be a point on segment 
and let 
meet the circle again at 
.
parallel to 
.
is parallel to 
.
,
be positive real numbers, such that 
.![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
.
?
,
.
and 
.
,
and multiply to 
,
.
,
and 
are complex numbers such that
with 
,
.
be the orthocenter of triangle 
,
be the midpoint of 
.
,
and 
at points 
and 
,
be the circumcenter of triangle 
.
are collinear.
and 
If each person remains there for 
minutes, what's the probability that all three people meet each other?
and solve for the area of a hexagon in that 
square.
,
.
.
.
and 
Then the number of elements in the set 
The altitudes 
, intersect
.
o
.
b
intersects 
,
intersects 
at 
lie on the same circle.
intersects the circle 
The line 
.
are collinear.
Y by
An acute triangle 
is given on the plane. Let the points 
, 
and 
be on the sides 
, 
and 
, respectively. (, and are different from the vertices 
, 
and 
) Also the points 
, 
and 
are taken such that 
is an equilateral hexagon which the opposite sides are parallel. Suppose that the circumcenters of and 
are coincident. Then determine the least possible value of:
Note: 
area of 
is given on the plane. Let the points 
, 
and 
be on the sides 
, 
and 
, respectively. (, and are different from the vertices 
, 
and 
) Also the points 
, 
and 
are taken such that 
is an equilateral hexagon which the opposite sides are parallel. Suppose that the circumcenters of and 
are coincident. Then determine the least possible value of:
Note: 
area of 
This post has been edited 1 time. Last edited by Ege_Saribass, Apr 27, 2025, 11:10 AM
Reason: missing information
Reason: missing information
be the circumcenter of 
.
.
.
,
.
.
and 
.
.
hence 
thus 
.
.
.
via inversion.
,
,
.
and 
and 
.
.
are linear.![$[E'F']$](http://latex.artofproblemsolving.com/b/8/1/b8142d200cf7f09b6a3c009b17fe198a406538ec.png)
intersects 
,
and 
.
,
and 
.
which is a contradiction.
.
and 
.
.
and 
.
and circumradius of the 
.
and 
,
hence 
.
