and 
be isogonal conjugates with respect to 
.
and 
be their respective pedal triangles. Let![\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]](http://latex.artofproblemsolving.com/7/f/e/7fe3ec4f69ec074635b2202daad9371e3bb8c72d.png)
Prove that the points 
,
and 
lie on the line 
.
and 
of the triangle 
,
.
and 
intersect at point 
.
is symmetric to point 
relative to line 
.
intersects the circumcircle 
of the triangle 
at the point 
.
is tangent to the circle 
Y by Davi-8191, bobcats4thewin, Adventure10, HWenslawski, ImSh95, and 3 other users
Let the incircle of triangle 
touch sides 
and 
at 
and 
respectively. Let 
and 
denote the circumcircles of triangle 
and 
respectively.
Let
and 
intersect at 
and 
and 
intersect at 
and 
and intersect at 
and 
Prove that 
and intersect in a common point.
Show that 
and 
are concurrent.
touch sides 
and 
at 
and 
respectively. Let 
and 
denote the circumcircles of triangle 
and 
respectively.Let
and 
intersect at 
and 
and 
intersect at 
and 
and intersect at 
and 
Prove that 
and intersect in a common point.
Show that 
and 
are concurrent.This post has been edited 1 time. Last edited by April, Jul 27, 2007, 12:46 AM
we have that 
we have that 
and 
we conclude the similarity of the triangles 
and so, we have 
we conclude that the segment line 
is the angle bisector of 
and so, it passes through the point 
as the intersection point of 
from 
where 
is the incenter of 
passes through the point 
We denote the points 
and we will prove that 
and similarly 
because of 
and 
we conclude that the intersection point of the segment lines 
lies the constant segment line 
lies also on 
is the incenter 
and 
are the incircle and circumcircle of 
respectively. 
are the tangent points.
is the circumcirlce of 
,
.
,
.
are concurrent. (see image below)
that is tangent to the incircle at 
and tangent to 
at a point, say 
.
,
.
is a line. We also have that 
and 
,
.
,
bisects 
is a line, which means that 
.
is the center of homothety between 
and 
and 
,
with respect to the incircle of triangle 
,
,
of the segments 
,
,
,
.
passes through the pole of inversion 
of 
(the one different from 
,
of 
,
under 
,
to 
is the orthic triangle of 
,
and 
,
,
to the sides of 
respectively). In fact, it is not even necessary to use the incenter. We could just use the well-known Miquel point of the triangle with respect to the points 
are the radical axes of the circles taken two at a time. Finally, the three lines concur at the radical center of the three circles!
is cyclic?
and 
is the angle bisector of 
be the midpoint of arc 
collinear.
similarly.
are collinear which is easy 
is parallel to the line joining midpoint of arcs 
,
collinear etc. we can deduce that those 
points are concyclic.
,
,
are concurrent at the incenter 
,
,
,
are concurrent at the Gergonne point. Hence, by the Cevian Nest theorem, we are done.
All angles directed. With this, of course, the problem falls to radical axis.
becomes the circle with diameter 
.
passes through the midpoint af arc 
.
are in verses under inversion about 
with radius 
)
.
,
,
,
.
the orthic triangle of 
.
,
,
are coaxial.
,
,
are coaxial.
be the orthocenter of 
is cyclic, we have that ![\[HB\cdot HE=HC\cdot HF.\]](http://latex.artofproblemsolving.com/9/c/e/9ce63cba8642022f88bd13974bb2255a4644e2d0.png)
Similarly, ![\[HA\cdot HD=HB\cdot HE,\]](http://latex.artofproblemsolving.com/5/0/3/50367e9f0ee173773c389d61299f8b7213ad7e91.png)
so ![\[HA\cdot HD=HB\cdot HE=HC\cdot HF.\]](http://latex.artofproblemsolving.com/c/1/4/c14c0ac13875cd757d89284d57f459cbab67dc37.png)
This implies that 
and the minor arc midpoint of 
and 
,
.
are coaxial.
are the 
.
are the midpoints of minor arcs 
respectively, it is well known that, 
and 
.
we have 
.
is cyclic, by converse of Reim's Theorem 
is cyclic, effectively proving that both 
and 
are cyclic. Thus by the existence of the radical center, we prove the desired concurrency as was necessary in the problem.
:
,
and cyclic variations are cyclic, whence ![\[\measuredangle PQD=\measuredangle PQI+\measuredangle IQD=\measuredangle PQI+\measuredangle IBD\]](http://latex.artofproblemsolving.com/7/3/7/737cb2fb3b297eae411683c16ba83439352a326e.png)
and![\[\measuredangle PED=\measuredangle PEI+\measuredangle IED=\measuredangle PAI+\measuredangle ICD=\measuredangle PAB+\measuredangle FAI+\measuredangle ICD,\]](http://latex.artofproblemsolving.com/c/e/3/ce3c2537db90fea3651f730be1237441762faa82.png)
which, after rearranging, reduces the condition to![\[\measuredangle PAB=\measuredangle PQI+\measuredangle IBD+\measuredangle ICE+\measuredangle IAF=\measuredangle PQI+\tfrac{\pi}{2}.\]](http://latex.artofproblemsolving.com/1/b/3/1b3ea6e80ed4874251ed3ffa65c92ebed3176cc9.png)
The above is true because 
.
be the midpoints of the minor arcs 
,
,
.
and 
concur at one point. This is true since the triangles 
are directly similar and the sides of the triangles are parallel to each other, hence 
to form a big triangle, by trig ceva and cyclic quads it is easy to see 
are concurrent, and obviously 
concur at 
as the 
-
-
,
each pass through the exsimilicenter of the incircle and circumcircle. 
are the incenter Miquel points for each of the vertices. It follows that 
,
,
concur at the exsimilicenter of the incircle and circumcircle.
passes through 
is the diameter. Analogously, the other circles pass through 
and we are done. (b) It is well-known that 
,
,
go to feet of altitudes in 
so 
and variants are cyclic. Inverting back and applying radical center theorem we are done.
.
