,
and 
,
an obtuse triangle can be formed?
and 
.
is an integer, and let 
be 
integers such that 
for any two integers 
and 
satisfying 
.
.
of the form 
with 
,
Determine all ordered pairs of polynomials 
with integer coefficients such that 
.
Y by narutomath96, Viswanath, lephuongnam1568, anantmudgal09, Adventure10, GeoKing, Funcshun840, Rounak_iitr, and 5 other users
Let 
be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from 
; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
(Russia) Fedor Ivlev
be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from 
; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.(Russia) Fedor Ivlev
and 
, 
similarly . Also suppose that 
is tangent to 
at 
. Solution consist of two steps , first I'm going to show 
are concurrent at 
which is lie on 
.And then show that 
which finish the proof .
cut 
at 
, then 
and so 
which is a fix line. On the other hand 
WRT 
, according to properties of pole and polar we deduce that 
and 
.
. It's well know that 
are midpoint of their corresponding arc .So 
.Besides , As we know 
are concurrent we deduce 
and 
are perspective 
)
or equivalently 
are concurrent at radical axis of 
it suffice to show that 
is cyclic .
be a line and 
be circle.Suppose 
be inscribed in 
be the circumcevian triangle of 
.
we get 
are collinear.
and 
are perspective.So 
are collinear.Hence we conclude that 
are collinear.But this line is nothing but the polar of 
.
,
is the radical axis of 
and 
.
are concurrent.
is its intouch triangle
respectively.Now let tangents at 
at 
respectively.Then its easy to show that 
is the radical axis of 
.
wrt 
concur at the pole of 
.
are concurrent ,say at 
.
wrt 
are collinear, their polars i.e 
are concurrent.So 
are perspective wrt the perspective axis of 
.
.
we conclude that the pole of the radical axis of 
of the three circles . To show that this radical centre is on 
, consider the inversion w.r.t to a circle centered at 
and then the reflection w.r.t 
are themselves and the incircle becomes the larger circle tangent to them ( internally ) . From the property of this transformation ( inversion + homothety ) , 
and the centre of this circle are collinear . We now show that this centre is actually 
!
be the mid-pt of arc 
that does not intersect the incircle . 
are defined in similar mannar . It is well-known that the power at 
to the incircle is equal to 
and 
, so 
and 'circle' 
lie on the radical axis of the incircle and 'circle' 
be the mid-pts of arc 
( not containing the third vertex of 
. ) of 
respectively . Then , we deduce that 
is homothetic to 
and the centre of homothety is the intersection of the perpendicular bisectors of 
, which is 
and the centres of these two circles lie on the perpendicular bisector of 
which passes through 
. In other words , 
is pedal triangle of 
is antipedal triangle of 
cut pedal circle 
again at 
.
intersect again at 
.
.
are concurrent.
be the intersections of 
with 
are concurrent.
on 
are concurrent if and only if 
. It can easily proved by trigonometric Ceva theorem .
,
,
,
.
,
at 
,
,
,
passes through 
,
,
are concurrent.
.
,
intersect at 
.
and 
intersect at 
,
is the polar of 
are all concurrent. Call this point 
.
,
and 
the same way. Since 
is cyclic, 
.
and 
,
is homothetic to 
wrt 
to 
bisects arc 
and 
similarly; then note that from the homothety from 
is the incircle of the triangle formed by the intersections of 
be the circumcenter of 
,
is the perpendicular bisector of 
and 
bisect 
.
passes through 
and 
and 
intersect on the radical axis of (O) and (I), which means that if M is the pole of this radical axis with respect to (I), then M (which clearly lies on 
.
,
and 
,
is on 
.
,
be the points where the incircle touches 
.
