MHB 231.13.3.75 top vertex of a regular tetrahedron

Click For Summary
The discussion centers on determining the center of a unit sphere placed symmetrically atop three unit spheres with specified centers. The coordinates of the center R are derived using the relationship between the height of the tetrahedron formed by the spheres and the side length. The height is calculated as h = (√3/2) * a, with a set to 2, leading to the coordinates R = (2√3/3, √3). The conversation highlights the challenge of finding a comprehensive equation for this geometric configuration. Ultimately, the solution is reached through estimation rather than a formal derivation.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{231.13.3.75}$
$\textrm{Imagine $3$ unit spheres
(radius equal to 1) with centers at,}\\$
$\textrm{$O(0,0,0)$, $P(\sqrt{3},-1,0)$ and $Q(\sqrt{3},1,0)$.} \\$
$\textrm{Now place another unit sphere symmetrically on top of these spheres with its center at R.} \\$
$\textrm{a Find the center of R.} \\$
 
Last edited:
Physics news on Phys.org
The following post may give you some insight:

http://mathhelpboards.com/challenge-questions-puzzles-28/tetrahedral-stack-spheres-5676.html#post26011
 
View attachment 7108

ok from this base

$\textrm{so if hieght of $h=\frac{\sqrt{3}}{2} a$ and $a=2$ then:}\\$
\begin{align*}\displaystyle
R&=\left(\sec \left(\frac{\pi }{6}\right),\frac{\sqrt{3}}{2} (2)\right)\\
&=\left( \frac{2\sqrt{3}}{3},\sqrt{3}\right)
\end{align*}

OK couldn't find some comprehensive equation for this so eyeballed it...
 

Attachments

  • s4.13.t.71.PNG
    s4.13.t.71.PNG
    3.8 KB · Views: 119
Last edited:
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
11K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
8
Views
10K
  • · Replies 1 ·
Replies
1
Views
17K