such that for all real numbers 
and 
,
be distinct real numbers such that 
and 
Evaluate all the possible values of 
.
be a pointed triangle and altitudes 
.
be an acute scalene triangle with orthocenter 
.
intersects 
at 
and line 
intersects 
at 
.
be the foot of the perpendicular from 
parallel to 
.
lies on line 
such that 
is parallel to 
lies on line 
such that 
is parallel to 
,
,
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 
at 
and 
,
at 
and 
,
.
intersects 
and 
,
,
and analogously 
.
and 
and equidistant from them. The midpoints of 
,
,
.
passes through the common point 
of 
.
of 
.
of the line segment 
. Since 
do not depend on 
,
meet 
at 
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 
and origin 
.
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 
and 
.
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 
with respect to circle 
.
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 
lies on the radical axis of 
.
,
,
,
,
.
works, which is independent of the position of 
.
,
.
and 
fixed points that lie on 
(line connecting midpoints of 
.
be the midpoint of 
.
,
and 
be the points of tangency of the incircle (centered at 
)
and 
respectively and let 
and 
.
and 
,
are collinear and 
.
be the point on 
.![\[
AY+AX-YX=\frac{AX+AC-CX}{2}+\frac{AX+AB-BX}{2}
\]](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 
be the midpoint of 
We will show that 
to prove the result.
is the mid-parallel of the 
lie on the 
lies on 
is perpendicular to 
which is immediate.
touch AX,BC in 
.
.
on C- outer angle bisector and hence they're fixed
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 
be the intersections 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 
;
,
be the 
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 
.
.
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.
)
)