.
and similar, so 
are homothetic about 
concur at 
and similar.
the inverted image of 
is the 
.
is a line through 
,
.
are tangent to this mixtilinear excircle at 
respectively, so by Brianchon's theorem on circumscribed hexagon 
,
intersect at some point ![\[\frac{\sin BAA'}{\sin CAA'}=\frac{\sin (F)_V(A)_VP_A}{\sin (E)_V(A)_VP_A}=\frac{\sin A(F)_VC}{\sin (E)_V(F)_VC}\frac{\sin (F)_V(E)_VB}{\sin A(E)_VB}=\frac{\sin ACF}{\sin EFC}\frac{\sin FEB}{\sin ABE}=\frac{EF}{EC}\frac{FB}{EF}=\frac{y}{z}.\]](http://latex.artofproblemsolving.com/c/a/a/caa98e4c813c9b366939e527fc042b116b3ca9cc.png)
.
be the foot of the altitude from 
to 
.
so by the homothety, 
are collinear. Now it suffices to show that 
.![\[\frac{FD'}{ED'}=\frac{DF}{DE}\frac{\cos EFD}{\cos FED}=\frac{DF}{DE}\frac{\sin \frac{C}{2}}{\sin \frac{B}{2}}=\frac{2y}{2z}=\frac{y}{z}\]](http://latex.artofproblemsolving.com/e/7/3/e738d2bd1ec222e2ed2137fdeede4cb45b36a560.png)
,
collinear (by, for example, the ratio lemma and phantom pointing), so 
pass through 
.
,
be the tangency points of those circles with the incircle. Let the tangent 
through 
similarly. (Note that 
.
wrt incircle. Then, by radical axis theorem, 
.
are collinear. But we get 
are also collinear by pascals, so 
are collinear. We conclude 
.
.
.
is the angle bisector of 
.
,
(because 
). Likewise, 
is the radical axis of 
.
,
be the orthocenter of 
,
is the common radical axis of 3 circles 
,
lies on the Euler line of 
and 
respectively. Let 
,
,
.
,
is the midpoint of the arc 
not containing 
is on the radical axis of 
,
.
with center 
.
,
is cyclic. Noting 
gives 
,
.
pass through 
and 
,
. Hence 
in 
are tangent to the incircle at 
intersects 
similarly. Clearly 
is homothetic to the intouch triangle. But 
,
is tangent to 
and 
,
,
be the exsimilicenter of the two circles. Then 
and 
.![\[
\frac{SF}{SE}=\frac{SP}{SN}
\]](http://latex.artofproblemsolving.com/3/4/8/34884ab910d1690e8da3dacf6eacd7a6963942b3.png)
![\[
SY\cdot SF = SE\cdot SZ
\]](http://latex.artofproblemsolving.com/5/4/0/540bbcfbcf6cec63bcdf7dacfb5bc75a95651e44.png)
![\[
SY\cdot SP = SZ\cdot SN
\]](http://latex.artofproblemsolving.com/5/8/4/5842d7fc20511563f02883eff820327c18294aa7.png)
.
.
.
be the power of a point 
with respect to a circle 
( maps 
) 
is the midpoint of arc 
is the midpoint of arc 
,
in 
,
,
is the radical axis of 
pass through the midpoint of 
and 
.
is B-midline, C-midline of 
,
are homothetic .
be the homothety center of 
and 
.
are concyclic ,
is the radical center of 
.
and 
are concurrent at 
and 
are homothetic 
.
pass through the A-excenter 
, B-excenter 
, C-excenter 
of 
,
is 
( in 
Denote 
the tangents of 
and 
Since 
then 
and similarly 
and 
are perspective through the radical center 
of 
of 
then there exists a homology sendind 
Thus in this figure, 
concur at 
their common midpoint. Hence, in the original figure 
concur at 
lying on 
(incircle) with 
be midpoints of big arcs 
.
,
be intersection of tangents to 
and 
.
,
are collinear and by Pascal at 
,
are collinear, so 
are collinear. But 
have the same power of a point to 
and so that line is the radical axis, so 
.
are collinear! So 
but 
so 
.
are homothetic! At 
too.
are the perpendicular bisectors of 
.
,
.
.
are collinear!
are concurrent, which is true!
,
.
,
be the intouch triangle of 
be the incircle. Now 
and 
touch to 
and 
respectively, and the common tangents from this points cut at 
,
we get 
and 
and 
,
,
we get 
,
(defined analogously) with respect to 
,
,
,
,
.
intersect 
,
and 
similarly. By homothety, we have that 
swaps line 
.
is homothetic with 
be the center of homothety of 
,
and because it's the center of homothety 
,
as the excenters of triangle 
.
at 
bisects angle 
.
and 
.
and from Fact 5, we know 
which is symmetric. It follows that 
denote the incircle, circumcircle of triangle 
,
be the intersections of 
respectively.
