# Side length of a regular tetrahedron given the radius of the sphere tangent to its edges

I am working on a physics project where we have to build a container to protect a glass ornament when dropped from a high place. My design involves building a tetrahedron out of straws and putting the glass ornament inside of it. I'm not sure what length to cut each side of the straw tetrahedron. I did some measurements and I estimated the radius of the sphere to be about 1.3. The tetrahedron has no "walls" where the ornament will touch at a single point. It is simply an empty frame made out of regular McDonald's straws, and the ornament will protrude out of each side a little bit, thus there are no faces, only edges. How can I find the best side length?

• Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron – Daniel Mathias Dec 31 '18 at 23:38
• The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over. – Will Jagy Dec 31 '18 at 23:56

## 2 Answers

Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:

$$r=\frac{a}{\sqrt8}\iff a=2r\sqrt2\approx2.82r$$

Proving the formula is a fairly simple exercise.

https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron

• is 'a' the side length of the tetrahedron? – user22333 Jan 1 at 0:42
• @user22333 Exactly. – Oldboy Jan 1 at 6:32

Added: in the construction below, the point on an edge closest to the origin is at $$(0,0,1),$$ radius for this is exactly $$1,$$ therefore the edge length divided by $$\sqrt 8$$

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one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $$(\pm1,\pm1,\pm1).$$ For example $$(-1,-1,-1) \; , \; \; \; (-1,1,1) \; , \; \; \; (1,-1,1) \; , \; \; \; (1,1,-1) \; . \; \; \;$$ The edges are all length $$\sqrt 8$$

One triangle side is $$x+y+z=1.$$ The closest point to the origin in that plane is $$\left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right)$$ with distance from the origin $$\frac{\sqrt 3}{3} = \frac{1}{\sqrt 3},$$ which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$\sqrt{24}$$

• OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges. – Oldboy Jan 1 at 0:35
• @Oldboy I see what you mean. Added a note at the beginning. – Will Jagy Jan 1 at 0:40