[/quote]
,
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
be the incenter of triangle 
.
,
,
be the orthocenters of triangles 
,
,
,
,
,
,
and 
have a common tangent that meets 
and 
.
and 
where 
than 
meets circle 
.
bisects angle 
.
and 
be two triangles inscribed in the same circle, centered at 
,
.
with respect to triangle 
.
.
lie in this order on the circle and form a convex, non-degenerate hexagon.
be a triangle with circumcircle 
and 
a line without common points with 
intersect 
different from 
,
and 
have a common point different from 
,
be the foot of the altitude from 
.
be a point in the interior of the segment 
.
be the point on the segment 
such that 
.
be the point on the segment 
such that 
.
and 
.
.
be a point on segment 
.
be a fixed circle passing through 
be a variable point on 
be the intersection of the tangent to the circumcircle of 
at 
and the tangent to the circumcircle of 
at 
.
.
and let the angle bisector of 
intersect 
and 
.
and 
intersect at
is tangent to segment
)
,
,
intersect 
,
.
lie on lines 
and 
is a parallelogram. Point 
.
intersects 
at 
intersects 
at 
.
intersects 
at points 
respectively.
.
be a general quadrangle which does not form an orthocentric system. Let 
)
)
)
)
be the orthocenter of 
,
,
,
cyclically.
)
)
,
,
,
,
,
,
,
,
,
,
,
lie on the same conic, here denoted by 
.
be the nine-point conic of 
,
,
be the intersections, other than 
passes through 
be inscribed in circle 
.
be the 
,
,
.
and 
be the angle bisectors of triangle 
intersect 
and 
.
and 
.
such that 
.
at points 
respectively. Let 
respectively such that 
is parallel to 
.
be the circle that is externally tangent with the excircle
,
Y by doxuanlong15052000, Adventure10, Mango247, cubres
For a given triangle 
, let 
be a variable point on the line 
such that 
lies between 
and and the incircles of the triangles 
and 
intersect at two distinct points 
and 
Prove that the line 
passes through a point independent of .
, let 
be a variable point on the line 
such that 
lies between 
and and the incircles of the triangles 
and 
intersect at two distinct points 
and 
Prove that the line 
passes through a point independent of .This post has been edited 2 times. Last edited by djmathman, Sep 27, 2015, 2:19 PM
and 
touch 
,
,
.
,
and analogously 
.
and 
and equidistant from them. The midpoints of 
,
,
passes through the common point 
.
.
. Since 
do not depend on 
meet 
and have inradius 
.
and so on, we have 
.
at 
and let 
at 
.
.
and 
.
,
.
,![\[\angle{NMX}=\frac\pi2-\frac{\angle{AXB}}{2}=\frac{\angle{D_1BF_1}}{2}+\frac{\angle{E_2AF_2}}{2},\]](http://latex.artofproblemsolving.com/5/1/3/51306c490e80cbfc5133acfb02303a00b55eb928.png)
and the tangent addition formula combined with some simple ratios and lengths gives us
Also,![\[BM=BD_2+MD_2=\frac{sv}{r}+\frac{w}{2}.\]](http://latex.artofproblemsolving.com/9/4/c/94c447c172edef118399445530fcdff48892b684.png)
Now consider the coordinate plane with 
-
.
is![\[y=\frac{uw}{r(u+v+w)-s(v+w)}\left(x-\frac{sv}{r}-\frac{w}{2}\right).\]](http://latex.artofproblemsolving.com/8/d/3/8d370602e967411e3da2171d0dc9bf9728c442d6.png)
But it's easy to check that the point![\[\left(\frac{w}{2}+\frac{v(u+v+w)}{v+w},\frac{uvw}{r(v+w)}\right)\]](http://latex.artofproblemsolving.com/6/a/2/6a24f4c8ad53d53bbb4d4b71c73ff81d09d08148.png)
always lies on this line, so we're done.
![[asy]
import olympiad;
unitsize(2cm);
pair A=(3,4);
pair B=(0,0);
pair C=(4,0);
pair X=(10,0);
path C1=incircle(A,B,X);
path C2=incircle(A,C,X);
pair O1=incenter(A,B,X);
pair O2=incenter(A,C,X);
pair B1=foot(O1,B,X);
pair B2=foot(O1,A,X);
pair C_1=foot(O2,C,X);
pair C_2=foot(O2,A,X);
pair M1=(B1+C_1)/2;
pair M2=(B2+C_2)/2;
pair D=(A+B)/2;
pair E=(A+C)/2;
pair F=(A+X)/2;
pair k[]=intersectionpoints(D--F,M1--M2);
pair K=k[0];
pair p[]=intersectionpoints(C1,C2);
pair P=p[0];
pair Q=p[1];
draw(A--B--C--X--A--C);
draw(D--F);
draw(M1--M2);
draw(C1); draw(C2);
dot(A); dot(B); dot(C); dot(X); dot(P); dot(Q); dot(B1); dot(B2); dot(C_1); dot(C_2); dot(M1); dot(M2); dot(D); dot(E); dot(F); dot(K);
label("$A$",A,dir(90));
label("$B$",B,dir(-90));
label("$C$",C,dir(-90));
label("$X$",X,dir(-90));
label("$P$",P,dir(-135));
label("$Q$",Q,dir(110));
label("$B_1$",B1,dir(-90));
label("$B_2$",B2,dir(90));
label("$C_1$",C_1,dir(-90));
label("$C_2$",C_2,dir(90));
label("$M_1$",M1,dir(-45));
label("$M_2$",M2,dir(90));
label("$D$",D,dir(135));
label("$E$",E,dir(45));
label("$F$",F,dir(90));
label("$K$",K,dir(140));
[/asy]](http://latex.artofproblemsolving.com/8/b/9/8b9de5d7b405afe8023cb6b0657c24ac430e6129.png)
be tangent to 
and 
.
touch 
at 
be the midpoint of 
be the midpoint of 
always passes through.
does not change as 
is a constant.
and 
and 
.![\[M_1X=M_2X=\frac{BX+CX+2AX-AC-AB}{4}\]](http://latex.artofproblemsolving.com/e/c/8/ec8c8a83f26c8ff658b46607643399d15afb0f69.png)
Since 
,
.
