across 
and 
meet on 
and 
.
and 
,
.
Therefore 
,
are concurrent
and 
concur at the point at infinity along line 
and 
.
.
and 
.
and this implies 
,
,
and sends 
proving the claim.
is harmonic.
line 
gives 
lies on 
lives 
is on 
gives 
is on 
intersect on 
But as 
ray 
bisects 
hence 
are collinear.
gives 
Also, Pascal on 
where 
is the intersection of 
collinear.
Then,![\[-1=(AT;GM)\overset{F}{=}(TA;HN)\overset{K}{=}(AT;PD)\]](http://latex.artofproblemsolving.com/f/8/8/f885ac755d17ed36da3da4679ed8f4ebd5812576.png)
Now, let 
be the reflection of 
which by symmetry lies on line 
Then, lines 
and 
are symmetric about 
Thus, if 
we have 
so![\[\angle LAI=\angle GAB+\angle BAI=\angle CAI+\angle CAX'=\angle IAX'.\]](http://latex.artofproblemsolving.com/d/2/6/d265b4d8bfc7ceb0295c0cfcbb02a5660b6f88a5.png)
Then, as 
so 
Now, from the above harmonic bundle we have 
implying 
collinear. Thus lines 
intersect at 
which is on 
and 
,
,
is the A-mixtilinear intouch point of 
are colinear, also 
so 
.
and 
are colinear.
so 
is cyclic hence by Reim's on 
we get 
colinear:
Thus we are done 
.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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draw((-1.1764612326917248,2.1702199410229266)--(-1.1869019636037623,-1.253594116494785), linewidth(0.9) + wwqqcc);
draw((-1.1869019636037623,-1.253594116494785)--(-1.192683562367836,-3.1495455867639897), linewidth(0.9) + qqttcc);
draw((-1.1764612326917248,2.1702199410229266)--(2.2959769984063434,-3.1398950734597455), linewidth(0.9) + qqttcc);
/* dots and labels */
dot((-1.1764612326917248,2.1702199410229266),dotstyle);
label("$A$", (-1.2543352321169878,2.284937672900228), NE * labelscalefactor);
dot((-2.2922035259762987,-1.9155969061141573),dotstyle);
label("$B$", (-2.4901969954542125,-2.0686662645189635), NE * labelscalefactor);
dot((3.388653361662767,-1.8998822259328612),dotstyle);
label("$C$", (3.4644096824433257,-2.0031281407298573), NE * labelscalefactor);
dot((-0.3960858022931656,-0.46368262400914606),linewidth(4pt) + dotstyle);
label("$I$", (-0.3461640878464513,-0.6268275411586292), NE * labelscalefactor);
dot((0.5355411187616752,2.677453803655398),linewidth(4pt) + dotstyle);
label("$M$", (0.4871063434945564,2.7717923067621593), NE * labelscalefactor);
dot((-1.192683562367836,-3.1495455867639897),linewidth(4pt) + dotstyle);
label("$T$", (-1.3011481776979432,-3.3700661511883565), NE * labelscalefactor);
dot((2.2959769984063434,-3.1398950734597455),linewidth(4pt) + dotstyle);
label("$T'$", (2.3596241667327758,-3.323253205624709), NE * labelscalefactor);
dot((-1.1844998230464998,-0.4658635756563433),linewidth(4pt) + dotstyle);
label("$F$", (-1.1232589844903125,-0.3646750460022047), NE * labelscalefactor);
dot((0.553092370638971,-3.6673237523332007),linewidth(4pt) + dotstyle);
label("$N$", (0.5151941108431297,-3.866283374163017), NE * labelscalefactor);
dot((-2.110941040256875,1.241075874065847),linewidth(4pt) + dotstyle);
label("$G$", (-2.2842200348980084,1.3018658160636365), NE * labelscalefactor);
dot((2.2503235210467407,2.179699290269044),dotstyle);
label("$D$", (2.284723453803247,2.2755750837874986), NE * labelscalefactor);
dot((-3.8350570442277574,-0.4731956841492926),linewidth(4pt) + dotstyle);
label("$L$", (-4.016299021393362,-0.6551138826931464), NE * labelscalefactor);
dot((-1.1869019636037623,-1.253594116494785),linewidth(4pt) + dotstyle);
label("$K$", (-1.476048890627472,-1.4571094919515259), NE * labelscalefactor);
dot((-2.4626528494792534,0.5161344743313137),linewidth(4pt) + dotstyle);
label("$X$", (-2.6680861886618437,0.49668315236890426), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/c/0/b/c0b69a0c295f22287dbaca8540da28e71692ca82.png)
.
we find that 
.
