Finding a tetrahedron in a unit sphere

Question

I am trying to solve an exercise where I have to find a tetrahedron in a unit sphere. My first thought and i think it is not a bad idea is to let one point be at $(0,1,0)$ (assume this is $A$) and then find a plane

$y = -d$

so that

$|\overline{AB}|$ = $|\overline{BC}|$

using a new point $O = (0, d, 0)$ which will be the centroid of the triangle $BCD$. Now i know that the points $B, C, D$ will be on the sphere so that the equation

$x^2 + z^2 = 1 - d^2$

holds true. Furthermore, I know that

$2 * sin(60) * \overline{OB} = \overline{BC}$.

I feel like i have collected all the parts of the puzzle, but i still cannot solve it. Any suggestions?

EDIT: Since somone mentioned it in the comments, it should be a regular tetrahedron

Answer 1

Plane $y = d$ will cut the sphere $x^2 + y^2 + z^2 = 1$ such that you get a circle

$x^2 + z^2 = 1 - d^2 \,$ (so radius is $\sqrt{1-d^2}$) as you rightly found.

Now this circle is circumcircle of the base of our tetrahedron (equilateral triangle). So the side of the equilateral triangle is $\sqrt{3(1-d^2)} \,$.

Now the distance from the vertex $P (0,1,0)$ to the center of the circle at $y = d$ will be $(1-d)$.

So the distance from the vertex $P$ to any of the three vertices on the base will be

$ = \sqrt{1-d^2 + (1-d)^2} = \sqrt{2 - 2d} \,$ (by Pythagoras)

Equating both as the tetrahedron is regular $\sqrt{2 - 2d} = \sqrt{3(1-d^2)}$

Solving it, you get value of $d = 1, -\frac{1}{3}$.

Now we take $d = - \frac{1}{3}$ and find $3$ points on the circle $x^2 + z^2 = \frac{8}{9}$ at $y = -\frac{1}{3}$ such that it is an equilateral triangle. As it is symmetric around $y$ axis, you will have multiple such triples. Pick one for example where $x = 0$ and then the rest of the two points.

Answer 2

You have better work in spherical coordinates for instance with the mathematical convention $$ \left\{ \matrix{ x = r\sin \varphi \cos \theta \hfill \cr y = r\sin \varphi \sin \theta \hfill \cr z = r\cos \varphi \hfill \cr} \right. $$ polar angle $\varphi$, azimuthal angle $\theta$, and $r = 1$ for the unitary sphere.

Take a vertex in the pole $(0,0,1)$.

The other three vertices will be, by symmetry, on azimuthal planes at $120^{\circ}$ from each other.
So $\theta = 0, \pm 2/3 \pi$.

For the angle $\varphi$ consider that it is the same for all the three vertices at the base, and that their position vectors, summed up shall give a null resultant with the polar vertex.
So they are at $z= -1/3$ or $\cos \varphi = -1/3$, i.e. $\sin \varphi =2 \sqrt{2}/3$.

And you have all the elements to fix the other three vertices.