has incentre 
and incircle touching 
be the antipode of 
is chosen on the internal angle bisector of 
such that 
.
be the midpoint of arc 
,
be the midpoint of 
.
x
to 
row and 
column get symbolized by square 
. Subset 
belong to 
with 
or 
are not belong to 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle 
and 
are defined similarly. The circles 
distinct from 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
Y by Davi-8191, Amir Hossein, HamstPan38825, megarnie, Adventure10, Mango247, Rounak_iitr, GeoKing, Nari_Tom
In 
, a point 
lies on line 
. The circumcircle of 
meets 
at 
(other than 
), and the circumcircle of 
meets 
at 
(other than ). Prove that as varies, the circumcircle of 
always passes through a fixed point other than , and that this point lies on the median from to .
Proposed by Allen Liu
, a point 
lies on line 
. The circumcircle of 
meets 
at 
(other than 
), and the circumcircle of 
meets 
at 
(other than ). Prove that as varies, the circumcircle of 
always passes through a fixed point other than , and that this point lies on the median from to .Proposed by Allen Liu
are concyclic. From the complete cyclic 
it follows that the polar of 
goes through 
are conjugate points WRT 
with diameter 
Now all circles through a fixed point 
with the polar of 
which in fact is the projection of the orthocenter of 
cuts 
at 
then 
.
be the orthocenter of 
with altitudes 
and 
,
are concyclic. Easy to see that the reflection of 
.
and 
are cyclic. Let 
.
,
are collinear.
cut 
,
.
,
is the reflection of 
meets 
;
,
,
is a fixed point on the 
)
,
.
and 
is the center of the spiral similarity sending ![\[\dfrac{\sin\angle QAB}{\sin\angle QAC}=\dfrac{QE}{QF}=\dfrac{CE}{BF}=\dfrac{CA\cdot\dfrac{\sin A}{\sin\angle AEP}}{AB\cdot\dfrac{\sin A}{\sin\angle AFP}}=\dfrac{CA}{AB}\]](http://latex.artofproblemsolving.com/5/8/2/582f478840fa669cec320db2ffffa9dd7677b41d.png)
and consequently 
is fixed, it follows that 
creativity xD 
be on 
and let 
be on circle 
on arc 
.
now 
and 
so we get 
now along with 
we have that 
yeilding 
means 
is cyclic.
and just check that 
is fixed.
where 
are the powers of 
WRT 
,
.
in the equation, which gives 
since 
.
so we're done.
and let 
.
for some real 
.
is given by: 
and it is easy to see that this intersects line 
at the point 
.
.
has equation 
.
lies on this circumcircle, and moreover lies on the 
-
so we are done.
again, since 
,
and 
analogously. Let 
,
,
are concyclic 
.
,
,
M is the midpoint of 
dictates that 
and reflection WRT bisector of 
(
is inverse of 
under the inversion and 
lies on the arc 
and 
, 
are collinear. let 
and the tangents from 
meet at 
because 
is a polar of 
passes throw the inverse of 
DONE
be the composition of an inversion with center 
,
and 
Hence, 
takes line 
It follows that 
,
and 
is the line 
at 
,
are collinear, i.e. 
This completes the proof. 
,
denotes the power of a point. It is not hard to check that this function is linear.
denote the circumcircles of 
.
is constant. We compute
So 
Therefore, 
,
,
.
and 
and 
.
which is fixed. So Done!!!
and 
.
passes through a fixed point on the 
.
,
.
concur so by the converse of Radical Center 
;
the inverse of a point 
;
and 
.
and 
.
,
;
.
be a triangle and 
,
such that 
.
,![\[ \text{Pow}(\bullet, (AEF))-\text{Pow}(\bullet, (AEC))=\text{Pow}(\bullet, (AFB))-\text{Pow}(\bullet, (ABC)). \]](http://latex.artofproblemsolving.com/8/4/3/8436046e0c596acfb7d79228637aba0f4b89432c.png)
Proof. Let 
denote LHS and 
denote the RHS. Note that 
and 
are linear functions. It suffices to show that 
holds for three distinct non-collinear positions of 
.
is constant as point 
is constant. Let 
and 
.
which is indeed constant.