,
.
and 
.
.
be the intersection of 
and 
,
are collinear. Thus we only need to show that 
are concurrent, where 
are defined similarly.
.
.
be the intersection of lines 
.
similarly. Note that 
.
concur at 
to be the 
,
to be the touchpoint of 
to be the midpoint of 
.
similarly.
colinear
,
concur at a point 
,
to 
.
are collinear (they all lie on the Euler line of 
)
)
are colinear. So together, we have 
colinear, or 
.
is symmetric in 
.
for homothety reasons.
, which is evidently symmetric in 
and 
is tangent to 
similarly. Now, define 
,
,
.
and its cyclic counterparts are concurrent.
and 
as 
and the intersection of 
and 
as 
.
and 
we have that 
.
and 
.
.
.
,
.
of ![\[PX\cdot PX' = PX\cdot PD\cdot\frac{k+1}{2} = \frac{k+1}{2}\mathrm{Pow}_{(DEF)}(P),\]](http://latex.artofproblemsolving.com/a/2/2/a22bab6cff916315bdb60f0c3d392fb769422774.png)
which is symmetric in 
are collinear (here 
)
collinear, it suffices to show that 
collinear, or that 
,
to the intersection of the polar of 
with respect to ![\[-1=(AU;M\infty) = (SD;T'\infty),\]](http://latex.artofproblemsolving.com/8/7/1/8715a013e2e092d69567dcb94f0f1d2809d2dd9e.png)
where 
,
is the intersection of the polar of 
,
.![\[TD^2=TX^2=TB\cdot TC,\]](http://latex.artofproblemsolving.com/a/b/8/ab8ada06f7cd054cb579a50a71bbb64604370397.png)
so 
,
.
,
,
.
are the internal and external angle bisectors of 
.
is the perpendicular bisector of 
,
,
.
similarly. Furthermore, let 
and 
the midpoint of 
are defined similarly.
are collinear, where 
satisfies 
,
,
,
.
(all of them have same power of point WRT 
)
,
are concurrent. Let 
.
,
.
,
are collinear. By Pascal's Theorem on 
and 
,
lies on 
.
,
lies on 
.
similarly. It is enough to show that 
concur. Clearly 
lies on 
concur. Now the problem ought to be true no matter what configuration 
form: take a homography sending 
to the center of 
and 
.
respectively, and let 
and 
and 
respectively. Also, let 
and 
noticing 
by Radical Axis. By ISL 2002/G7, 
and 
and by Vietnam TST 2003/2, 
and 
Also, 
by Pascal on 
and 
respectively. Similarly, 
and 
![[asy]
//setup
size(13cm);
pen blu,grn,lightblu; blu=RGB(102,153,255); grn=RGB(0,204,0); lightblu=RGB(196,216,255);
//defns
pair A,B,C,I,O,D,E,F,H; A=(-1,10); B=(0,0); C=(14,0); I=incenter(A,B,C); O=circumcenter(A,B,C); D=foot(I,B,C); E=foot(I,C,A); F=foot(I,A,B); H=D+E+F-2*I;
path w=incircle(A,B,C);
pair Ka,Kb,Kc; Ka=extension(E,F,B,C); Kb=extension(D,F,A,C); Kc=extension(D,E,A,B);
path ya,yb,yc; ya=circle((Ka+D)/2,distance(D,Ka)/2); yb=circle((Kb+E)/2,distance(E,Kb)/2); yc=circle((Kc+F)/2,distance(F,Kc)/2);
pair Ja,Jb,Jc; Ja=foot(D,Ka,2I-D); Jb=foot(E,Kb,2I-E); Jc=foot(F,Kc,2I-F);
path wa,wb,wc; wa=circumcircle(B,C,Ja); wb=circumcircle(C,A,Jb); wc=circumcircle(A,B,Jc);
pair Ta=foot(D,E,F); pair U=2*circumcenter(I,Jb,Jc)-I;
pair X=extension(D,Ja,E,Jb);
//draw
filldraw(D--E--F--cycle,lightblu,blu);
draw(A--B--C--A--D--E--F--D--Ka--F,blu); draw(w,blu+dashdotted);
draw(D--Ta,purple+dotted);
