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Let there be a unit sphere. Centered at the origin, there also is a regular tetrahedron whose faces are tangent to the circumference of the sphere. By this, I mean the tetrahedron completely encompasses the sphere. What is the distance between the origin and the vertices of the tetrahedron in this case? What is the edge length of this tetrahedron?

I looked up the features of a regular tetrahedron but I am not sure whether the origin in question would be equivalent to the centroid of the tetrahedron. I can see that you can use Pythagoras' Theorem to find the distances, where (assuming the tetrahedron's centroid = origin) there would be a right triangle with the following vertices: the origin, a vertex of the tetrahedron, the centroid of an equilateral triangle (one of the 4 equal faces of the tetrahedron). The angle between the origin and the centroid of the equilateral triangle is 90°, so the distance between the vertex and the origin is the hypotenuse of the triangle formed. I don't know about the other angles. enter image description here

Christian Blatter
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abtoiew
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  • Do you mean that the faces of the tetrahedron are tangent to the sphere, so that sphere is inscribed in the tetrahedron? What have you tried so far? – saulspatz Apr 05 '21 at 09:36
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    What have you tried? Can you please edit the question to show your attempt? – Math Lover Apr 05 '21 at 09:48
  • @saulspatz Yes, I meant to say faces, thank you. I updated the question. – abtoiew Apr 05 '21 at 09:56
  • @MathLover I added in what I tried along with 2 diagrams. – abtoiew Apr 05 '21 at 10:12
  • The centroid of a regular tetrahedron divides the altitude into $1:3$ ratio. Altitude is $\sqrt{\frac{2}{3}} \ a$ if $a$ is the side length. So radius of inscribed sphere will be $\frac{1}{4}$ of the altitude (radius is distance from centroid to the face) with center at the centroid. – Math Lover Apr 05 '21 at 10:14
  • You may want to refer to this: https://en.wikipedia.org/wiki/Tetrahedron – Math Lover Apr 05 '21 at 10:15
  • @MathLover Does it mean that a regular tetrahedron centering at the origin will have its centroid at the origin as well? – abtoiew Apr 05 '21 at 10:17
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    Yes, a regular tetrahedron circumscribing a sphere centered at the origin will have its centroid also at the origin. – Math Lover Apr 05 '21 at 10:19
  • @MathLover Isn't it so that if the tetrahedron is circumscribing a sphere, the sphere is touching all of its vertices? In this case, the sphere is touching its faces only; therefore, the tetrahedron is inscribing the sphere. – abtoiew Apr 05 '21 at 16:39

1 Answers1

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Here's an alternative approach. Let $r > 0$. The points $p_{0} = (r, r, r)$, $p_{1} = (r, -r, -r)$, $p_{2} = (-r, r, -r)$, and $p_{3} = (-r, -r, r)$ are vertices of a regular tetrahedron. The rest can be handled with coordinate geometry.

Barycenters of faces and radius of inscribed sphere:

Each lies at distance $\sqrt{3}r$ from the origin, and their mutual separation is $2\sqrt{2}r$. The barycenter of each face is a point of tangency for the inscribed sphere centered at the origin. Particularly, $\frac{1}{3}(p_{1} + p_{2} + p_{3}) = -\frac{1}{3}(r, r, r)$ lies on the inscribed sphere, which therefore has radius $R = \frac{1}{\sqrt{3}}r$.

Requested dimensions for unit inscribed sphere:

If $R = 1$, then $r = \sqrt{3}$, so each vertex is at distance $3$ from the origin, and the edge length is $2\sqrt{6}$.

Andrew D. Hwang
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