AoPS Academy
satisfy that: 
and 
. Find the maximum: 
, such that 
Prove that :
equality cases : ?
(
), 
is its incenter and 
is the 
-excenter. 
are on minor arcs 
and 
respectively such that 
. 
is on line 
such that 
.
bisects 
.
and the line through perp to 
) is concurrent. Similarly 
and the line through perp to 
are concurrent. These concurrencies can be proved by radical axis theorem with circles 
. Now the conclusion follows from the fact that 
and 
are parallel where 
and M is the midpoint of 
.
. If 
is the midpoint of arc and 
is the midpoint of are 
then 
. Let 
be the -antipode on 
and be the midpoint of the minor arc 
. Since 
and 
, 
and 
are collinear. Let 
be the orthocenter of 
and be the midpoint of . It suffices to proof that 
are collinear as is the reflection of across (which imply 
would be collinear). is actually the orthocenter of 
. As 
. The observation then follows from 
and 
.
and point with the Gauss-Bodenmiller theorem, or by dropping altitudes and calculating powers of points, we can show that , are on the radical axis of 
and 
. Let 
be the foot from 
to 
respectively, then clearly the reflection of 
over is 
due to 
. Finally, is on and is on yields 
which gives that is also on the radical axis of and . Therefore, 
are collinear.
and two points 
, 
such that 
. 
, 
are two points lying on 
such that 
. 
satisfies 
. Then if the pedal circle of with respect to is tangent to , bisects . is the incenter of , the previous statement holds.
at 
respectively at 
respectively
and 
are cyclic of circumcircle of 
and hence 
collinear 2) 
collinear
or equivalently collinear
or equivalently collinear
hence by symmetry 
be midpoint of 
and 
intersect at 
are collinear
collinear
is cyclic
cyclic
cyclic
collinear
collinear
be the 
antipode of .Let 
.Then due to the right angles , 
.Now Let 
.And 
.Then 
thus 
is the symmedian of 
. Also 
.Also 
Thus 
is the bisector of 
.This means that 
.
Thus is the reflection of the symmedian across , the angle bisector , which means that is the median , as desired.
(), is its incenter and is the -excenter. are on minor arcs and respectively such that . is on line such that . bisects . be the -antipode in and 
. Move a point 
on line 
and a point 
on such that lies on the circle with diameter . Note that 
is a perspectivity at 
, hence is a projective mapping. Let 
and note that 
by the above mapping. Finally, the image of under a 
inversion at followed by reflection in is the point ; hence 
is a median in as 
is a symmedian, proving the claim.
be the incenter, 
-excenter and 
-excenter of . Let 
be point on such that 
. Let be the antipode of wrt . Let 
be midpoint of arcs 
, respectively.
.
. Observe that lies on the perpendicular bisector of 
. By DDIT, 
are reciprocal pairs of some involution. Hence, we conclude the angle bisector of 
coincide, as desired. 
, let 
, let 
.
.
with the choice of signs as usual. By ratio lemma, 
which yields that .
which gives bisects , as desired. 
and let be the midpoint of 
inversion along with the reflection across the angle bisector. Note that this swaps 
and 
maps to 
which lies on such that 
and maps to 
which also lies on such that 
. taking to and 
to . This means 
where the last step follows because of sine law on 
and 
.
and 
Now just notice that this homothety also takes to , that is the midpoint of to which means that must be the midpoint of 
which by homothety at taking to 
means that must bisect as desired 
. Solved with BVKRB- , this was my first time g-solving, and it was really fun lol . My solution has the proof of the first part pretty small and easy, and finish similar with that of BVKRB-. (), is its incenter and is the -excenter. are on minor arcs and respectively such that . is on line such that . bisects . passing through intersect at , the perpendicular to 
passing through intersect at , 
be the midpoint of the minor arc of 
, 
, 
. with radius followed by a reflection wrt the Angle Bisector of 
.
we get that 
is tangent to 
and similarly 
.
and 
are collinear triples, so, 
and since 
we get that .
, so lies on the perpendicular bisector of , we get is midpoint of (simple angle and similar triangle chase). that maps to 
, also maps 
to , thus is the midpoint of and we are done 
.
. By homothety ( by angle chasing), we want to be the midpoint of . Add 
; by radical axis 
(say, by taking 
for the first one and similarly for the second one) and in addition, 
, so 
and we are done since is midpoint of .
be the midpoint of . By the Inscribed Angle Theorem, 
. Therefore, the angles 
and 
are supplementary, so 
.![[asy]
unitsize(1 cm);
pair A, B, C, I, J, K, M, X, Y;
A = (0,0);
B = (15,0);
C = (2.5,15);
I = (5,5);
J = extension(A, I, B, C);
X = (10,0);
Y = (2.5,10);
M = (3.725,5);
K = extension(I, J, X, Y);
draw(A--B--C--cycle);
draw(I--J);
draw(X--Y);
draw(A--K);
label("$A$", A, S);
label("$B$", B, S);
label("$C$", C, NW);
label("$I$", I, NE);
label("$J$", J, NW);
label("$X$", X, S);
label("$Y$", Y, NE);
label("$K$", K, NW);
label("$M$", M, N);
[/asy]](http://latex.artofproblemsolving.com/1/3/5/1359a89293f872b5253f007e464d3022deb0b7c4.png)
. Since is the incenter of , 
is an angle bisector of . Therefore, 
. Hence,
This means that 
is an angle bisector of 
, so bisects . bisects .
be the antipode of in . Let be the midpoint of . is the intersection of 
and .
are collinear.
are collinear.
and 
. is the orthocenter of 
hence 
and 
are collinear. with radius 
. goes to the line thus 
.
goes to 
with diameter 
hence 
.
goes to 
with diameter hence 
. at 
and excircle is tangent to at 
.
.
.
which is the altitude from to and 
which is the altitude from to . Let be the midpoint of , 
be the altitude from to , 
. 
where 
. Let 
be the midpoint of .
thus we have 
.![$$\left. \setlength{\arraycolsep}{2pt}
\begin{array}{rcl}
-1=(G,D';I',J')=(R,D';F,E) \\[4pt]
QF^2=QD'.QR \\[4pt]
Pow_Q(EMF)=Pow_Q(DMSR) \\[4pt]
T^*\in (DMS) \\[4pt]
\end{array}\right\} \implies A,T,K
\text{ are collinear.}$$](http://latex.artofproblemsolving.com/c/b/9/cb9843a619d43a43af7b05ee9da44e8c53f1f79e.png)
and 
Prove that
,
where 
is the centre of 
,
.
with the centre 
,
and 
.
be the isogonal conjugate point of 
be the pedal triangle of 
is tangent to 
,
.
by 
and 
.
at 
Denote the second intersection between 
and 
.
.
and 
.
by 
.
then 
.
by 
,
respectively. We have
Thus 
,
.
,
collinear, 
collinear.
.
.
.