and 
intersect at two points 
and 
.
be the line tangent to these circles at 
and 
,
be the line parallel to 
on 
on 
and 
meet at 
;
and 
;
and 
.
.
Y by PRMOisTheHardestExam, Danielzh, LoloChen, PeterZeus
I plan to use these two problems to train the Vietnam IMO team this afternoon. Enjoy!
Problem 1. Let
be a triangle inscribed in circle 
. Bisector of 
meets again at 
. Let 
be the symmedian of triangle (
lies on 
). Let 
and 
be the point on lines such that 
and 
. Prove that line 
goes through circumcenter of triangle 
.
Problem 2. Let be a triangle with incircle 
. touches , 
, 
at , 
, 
, respectively. meets again at . Circle 
passes through 
and is tangent to at . Define similarly circles 
and . Prove that centroid of triangle 
lies on the line 
where is the nine-point center of triangle .
Problem 1. Let
be a triangle inscribed in circle 
. Bisector of 
meets again at 
. Let 
be the symmedian of triangle (
lies on 
). Let 
and 
be the point on lines such that 
and 
. Prove that line 
goes through circumcenter of triangle 
.Problem 2. Let
be a triangle with incircle 
. touches , 
, 
at , 
, 
, respectively. meets again at . Circle 
passes through 
and is tangent to at . Define similarly circles 
and . Prove that centroid of triangle 
lies on the line 
where is the nine-point center of triangle .Attachments:
goes through the circumcenter of triangle 
be the midpoint of arc 
,
intersects 
at 
and 
.
.
.
and 
where we have used 
.
be the midpoint of 
and 
.
be the midpoint of minor arc 
.
so 
is the midpoint of 
- that is the circumcenter of 
then reflects over 
we get part 
of the below problem:
cut 
at 
.
with 
cut 
.
be the circumcenter. 
be the orthocenter, by homothety we can easily prove 
concur at 
)
.So as proved in the link ,if we let 
so 
.
.
or 
which is true by 
.
so 
and 
we will prove 
is a parallelogram in order to prove the problem (from the centroid statement).
so we need 
.
similar with 
similar with 
.
.
so 
,
and 
so combining 
so we are done.
.
.
be the diameter of 
.
and 
of 
.
cuts 
to triangle 
are collinear. By Thales theorem and using the signed length of segments, we have
From 
are are collinear.
We get 
.
and
Circle 
We get intersections 
,
,
and
From this 
and the perpendicular bisector of 
,
Similarly, we get
and
Now centroid 
Thus, we get
which is a real number. This means 
.
be a quadrilateral inscribed in circle 
.
and 
meet at 
and 
outside 
)
and 
be the incircles of 
and 
,
.
cuts 
at 
.
meets 
.
.
be the centers of the two pinky circles called 
.
cut 
at 
.
at 
.
.
at 
so 
.
which is true since 
with 
is the Miquel point of 
and 
.
be the centers of 
and 
,![$$\frac{d(X,OK)}{d(X,OL)}=\frac{2[XOK]/OK}{2[XOL]/OL}=\frac{OL}{OK}\cdot\frac{[XOK]}{[XOL]}=\frac{OL}{OK}\cdot\frac{XK}{XL}=\frac{OL}{OK}\cdot\frac{R_K}{R_L}=\frac{OL}{OK}\cdot\frac{KQ}{LR}=\frac{OB}{BN}\cdot\frac{AM}{OA}=\frac{AD}{BC}=\frac{SM}{TN}=\frac{d(P,OK)}{d(P,OL)}.$$](http://latex.artofproblemsolving.com/5/1/0/510cdf9aee1221882fa845ca85864e06eb5cd197.png)
Where 
and 
are radius of 
and 
.
and 
such that 
.
and 
.
.
are concurrent.
is the symmedian, easily seen 
.
(because 
)
(a.a), so
We are done.
and 
.
be the orthocenter and Gergonne point of triangle 
cuts line 
at 
.
and 
are inscribed in the angle 
and 
,
lie on the same straight line.
be a circle inscribed in the angle 
is the insimimilcenter to the circles 
lie on the same straight line. 
The point 
be the medial triangle of 
and the symmetric ones. Let 
be the reflection of 
.
.
and 
are perpendicular to 
and 
are parellel iff ratio of the length of the projections of these segments onto 
and 
be the length of the projections of 
.
Note that 
.
and 
as the intersection of 
with 
is tangent to 
and 
.
.
.
are cyclic quads which gives 
.
? thnx.
.
with circle 
.
Orthocenter 
So feet of altitude
and
Also 
on the line 
Similar, tangent point 
Gergonne point is the intersection of 
Point 
is
We get the intersection of 
From this center 
of 
Similarly, we get centers 
and center 
of 
and
From center 
.
From (1), (2), (3) we get
This means 
in the lines 
and 
meet at 
.
and 
meet on the circumcircle of 
,
,
again at 
,
.
is tangent to nine-point circle of 
,
cut 
,
by simple angle chasing we can prove 
so 
e
where 
and arc 
be the orthocenter of 
where 
cut 
.
with 
(Notice that 
and the reflections over bisectors 
)
is a rhombus so 
are reflection over 
cut 
(The well known lemma applied for 
and since 
we need 
but since 
,
we are done.)
.
.
.
Then,
where 
and 
,
with circumradius 
an note that 
so
where 
From (2), (3), we get
We see that (4) is true because two triangles 
and 
are similar, they have circumradii 
and 
,
,
s.t. 
.
be the centroid and circumcenter and 
.
of (O) are 
and define 
as the length of BC CA AB and radius of (O).Let 
be 
.
.
,
,
.
and 
,
,
