[/quote]
,
is the foot of the altitude from 
of triangle 
and 
points 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles 
.
.
![\[{x^3}{y^2}\left( {2y - x} \right) = {x^2}{y^4} - 36\]](http://latex.artofproblemsolving.com/d/5/6/d5668b2a486f5264e1cf2faeb363ac69bf91e3f0.png)
rectangles that can be covered without gaps and without overlaps with hooks such that
.
and 
is chosen on 
at 
.
be an acute triangle with 
,
be the center of its circumcircle. Let 
be the reflection of 
with respect to 
.
at 
,
intersects the line through 
.
be a point on the ray 
,
.
.
be a triangle with side‐lengths 
,
.
be the areas of triangles 
,
,
,![\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]](http://latex.artofproblemsolving.com/1/3/9/139a3e15f08afb78ca0b8301f24d30db8f202207.png)
lie the following points: 
and 
on 
,
on 
,
on 
.![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet 
.
,
are collinear.
and 
be two fixed points in the plane. For each point 
.
with 
and 
be points on segments 
and 
,
.
be the circumcircle of 
,
the circumcircle of 
.
to 
and 
,
intersect 
,
intersect 
.
is tangent to both 
.
where 
and 
are integers and 
.
intersects 
,
intersects 
.
intersects 
and 
,
and 
intersect at 
,
and 
.
,
is circumcircle. Let 
.
and 
,
(opposite to 
and tangents to 
)
.
be two triangles inscribed in the same circle, centered at 
.
with respect to triangle 
.
.
lie in this order on the circle and form a convex, non-degenerate hexagon.
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let 
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
and 
,
be midpoint of arc 
on 
,
be 
respectively now let reflection of 
over 
and reflection of 
over 
.
and 
are diameters on 
respectively, now let 
be the 
excenters of 
which implies 
cyclic and similarily 
is cyclic however since 
we also get that 
is cyclic and similarily 
is cyclic, however it doesn't end here...
so 
on 
however 
are symetric in 
are reflections of 
over 
are colinear, and similarily 
are colinear.
which means 
is cyclic and similarily we have 
cyclic, but also note that 
which means that 
are colinear.
which does in fact show that 
is isosceles and therefore 
,
,
where 
(clearly then 
)
which means that 
is cyclic but by summing arcs we end up realising 
is diameter and in fact now this means 
is 
be the miquelpoint of 
then 
colinear and then Reim's twice gives 
and from double Reim's once again we have that 
and this is excellent news because now we can note that 
which shows that 
are tangent at 
and 
.
and 
be the projections of 
and 
be the projections of 
is tangent to 
is a parallelogram, by projecting onto 
.
Simplifying yields 
,
But 
With Thales' Theorem, 
so 
thus getting 
and the conclusion follows.
are in centres, 
meets 
in 
and tha
.
