[/quote]
,
be complex numbers such that [center]
for any 
with 
[/center]
and 
for any 
,
,
we have ![\[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]](http://latex.artofproblemsolving.com/a/9/8/a98b6c29802355c491016bcce58228a020423526.png)
and two lines 
and 
are given in the space. For each nonnegative integer 
we denote by 
the projection of 
on 
and by 
the projection of 
Prove that there exist two segments ![$[A'A''] \subset d_1$](http://latex.artofproblemsolving.com/8/8/6/88667ee717ddd630efd34a9c4c83ee359c4897ed.png)
and ![$[B'B''] \subset d_2$](http://latex.artofproblemsolving.com/3/7/4/374927492266c8adaed5d34e3871d250baba9c1b.png)
of length 
and a nonnegative integer 
such that ![$A_n \in [A'A'']$](http://latex.artofproblemsolving.com/4/8/5/4851110dd8f0bf0ee7e42239ee051da76f5fbd65.png)
and ![$B_n \in [B'B'']$](http://latex.artofproblemsolving.com/5/e/7/5e7d78dd940f5c81673b779a9b45e83f4dc05b4d.png)
for any 
,
,
,
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
.
for all 
.
opposite the vertex 
be tangent to the side 
at the point 
.
on 
and 
on 
analogously, using the excircles opposite 
and 
,
lies on the circumcircle of triangle 
on the side 
and 
.
be the circumcenter of triangle 
.
.
Let 
and 
be the midpoints of 
respectively. Assuming that 
is a cyclic quadrilateral, find 
be a triangle, and let 
,
and 
as the orthocenters of triangles 
and 
.
passes through a fixed point which is independent of the choice of 
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
.
intersect at point 
be the reflection of 
and 
intersect at point 
.
,
be the orthocenter of 
at 
be the midpoint of 
,
meet 
at 
is tangent to 
,
and 
,
is the intersection point of the diagonals 
.
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
.
.
.
and 
.
Y by sixoneeight, OronSH, goaoat
Quadrilateral 
has an incenter 
Suppose 
. Let 
be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
The angle bisectors of
and 
meet 
at 
and 
. Prove that 
.
has an incenter 
Suppose 
. Let 
be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.The angle bisectors of
and 
meet 
at 
and 
. Prove that 
.This post has been edited 1 time. Last edited by popop614, Apr 3, 2025, 12:42 AM
Reason: asfdasdf
Reason: asfdasdf
Furthermore, the 
relation is the same as 
by angle bisector theorem, which we can simplify to 
using the previous equation. Thus, it suffices to prove this last relation.
implies 
.
lies on the rectangular hyperbola with foci 
.
homothety at 
,
,
,
.
be the harmonic conjugate of 
.
Thus, 
.
so 
or 
,
.
is the Newton line of 
.
lies on 
,
satisfying 
.
.
,
,
be the reflection of 
with 
,
