such that 
divides 
.
.
be the foot of the altitude from 
.
and 
be points on the circumcircles of triangles 
and 
,
is tangent to 
,
.
.
such that for all 
,![\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]](http://latex.artofproblemsolving.com/f/9/3/f93b7526ff4338eea89a1c59bc009aa001ffc15d.png)
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
.
be the second intersection of the circumcircles of triangles 
and 
.
,
,
are concurrent.
,
,
is not greater than 
.
.
.
,
is a positive integer. Let 
be the set of all points 
,
is a real number. The path that 
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
,
and 
although its supposed to be a quick comparison.
![$ a,b \in [0 ,1] . $](http://latex.artofproblemsolving.com/9/1/9/91980d5ca5d97fe4b61797e6fbb378211674d9cd.png)
Prove 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 
.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 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 
be the contact triangle of 
be the excentral triangle. We claim the desired point is 
and 
.
.
.
,
.
passes through 
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 ![\[-1=(AU;M\infty) = (SD;T'\infty),\]](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 
,
.
,
.
,
.
and 
are the internal and external angle bisectors of 
.
is the perpendicular bisector of 
is the arc midpoint of 
,
,
,
be the tangency points of 
,
is tangent to 
and 
similarly. Now, define 
,
,
.
and its cyclic counterparts are concurrent.
- note that since the bevan point lies on 
and 
as 
and the intersection of 
and 
as 
.
and 
we have that 
.
the touching points 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, 
lie on 
and 
and by Vietnam TST 2003/2, 
and 
Also, 
by Pascal on 
and 
respectively. Similarly, 
and 
![[asy]
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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);
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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);
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//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 
.
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 
Let 
be midpoint of 
.
and 
similarly.
and analogous. Obviously 
and we will prove that 
bisector of 
then 
is 
and so 
and 
.
lies of 
as required. 
is also homothetic to 
;![\[\text{Pow}(T,\omega_A)=TX \cdot TD'=\frac{\text{Pow}(T,\omega)}{TD} \cdot TD'=\lambda\text{Pow}(T,\omega)\]](http://latex.artofproblemsolving.com/b/9/b/b9bfe92a2addb17b18f3b1bc491347297402b4a1.png)
Which is symmetric in 
be the orthic triangle.
Define 
similarly.
with the incircle. 2002 ISLG 7 implies 
similarly.
be the second intersection of 
with 
By curvilinear properties, 
Define 
similarly.
denote the midpoint of arc 
Let 
are collinear.
have circumcenter 
This moves 
Thus, 
which is fixed. Thus, 
are homothetic, but since 
is the midpoint of respective sides, it follows 
Define 
are collinear due to them being the circumcenters of homothetic triangles.
where the third equality follows from radax on the incircle as well as similar triangles. This implies 
Similar reasoning finishes.![[asy]
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