# Formula to determine multi-faceted 3-D figures **URL:** \ **Category:** Factual Questions\ **Created:** [January 30, 2003, 10:42pm UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740 "2003-01-30T22:42:04Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![Mr.Blue\_Sky](https://avatars.discourse-cdn.com/v4/letter/m/d26b3c/32.png) [@Mr.Blue\_Sky](https://boards.straightdope.com/u/Mr.Blue_Sky)\ **Post date:** [January 30, 2003, 10:42pm UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/1 "2003-01-30T22:42:04Z")
Let’s say I want to make a dodecahedron. How would I determine the shape and size of the facets? Let’s also assume I want the finished shape to be 12" in diameter. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 30, 2003, 11:30pm UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/2 "2003-01-30T23:30:12Z")
That’s a good question. For simplicity’s sake, I’ll look into a regular dodecahedron, which has 12 regular pentagons as its faces. More later. ---
**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\ **Post date:** [January 30, 2003, 11:35pm UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/3 "2003-01-30T23:35:15Z")
> [@](#): > > \*Originally posted by Mr. Blue Sky \* > Let’s also assume I want the finished shape to be 12" in diameter. Across its widest point or its narrowest? Or average? ---
**Author:** ![Mr.Blue\_Sky](https://avatars.discourse-cdn.com/v4/letter/m/d26b3c/32.png) [@Mr.Blue\_Sky](https://boards.straightdope.com/u/Mr.Blue_Sky)\ **Post date:** [January 31, 2003, 12:45am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/4 "2003-01-31T00:45:04Z")
> [@](#): > > \*Originally posted by KneadToKnow \* > \*\*Across its widest point or its narrowest? Or average? \*\* Widest. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [January 31, 2003, 1:02am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/5 "2003-01-31T01:02:15Z")
From [MathWorld](http://mathworld.wolfram.com/Dodecahedron.html), the circumradius of a dodecahedron with edge length e is 1/4 \* (sqrt(15 - sqrt(3)) \* e. You’re looking for the diameter of a circumscribed sphere to be 12 inches, so set 1/2 \* (sqrt(15 - sqrt(3)) \* e = 12, and solve for e. ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [January 31, 2003, 1:20am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/6 "2003-01-31T01:20:42Z")
If you have access to a university library, you might try to look up the book “Regular Polytopes” by H.S.M. Coxeter, which should contain (among other things) the ratios of the edge lengths to the inradii/outradii/midradii of all five regular Platonic solids. **Tyrrell** , I think you mean 1/4 \* (sqrt(15)-sqrt(3))\*e. ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [January 31, 2003, 1:34am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/7 "2003-01-31T01:34:34Z")
Wait a cotton-pickin’ minute here. That MathWorld link says that the circumradius is 1/4 \* (sqrt(15) + sqrt(3))\*e (Not “-” as I originally typed…sorry…) But it also says that you can construct a dodecahedron such that two adjacent vertices will have coordinates (1,1,1) and (1/phi, phi, 0) where phi=(sqrt(5)+1)/2. In which case the edge length is sqrt(5)-1 and the circumradius is sqrt(3), giving a ratio of (sqrt(15)-sqrt(3))/3, not (sqrt(15)+sqrt(3))/4! I think MathWorld is smoking something here…when I worked out the ratio myself by a different technique, I got (sqrt(15)-sqrt(3))/3 for the circumradius/edge-length ratio as well. I think the correct answer to the OP should be 2\*(sqrt(15)-sqrt(3)) inches. ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [January 31, 2003, 1:39am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/8 "2003-01-31T01:39:29Z")
Correction: I’m the one who’s smoking something. I divided when I should have multiplied. (sqrt(15)-sqrt(3))/3 is the ratio of _edge length_ to _circumradius_. The ratio of _circumradius_ to _edge length_ is one over that, or (sqrt(15)+sqrt(3))/4, which is exactly what MathWorld says. I’ll shut up now… ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [January 31, 2003, 1:47am UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/9 "2003-01-31T01:47:24Z")
More generally, how does one extend what we were taught in plane geometry to three dimensions, to work out what these ratios are? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 31, 2003, 3:57pm UTC](https://boards.straightdope.com/t/formula-to-determine-multi-faceted-3-d-figures/151740/10 "2003-01-31T15:57:20Z")
> [@](#): > > \*Originally posted by Lumpy \* > \*\*More generally, how does one extend what we were taught in plane geometry to three dimensions, to work out what these ratios are? \*\* Very carefully. 😉