draw(ya,purple); draw(yb,purple); draw(yc,purple);
draw(4*I-3*H--3*H-2*I,linewidth(1));
draw(D--Ja^^E--Jb^^F--Jc,grn+linewidth(1));
draw(wb,red); draw(wc,red);
draw(Jb--U--Jc,red);
draw(A--U,magenta+linewidth(1));
clip((-7,-15)--(-7,12)--(13,12)--(15,10)--(15,-15)--cycle);
//label
label("$A$",A,dir(120)); label("$B$",B,-dir(80)); label("$C$",C,dir(-50));
label("$D$",D,dir(120)); label("$E$",E,dir(90)); label("$F$",F,dir(120));
label("$K_a$",Ka,-dir(20));label("$J_a$",Ja,dir(20));
label("$J_b$",Jb,-dir(80)); label("$J_c$",Jc,dir(70));
label("$U$",U,-dir(90)); label("$X$",X,dir(-90));
label("$\textcolor{red}{\omega_b}$",(14,1.2)); label("$\textcolor{red}{\omega_c}$",(-4,8));
label("$\textcolor[rgb]{.6,0,1}{\gamma_a}$",(-5,2.8));
label("$\textcolor[rgb]{.6,0,1}{\gamma_b}$",(6.5,10.5));
label("$\textcolor[rgb]{.6,0,1}{\gamma_c}$",(6,-4));
[/asy]](http://latex.artofproblemsolving.com/6/5/3/653514f9e4393016e2c01db39e6bbe8a987261a9.png)
,
,
denote the orthocenter and Euler line of 
is just 
,
(which exists since the former 3 circles are coaxial). We'll show this is the desired concurrency point. Also, define 
and variants. Clearly, 
is the raxis of 
,
.
denote the foot of 
,
( = variants) wrt the 
wrt all circles by orthogonality, hence also on raxis, done. 
meet @ 
is the raxis of 
.
,
,
and 
,
by a well-known lemma. Finally, let 
and 
similarly.
,
are all parallel, and cyclic variants hold. Thus 
;
are homothetic. Since 
is the perpendicular bisector of 
is on 
,
,
that 
,
,
.
,
be the reflection of 
is cyclic, 
,
is cyclic. Since 
,
,
,
is homothetic to 
,
is the ratio of homothety from 
as 
is a harmonic quadrilateral. Which implies that 
and take 
it becomes clear that 
it further implies that 
is tangent to 
is angle bisector of 
and midpoint of 
and 
as 
,
.
which comes from the fact that triangles 
is cyclic and
and 
being parallel implies the result. Now the homothety form 
is tangent to incircle 
at 
,
,
concur on 
.
and 
respectively, are collinear. Note however 
,
are the same. Hence 
and 
’
and 
to 
and 
)
e
e
and 
are collinear
is the homothetic center of the intouch triangle and the excentral triangl
passes through the midpoint of arc 
,
,
(
and we know 
is on the radical axis of 
are colinear 
and 
the excentral triangle. Let 
, 
and 
respectively.
(and similarly) are collinear and point 
and 
(and similarly) and also from 
and 
intersect at the 
E
which lies on 
is also homothetic to these two triangles. Thus, we only need to show the following claim.![\[X_{57}P \cdot X_{57}L=\frac{X_{57}P \cdot X_{57}I_a \cdot X_{57}L}{X_{57}I_a} = \frac{X_{57}B\cdot X_{57}E_{Ob}\cdot X_{57}L}{X_{57}I_a} = \frac{X_{57}B \cdot X_{57}E_{Ob}\cdot X_{57}N}{X_{57}I_c} = \frac{X_{57}R \cdot X_{57}I_c \cdot X_{57}N}{X_{57}I_c}= X_{57}R \cdot X_{57}N\]](http://latex.artofproblemsolving.com/5/f/d/5fdc97e79b03da98ac09be4b2ea7f8e758d0e293.png)
which implies that 
is the Radical Axis of circles 
and 
are the other two pairwise Radical Axes. Thus, these lines 