.
.
However this reduces to,
as desired.
and 
,
and 
,
be the reflection of 
,
.
is cyclic, as![\[\measuredangle GFI = \measuredangle GNN = \measuredangle GAI.\]](http://latex.artofproblemsolving.com/8/5/2/852d79f4bd140ad9d3a076e7926be2fae4db7420.png)
,![\[\measuredangle MXF = \measuredangle MIF = \measuredangle TNM = \measuredangle MNU = \measuredangle MXU,\]](http://latex.artofproblemsolving.com/1/9/4/194cb6e25ee03e2f54b9e68f8106a7948352e0f8.png)
,
,![\[(AN;UT) = \frac{UA}{TA} = \frac{TD}{TA} = \frac{PD}{PA} = (AD; \infty P) = (AK;FT).\]](http://latex.artofproblemsolving.com/a/d/d/adda9178e9b9ccedb005201805208ac81a874afd.png)
and 
is cyclic. Let 
is cyclic then we are done.
by Reim we only need 
on the same line. Let 
intersact the circle at another point 
we only need 
by angle chasing. This can be proved by using Menelaus on 
and 
.
.
.
AM, GD, FI and TN are concurrent at X. Now we will show that FI, MN, TD and T'A are concurrent. By Pascal on AMNGDT we get that 
.
and AD = TT', T'A, TD and MN are concurrent 
by phantom points. Let 
and 
.
.
.
.
by phantom points. Let 
and 
.
and 
,
,
,
.
and 
be the tangency points of the mixtilinear incircle to 
and 
be the arc midpoints of 
be the intersection of the tangents at 
and 
,
.
is an isosceles trapezoid, 
,
be the 
and 
be the intersections of 
with 
be the midpoint of 
be the midpoint of ![\[(R_AX_A;P'Q') = \frac{R_AP'}{X_AP'}\cdot \frac{X_AQ'}{R_AQ'} = \frac{AR_A}{EX_A}\cdot \frac{E'X_A}{AR_A} = -1.\]](http://latex.artofproblemsolving.com/1/0/e/10e9e3efe17c85da22eb161579ea3a79d3a1a9ff.png)
Also, 
.
that satisfy both properties, which is the projections of the incenter and 
.
and 
,
,
intersect on 
.
and 
,
,
are on 
and 
.
.
is cyclic, so 
.
times
,
(this is the 
gives 
,
,
collinear. This implies that 
lies on 
gives 
,
,
collinear. But 
passes through 
are collinear.
.
gives 
,
,
collinear. But 
,
collinear.
are collinear. Probably Pascal spam again (which is the motivation for everything really).
gives us that 
and 
intersect at a point 
.
.
gives that 
,
,
are collinear. But 
,
holds true.
.
holds. Using Pascal's on
gives 
,
,
are collinear. Note that 
passes through 
.
passes through 
has to be true.
is also 
.
and we are done.
are collinear. Also note that 
are collinear. 
.
.
are symetric to 
are collinear.
.
is an involution. Project onto 
is equavilent to showing the concurrency of 
and the tangent to 
,
is an orthogonal system thus, 
are concurrent.![\[(A,T;G,M)\overset{N}{=} (I,NT\cap IJ;F,P)\overset{A}{=} (N,M;T,U)=-1\]](http://latex.artofproblemsolving.com/8/5/5/855956006955b1ce9942683e558c1174e0255f7c.png)
Hence 
are concurrent. Apply pascal on 
to see that 
are collinear. Since the line passing through both 
and 
is 
,
are concurrent![\[\frac{NJ}{ND}.\frac{KD}{KI}.\frac{VI}{VJ}=\frac{NJ}{ND}.\frac{AD}{IF}.\frac{VF}{VI}=\frac{NJ}{IF}.\frac{IJ}{NJ}.\frac{VF}{IV}=\frac{IJ}{IV}.\frac{VF}{IF}=1\]](http://latex.artofproblemsolving.com/5/7/7/577bde3ac74b737f22e9d60b79506499d869c418.png)
Thus, 
are collinear. Let 
intersect 
to get 
are collinear. Hence 
are collinear which is equavilent to 
as desired
,
,
.
implies 
cyclic.
gives us: 
.
