= 2∠AMP.
,
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of 
(which may depend on 
)
.
be a positive integer. Consider 
points on the circumference of a circle.
}
and 
in this set at last in one of the permutation
Y by canhhoang30011999, harshmishra, Adventure10, Mango247, GeoKing
The incircle of a non-isosceles triangle 
with the center 
touches the sides 
at 
respectively. The line 
meets the circumcircle of at 
. The line 
meets the line 
at 
and the line 
meets the circumcircle of at 
. Define 
similarly. Prove that the lines 
are concurrent.
with the center 
touches the sides 
at 
respectively. The line 
meets the circumcircle of at 
. The line 
meets the line 
at 
and the line 
meets the circumcircle of at 
. Define 
similarly. Prove that the lines 
are concurrent.
is not harmonic in general. The concurrency point is not the Lemoine point of 
but its center 
i.e. the perspector of 
be the projections of 
on 
and 
and the arc 
of the circumcircle 
we get 
is cyclic 
Furthermore, 
implies that 
bisects 
externally or 
Since 
then 
therefore isosceles 
are similar with corresponding cevians 
and 
Hence since 
is cyclic on account of the right angles at 
we deduce that 
are collinear and analogously 
and 
By Cevian Nest Theorem the conclusion follows.
is circumcircle of triangle 
.
is center of spiral similar goes 
to 
.
is on the ray 
.
.
is tangent to the circle 
and 
is tangent to the circle 
.
and 
.
we have 
and 
.
.
to get that,
.
and using them in (4), we get that ,
.
are concurrent.
To jayme, yes indeed, my solution shows this directly. Let the perpendicular to 
,
.
are collinear and perpendicular to 
,
are concurrent at 
of 
,
are collinear . ... 
is the external bisector of 
,
we get 
.
are concurrent
) :
be the projection of 
on 
and 
be the midpoint of arc 
.
,
,
is the bisector of 
and 
.
,
is the bisector of 
,
.
,
,
we get 
.
of 
and 
,
from my post, we can do this:
be the radical axis of 
.
be 
and note that 
by harmonic conjugates (fact 1). So, note 
for obvious reasons, and because of the fact 1 we conclude 
meets 
contains the foot of the 
altitude of 
;
concurrent.
.
,
we get that 
is harmonic. This yields 
,
is the bisector of 
,
.
which is equivalent to the fact that 
are concurrent.
be a circle tangent to 
and tangent to 
at some point, say 
.
similarly. We claim that 
etc.
denote the midpoint of arc 
denotes the homothety centered at 
are collinear. Since 
is a diameter of 
.
is a harmonic pencil. But 
and hence 
is the bisector of 
,
.
gives 
are collinear, so 
.
etc. are concurrent at the 
lie on same line 
.
images of points 
wrt this projective transformation. So we have that 
,
,
.
,
,
concurrent at point 
.
,
,
concurrent at point which is isogonal to 
.
and hence 
so we're done.
denote lengths 
,
respectively, 
be the semiperimeter of 
denote the circumcenter and circumradius of 
,
,
is the polar of 
wrt 
,
or
so 
,
or
Now, by the Incenter-Excenter Lemma, we know that 
.
(tangents), we must have that 
.
be the point at infinity along 
or
and symmetry implies that 
and 
are also satisfied.
Which implies the concurrence by the trigonometric form of Ceva's Theorem, AWD.
are the midpoints of 
respectively and let 
be the midpoints of 
respectively. Then 
.
.
is the foot of perpendicular from 
concur at 
from this problem 
is the orthic triangle of 
,
.
and 
.
intersects the circle again at 
,
intersects at 
Thus, 
is an angle bisector, and 
.
,
![[asy]
import graph; size(15cm);
real a=105;
real b=205;
real c=335;
real r=10;
path omega=circle(origin, r);
pair A=r*dir(a);
pair B=r*dir(b);
pair C=r*dir(c);
pair I=incenter(A,B,C);
pair A1=foot(I,B,C);
pair B1=foot(I,C,A);
pair C1=foot(I,B,A);
dot(A1);
dot(B1);
dot(C1);
draw(A--B--C--cycle);
draw(omega,blue);
pair X=10*C1-9*B1;
pair P=10*B-9*C;
pair A3=intersectionpoints(B1--X,B--P)[0];
pair A2=r*dir((b+c)/2);
pair B2=r*dir((a+c)/2+180);
pair C2=r*dir((a+b)/2);
dot(A2);
dot(B2);
dot(C2);
pair A4=intersectionpoints(A2--A3,omega)[1];
draw(A2--A3);
dot(A4);
draw(B1--A3);
draw(A3--A2, magenta);
pair D=(B+C)/2;
dot(D);
draw(B--A3);
draw(A1--A4--A2--D--cycle,red+ linewidth(1));
draw(A--A4,green);
label("$A$",A,dir(90));
label("$B$",B,dir(-135));
label("$C$",C,dir(-45));
label("$D$",D,dir(-45));
label("$A_1$",A1,dir(90));
label("$A_2$",A2,dir(-90));
label("$A_3$",A3,dir(-90));
label("$A_4$",A4,dir(-90));
label("$B_1$",B1,dir(45));
label("$C_1$",C1,dir(-45));
draw(B--A4--C,dotted);
label("$B_2$",B2,dir(45));
label("$C_2$",C2,dir(135));
[/asy]](http://latex.artofproblemsolving.com/b/b/8/bb8a97e3f97d77c934ebfc5768d27d52bb75409d.png)
,
(
,
is cyclic. Because 
,
.
bisects 
.
(
by Ceva's theorem.
, 
and 
.
are collinear.
.
we get that 
is tangent to both 
and 
.
.
.
.
are concurrent by the converse of Ceva's Theorem. 
.
and 
so 
are collinear.
be the feet of the 
-
-
.
denote the orthocenter of 
are concurrent, which finishes. 
.
.
.
,![\[ -1=(BC;A_3A_1)\stackrel{A_2}{=}(BC;A_4X)\implies \frac{BX}{CX}=\frac{BA_4}{CA_4} \]](http://latex.artofproblemsolving.com/5/6/b/56b860053dc17eaece66d7e99a8c402033d04ab9.png)
Now we have ![\[ \frac{\sin \angle BAA_4}{\sin \angle CAA_4}=\frac{\sin \angle BCA_4}{\sin \angle CBA_4}=\frac{BA_4}{CA_4}=\frac{BX}{CX}=\frac{BA_1}{CA_1} \]](http://latex.artofproblemsolving.com/1/a/1/1a1336e3306e21f7c50b024e18ec02d1e8c07255.png)
Now by sine ceva we are done.
.
are concyclic which means 
.
are collinear where 
is a midpoint of arc 
and 
are collinear where 
is a midpoint of arc 
and 
is a midpoint of arc 
and 
are concurrent, as desired.
,
similarly. Now ![\[ (A_3, A_1; B, C) \overset{A_2}{=} (A_4, A_5; B, C). \]](http://latex.artofproblemsolving.com/d/e/4/de4b0d3bbe49d49fcdf737631c1348156375aae7.png)
Thus, if 
is the foot of the altitude from 
,
collinear (this is well-known), and Cevian Nest finishes.
.
are collinear. By spiral similarity, we know that 
maps to 
,
and 
are collinear. Since 
are concurrenct by Ceva, harmonic lemmas give 
harmonic, projecting over 
harmonic, projecting through 
harmonic, projecting through 
harmonic, so 
are collinear with 
.
,
