# 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

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