.
gives us gives us 
where 
(well known incenter config). S is where the tangent line from 
.
gives the intersections to the tangents of 
,
cyclic. So by radical axis 
.
.
,
So by prism lemma 
collinear. Finishing.
and 
concur on 
and 
be the second intersection of line 
with 
are equal by symmetry, it follows that points 
and 
,
, 
, 
and 
are collinear. Thus, the tangents to 
,
, 
and 
are collinear. Thus, 
passes through the intersection of the tangents to 
is harmonic. Thus,![\[-1=(AT;GM)\overset{I}{=}(N,M;,T,GI \cap \Omega)\]](http://latex.artofproblemsolving.com/9/2/a/92ab61dd819aee86b92367763025d9f60fd16b9b.png)
which implies that points 
to a rectangle while preserving its circumcircle.![[asy]
import geometry;
size(10cm);
defaultpen(fontsize(9pt));
pair foot(pair P, pair A, pair B) {
return foot(triangle(A,B,P).VC);
}
pen pri; pri=RGB(24, 105, 174);
pen sec; sec=RGB(217, 165, 179);
pen tri; tri=RGB(126, 123, 235);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pair A=dir(110);
pair T = dir(250);
pair N = dir(290);
pair M= dir(70);
pair I = intersectionpoint(line(A,N),line(T,M));
pair F = midpoint(A--T);
pair G = intersectionpoints(circle(A,T,N),line(N,F))[1];
pair L = intersectionpoint(line(A,G),line(F,I));
pair D = intersectionpoints(line(foot(G,M,N),G),circle(A,M,N))[1];
pair D_1= intersectionpoints(line(D,I),circle(A,M,N))[0];
pair K = intersectionpoint(line(A,T),line(D,D_1));
pair X = intersectionpoint(line(L,M),line(K,N));
filldraw((path)(A--T--N--M--cycle), white+0.1*pri, pri);
filldraw(circumcircle(A,N,M), tfil, tri);
draw(A--L,sec);
draw(L--I,sec);
draw(G--N,pri);
draw(D_1--D,pri);
draw(D_1--M,pri+dashed);
draw(L--M,dashed+tri);
draw(X--N,dashed+tri);
dot("$A$",A,dir(A));
dot("$T$",T,dir(T));
dot("$N$",N,dir(N));
dot("$M$",M,dir(M));
dot("$I$",I,dir(90));
dot("$F$",F,dir(170));
dot("$G$",G,dir(G));
dot("$L$",L,dir(L));
dot("$D$",D,dir(D));
dot("$D'$",D_1,dir(D_1));
dot("$K$",K,dir(340));
[/asy]](http://latex.artofproblemsolving.com/3/f/1/3f1ccc41e51d391d01cc8cb475550aca20e227b6.png)
,
and 
, 
and 
.
, 
concur.
.
.![\[\frac{GL}{AG}=\frac{FL}{F'F}= \frac{FL}{AM} = \frac{FS}{AS}\]](http://latex.artofproblemsolving.com/c/9/3/c932134e2cdbbba5690c3f2e1e77d7f06d761a63.png)
which implies that 
.
intersect at the center 
which implies that 
as well. This implies that points ![\[-1=(MN;AD)\overset{I}{=}(TA;ND')\overset{M}{=}(T,A;\infty , AT \cap MD')\]](http://latex.artofproblemsolving.com/9/9/2/99207950eec6bcacf84d6f6d0c7e81bef9291c4e.png)
which implies that 
bisects segment 
.
,
, 
and 
are collinear. Thus, points 
and 
,
,
.
,
,
,
concur at 
,
,
meet at a point 
;
by definition. On the other hand, setting 
and using the fact that 
passes through ![\[-1 = (AD; NM) \stackrel I= (NP; AQ) \stackrel A= (IR; F'U_2).\]](http://latex.artofproblemsolving.com/0/f/5/0f55c0c1342c2ed0014364feeb8775c032ad0dda.png)
So 
,
.
.
(or Reim's theorem 
is tangent to 
.
at 
,
is the radical center of 
,
.
.
.
.
.
so 
,
.
,
,
,
are collinear along the line 
.