.
passes through 
are collinear, where 
is the pole of 
,
of triangle 
.
,
with 
.
we can see that 
.
.
,
,
.
always lies on 
and 
are swapped. Furthermore, we have that the 
under the inversion, so after reflection 
below. Thus, it is equivalent to solve this following:
,
,
.
for any choice of 
.![[asy]
import geometry;
import olympiad;
size(8cm,0);
point A = (0.179,0.888);
point B = (0,0);
point C = (1,0);
point D = (0.648,-0.238);
circle w = circumcircle(A,B,C);
line AB = line(A,B);
line AC = line(A,C);
line BD = line(B,D);
line CD = line(C,D);
point EE = intersectionpoint(AB,CD);
point F = intersectionpoint(AC,BD);
line Btan = tangent(w,B);
line Ctan = tangent(w,C);
point P = intersectionpoint(Btan,Ctan);
line AP = line(A,P);
point M1 = intersectionpoints(w,AP)[0];
point M = midpoint(B--C);
point H = orthocenter(A,B,C);
point Q = foot(H,A,M);
draw(w);
draw(B--C);
draw(B--P--C--P--A);
draw(A--EE--C);
draw(A--F--B);
draw(EE--F,dashed);
draw(A--M,dashed);
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,E);
dot("$D$",D,S);
dot("$E$",EE,SW);
dot("$F$",F,SE);
dot("$P$",P,S);
dot("$M'$",M1,SW);
dot("$M$",M,S);
dot("$Q$",Q,NE);
[/asy]](http://latex.artofproblemsolving.com/0/2/3/02331c1a0c9ef59c09e68d1791f861b654d41720.png)
as it is simply the 
.
.
by the definition of the 
subtends arc 
in 
,![\[\angle AQC=180-\angle C=\angle A+\angle B=\angle ABT,\]](http://latex.artofproblemsolving.com/f/d/f/fdf513479e4d8db99bf5e962a7cadfee954e5d01.png)
Where the first equality follows as both 
and 
subtend major arc 
as 
is a tangent to 
since the 
.![\[\frac{AB}{AP}=\frac{AQ}{AC} \implies AB \cdot AC=AP \cdot AQ,\]](http://latex.artofproblemsolving.com/e/1/4/e14dbf94617b6ebc5f13ebbbe9b5bbad762426fe.png)
so we are done.
such that 
.
,
,
denotes the power of point 
.
and 
.
,
for all points 
is fixed. 
,
.
is a point on 
,
,
.
says 
,
meet at 
meet at 
be the intersection of the tangents at 
are collinear.
and 
,
are collinear.
to the center of the circle.
.
for 
.![\[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]](http://latex.artofproblemsolving.com/6/9/3/69317adbb0a4cf4bbc2c020f6f869d4693ddb3f0.png)
Since this circle passes through 
and 
.
Now, let 
.
Thus, 
.
.
with a similar calculation we can also see that 
.
.
Thus, 
.
and 
,
H
.![\[\measuredangle CPB = \measuredangle PCB + \measuredangle CBP = \measuredangle ECD + \measuredangle DBF = \measuredangle EAD + \measuredangle DAF = \measuredangle EAF \]](http://latex.artofproblemsolving.com/4/8/1/48109e4f884ef05a4fa2c7d7bab975313829b511.png)
Thus, 
,
.
which implies that 
.![\[\measuredangle BCH_A' = \measuredangle FPH_A' = \measuredangle FAH_A' = \measuredangle CAH_A'\]](http://latex.artofproblemsolving.com/0/f/a/0fa366421f39d5e642574303205203d11dc03e15.png)
and similarly, 
.
lies on the intersection of the two circles passing through 
.
)
.
(due to Brokard's theorem). Thus, 
is the symmedian of 
which implies line 
is median of 
meet 
.
meet 
in 
Let 
meet 
.
,
is cyclic. Hence, observe that
is concyclic
is tangent to 
is the midpoint of 
.
for some 
.
.
gives us 
.
and similarly 
.
for some 
.
and plugging in 
.
,
.
and intersected those circles to get the point to plug in.
