be an acute triangle with circumcircle 
.
be the midpoint of 
be the midpoint of 
.
be the centroid of the triangle 
be a circle through 
.
and 
are collinear.
with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
-
.
,
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
such that for every index 
,
,
is a multiple of 
.
such that
for all 
.
for all 
![\[\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)}.\]](http://latex.artofproblemsolving.com/7/d/6/7d6247fbcfd6eac156ce774f727d7e95c470de52.png)
be complex numbers with 
.
![$\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{lcm(1,2,3,...,n)}=e$](http://latex.artofproblemsolving.com/a/1/3/a139e5cb990f3386de109c701f8edc71b4054712.png)
digit numbers are there
?
?
and 
then is this uniformly convergence on 
comment on uniformly convergence on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
where in general it is should be uniformly convergence.
and trying to solv
integer matrices 
the entries of 
there exists a 
and the determinant of 
Y by Adventure10
A circle 
is given with a non-diameter chord 
. On the tangent at point 
select point 
and mark 
the intersection of the circle with the interior of the line 
(if any). Let 
a point in circle such that quadrilateral 
is a trapezoid . Determine the set of intersections of lines 
and 
belonging to all such trapezoids.
is given with a non-diameter chord 
. On the tangent at point 
select point 
and mark 
the intersection of the circle with the interior of the line 
(if any). Let 
a point in circle such that quadrilateral 
is a trapezoid . Determine the set of intersections of lines 
and 
belonging to all such trapezoids.This post has been edited 1 time. Last edited by parmenides51, Jan 31, 2020, 11:24 AM