,
.
and 
be the incircles of 
and 
,
the power of 
by![\[f(P) = (P, \omega_1) - (P, \omega_2).\]](http://latex.artofproblemsolving.com/2/a/b/2abf96a14d9dd11250c4bec73d839bbbbb8a596a.png)
This function is linear. We now use barycentrics with respect to 
be the constant point that lies on 
.
,
,
,
,
.
works, which is independent of the position of 
.
,
.
and 
fixed points that lie on 
(line connecting midpoints of 
.
.
and 
be the points of tangency of the incircle (centered at 
and 
and 
.
,
are collinear and 
.
.![\[
AY+AX-YX=\frac{AX+AC-CX}{2}+\frac{AX+AB-BX}{2}
\]](http://latex.artofproblemsolving.com/3/5/9/35984c190f84406d97c6d2c0168a68e258658043.png)
![\[
2AY+BY-CY=AB+AC
\]](http://latex.artofproblemsolving.com/9/5/2/95201aa64321713feb88c11161caa3203e9583e6.png)
at their tangency points with 
past 
suffices.
at 
at 
We will show that 
to prove the result.
is the mid-parallel of the 
lie on the 
lies on 
is perpendicular to 
touch AX,BC in 
.
.
and hence it bisects 
it passes thru the midpoint of 
centered at 
and touches 
resp. Let 
be the center of 
,
be the midpoint of 
and finally let 
.
,
.
.
.
and
so 
which implies 
and 
and intersect the radical axii.
and 
.
,
,
is the desired point. It suffices to show that 
,
are collinear. Call these midpoints 
and 
.
and 
,
.
The coordinates of 
,
To show that 
are collinear, we simply show that the displacement vectors 
and 
are proportional. In fact, because all of these coordinates are homogenized, we only to verify the proportion for the first two coordinates. Looking at the first coordinates, we receive a ratio of
But for the second coordinates we also have
We thus have the desired.
be the incenter of 
be the incenter of 
and 
with the 
and 
at 
at 
respectively, and let 
and 
.
and 
respectively. It is also easy to see that 
since both of them are perpendicular to 
.
and 
,
and 
be the incenters of 
,
touchpoints of incircle of 
and 
,
and 
and 
,
and 
and 
are fixed, since 
,
are fixed.
is cyclic. From angle chasing we also get, that 
.
is parallelogram. It's well-known, that 
;
,
of all the 3 triangles.Say 
and 
respectively and say 
and 
respectively.
are fixed.
and the external angle bisector of 
(
)
and say 
is the midpoint of 
,
or 
,
.
become the midpoints of 
and 
.
meet 
and 
respectively. Let also line homothetic to 
intersect internal bisector of and 
at 
respectively, so by Iran lemma 
But obviously 
is an isosceles trapezoid, therefore 
which is fixed.
be midpoints of 
.
at 
.
at 
.
and 
.
.
and 
,
is parallelogram
.
,
is constant, which implies 
.
intersection with 
is a parallelogram with center ![\[\angle AEF = \angle DEX = \angle EDX = \angle AFE\]](http://latex.artofproblemsolving.com/0/9/e/09eca4c46664ee54ce39484b09eb18a20832a41c.png)
so 
.
.
so 
.
,
so 
passes through 
,
and 
,
.
and 
.
.
being the intouch triangle of 
and 
be the midpoints of 
and 
be the midpoints of 
.
.
and 
the incircles of 
and 
be the centers of 
be the tangency points of the incircle of 
and 
due to the common tangents. Further, 
.
and 
concur (say at 
due to the above collinearities.Further notice that by Iran Lemma on 
and 
concur. But clearly the latter two of these lines intersect at 
also must pass through 
by Midpoint Theorem and 
since both these lines are perpendicular to 
.
and 
and 
be the tangency points of 
.
and 
are tangents to 
and 
denote that semiperimeters of 
denote the semiperimeter of 
Thus, indeed 
.
and thus 
must also lie on 
.
,
and 
,
.
and 
,
and 
by power of a point, and 
.
.
and 
are independent of 
vanishes on is independent of 
constant (proving ![\[ f(A) = \frac{1}{4} \left[ (AB+AX-BX)^2 - (AC+AX-CX)^2\right] \]](http://latex.artofproblemsolving.com/f/a/a/faa6d9c5f109f4fe061806ba46b8a537f198280f.png)
and ![\[ f(B) = \frac{1}{4} \left[ (AB+BX-AX)^2 - (2BC+AC+CX-AX)^2\right]. \]](http://latex.artofproblemsolving.com/3/1/5/3150dc4bf8f7c8989444fdab2acaf7370d09ad07.png)
Now we can bash out 
,
,
.
lies on the tangent chord 
.
lies on the tangent chord 
.
,
is the midpoint of 
.
denote the intersection of 
.
at 
at 
respectively. Thus, it suffice to show 
which is immediate due to Iran's Lemma.
)
)
denote the tangencies of 
incircle of 
respectively.
By radax, 
implying that 
Finally, note that 
implying that 
Since 
and 
.
,
,
Y and Z are fixed too and YZ is fixed. Let L be the midpoint of DF and J be the midpoint of EG. By radical axis properties it follows that L and J lie on the 
and 
.
,
and from angle chase we get 
and L is the midpoint of DF, by the parallelogram, if 
,
,
