.
and 
. Prove that
be a positive integer. Consider a figure of a equilateral triangle of side 
small equilateral triangles of side 
.
vertices of the small triangles, such that for every integer 
,
,
is the center of the inscribed circle, point 
.
is equal to the diameter of the circle inscribed in triangle 
Y by Aref, rashah76, samrocksnature, Adventure10, jhu08, mathematicsy, MathLuis, Mango247, drago.7437, farhad.fritl
Let 
be a triangle, and 
the midpoint of its side 
. Let 
be the incircle of triangle . The median 
of triangle intersects the incircle at two points 
and 
. Let the lines passing through and , parallel to , intersect the incircle again in two points 
and 
. Let the lines 
and 
intersect again at the points 
and 
. Prove that 
.
be a triangle, and 
the midpoint of its side 
. Let 
be the incircle of triangle . The median 
of triangle intersects the incircle at two points 
and 
. Let the lines passing through and , parallel to , intersect the incircle again in two points 
and 
. Let the lines 
and 
intersect again at the points 
and 
. Prove that 
.
.
is the line, parallel 
in
and 
in which maps the points 
.![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
;
where the incircle 
touches the side-lines 
respectively; the intersections 
between the incircle and the median 
.
and the division 
is harmonically.
.
.
and 
results that 
,
what means 
is the polar of the point 
w.r.t. the incircle).
is an isosceles trapezoid 
.Therefore, 
. From the product of the relations 
we obtain 
,
form a harmonical division. This property is one among the first problems in which the my mathe-teacher applied the harmonical division (approx. in 1958 !)
,
and 
.
and 
,
and 
.
.![\[ \sqrt{BP} - \sqrt{CP} = \frac{b - c}{\sqrt{a/2}}. \]](http://latex.artofproblemsolving.com/d/2/9/d29b6a3317ea022652616ab090bbae8118c66116.png)
![\[ \sqrt{CQ} - \sqrt{BQ} = \frac{b - c}{\sqrt{a/2}}. \]](http://latex.artofproblemsolving.com/6/b/5/6b519833c9a2d2de9b59e8793b7a0d6e61c6d9c9.png)
be the circle passing through 
and 
respectively. Let 
be the circle passing through 
and 
respectively. Notice that 
.
be the second intersection of 
intersect 
and 
respectively. It is sufficient to prove that 
,
and 
.
and 
respectively. Hence 
and 
,
.
,
.![\[ \frac{CE}{CM} = \frac{\sin \angle EMC}{\sin \angle CEM} = \frac{\sin \angle BMD}{\sin \angle BDM} = \frac{BD}{BM} \]](http://latex.artofproblemsolving.com/8/c/c/8ccfb841c67addb1f897a80debc4a24dc688b00d.png)
and we are done. (You can avoid the little bit of trigonometry I'm using, if you really want...)
is the point on 
.
be the incircle of triangle ABC. The line 
intersects the incircle 
(closer to A) and line 
intersects the incircle 
and 
.
.
A line cuts 
at 
respectively. We have:![\[BC\cdot\frac{AM}{AM'}=BM\cdot\frac{AC}{AC'}+CM\cdot\frac{AB}{AB'}\]](http://latex.artofproblemsolving.com/2/6/a/26ab8cfe7d4de76a12a9edc2b0c4ee46f878d1ec.png)
form a harmonic division (in Virgil's lemma)...
and 
the intersection of 
with 
and 
,
.
and 
is concurrented, say at 
and 
.
and 
collinear, we can conclude that 
be the intouch triangle, let 
and let 
.
are collinear. It's clear that 
and 
,
are collinear. Then
so 
and 
to 
Then 
be the center of 
and let 
it suffices to show that if the tangents to 
then 
Let 
then it is well-known that 
but by polars 
as well. Thus 
as claimed.
at points 
respectively, there exists an involution swapping pairs 
Projecting onto 
we see that there exists an involution swapping 
and 
However, this is just the involution reflecting across the perpendicular bisector of 
as desired.
.
,
denote the contact points of 
,
be a triangle with incenter 
and 
,
and 
so that 
and 
passes through 
,
,
and 
we must have 
is cyclic, and similarly 
is cyclic. Hence ![\[\angle AFX = \angle AEX = \angle B'IX = \angle XIC', \]](http://latex.artofproblemsolving.com/6/f/6/6f690a72c7bc1233a0bf289f2f1378e689129398.png)
so that 
is isosceles. Now as 
,
are collinear, where 
is the concurrency point of ![\[(Q, M; P, P_{\infty}) \overset{A}{\underset{KX}{\longrightarrow}} (Z, K; X, P_{\infty}) \overset{Y}{\underset{AM}{\longrightarrow}} (A, K; G, L) = -1, \]](http://latex.artofproblemsolving.com/5/7/4/57420d2ad238aa97a1a72276f412ddce967131e8.png)
where the final equality follows from the lemma. 
,
such that 
,
be the intouch points on 
.
are collinear so then 
.
and we can conclude 
be point at infinity on line 
Let 
.
,
are concurrent follows from homothethy and simson lines. It can alternativey be proven using homothethy and harmonic bundles. Since 
across 
we know that it is also concurrent.
are collinear.
and 
By Brocard, 
Since 
,
we get 
which implies 
and the desired follows.
be the intouch triangle of 
,
,
.
such that 
.![\[-1 = (A,R;G,L) \overset{\infty}{=} (S,R;X,Y) \overset{A}{=} (\infty,M,P,Q),\]](http://latex.artofproblemsolving.com/5/f/d/5fd1bcaa5baebb80d6d61cef64da5e83ac7173bb.png)
.
be the inotuch triangle of 
,
.
are collinear. Since 
,
and 
.
are collinear. Let 
be the point at infinity along 
.
It follows that 
,
.
.![\[-1=(EF;KL)\stackrel{F}{=}(ZA;KL)=(ZA';XY)\stackrel{A}{=}(M\infty_{BC};PQ),\]](http://latex.artofproblemsolving.com/a/8/d/a8dc8c3e59bc547c2c92e684b449e2ee119849e7.png)
as desired.
.
with 
.
is the Miquel point of 
is the inverse of 
,
or 
to 
,
intersecting 
again at 
and 
.
such that 
and 
,
such that 
and 
.
,
through 
,
through 
,
tangent to 
respectively.
are concurrent.
.
is degenerate which implies 
is the angle bisector of 
,
is an isosceles trapezoid.
.
with 
collinear. Notice that, as we get 
are symmetric with-respect to line 
:
Taking perspective into line 
Therefore 
where 
.
be a line through A that is parallel to 
Let 
Let 
and 
.
are concurrent and meet at 
we have that the intersection of 
and 
.
and since the 
also lies on the polar of 
we can observe that 
. Now since we have 
concurs at 
.
is an isosceles trapezoid, 
and the line parallel to 
