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May 17

Convex polyhedra with all faces having n edges

For what values of n does there exist a convex (not necessarily regular) polyhedron all of whose faces have n edges? If there is a systematic necessary and sufficient feature of n (or if all n are possible), why? (And ideally, source, please.) Loraof (talk) 16:11, 17 May 2017 (UTC)[reply]

'n' is 5. This follows from the Euler characteristic of a sphere so it is true for any polyhedron which doesn't intersect itself. Basically one just follows the usual argument for finding the number of possible regular polyhedrons but with an inequality stuck in. V-E+F=2, E=nF/2, 3V <= 2E. The inequality is saying every vertex must have at least 3 edges coming out of it. This gives 6F >= nF+12. Setting n=6 won't work, and for n=5 F >= 12 and in fact the dodecahedron has 12 faces. Dmcq (talk) 17:10, 17 May 2017 (UTC)[reply]

I think that the argument given here can be modified to show that n < 6: if all faces are n-gons then the average angle of a face is

180^{\circ }\cdot \left(1-{\frac {2}{n}}\right)

; therefore the average of the sum of the facial angles at each vertex is at least

3\cdot 180^{\circ }\cdot \left(1-{\frac {2}{n}}\right)=180^{\circ }\cdot \left(2+{\frac {n-6}{n}}\right)\geq 360^{\circ }

. But in a convex polyhedron this number must be strictly less than 360 degrees. --JBL (talk) 17:17, 17 May 2017 (UTC)[reply]

So does this mean that a polygon can have faces that are all Heptagons if it is Toroidal?Naraht (talk) 18:24, 17 May 2017 (UTC)[reply]

Yes: Szilassi polyhedron is the sort of thing you're after, I think (though hexagons in this case). --JBL (talk) 19:03, 17 May 2017 (UTC)[reply]

You'd need two holes at least for heptagons, I don't know of anything like that offhand but a search with google might turn one up. It is amazing the things people make. Dmcq (talk) 21:51, 17 May 2017 (UTC)[reply]

Even if more than 3 faces meet at some vertices? Anyhow, here is one: [1] JBL (talk) 00:02, 18 May 2017 (UTC)[reply]

Wow! Thanks. I think I'll try and actually make one out of cardboard this weekend! Dmcq (talk) 14:04, 18 May 2017 (UTC)[reply]

Please report back if successful -- the figures in that paper leave something to be desired :) --JBL (talk) 15:17, 18 May 2017 (UTC)[reply]

Great answers--thanks, everyone! Loraof (talk) 16:09, 18 May 2017 (UTC)[reply]

I found [2] which gives a number of different arrangements of 12 heptagons in a genus 2 surface. One can easily enlarge the patterns on the pages and print them on thin cardboard. Dmcq (talk) 23:13, 21 May 2017 (UTC)[reply]

Inferring a polyhedron from its vertex coordinates

Given only the V vertices of a polyhedron in Cartesian coordinates, how can we determine the vertices of each face? Loraof (talk) 16:14, 17 May 2017 (UTC)[reply]

See Convex hull algorithms for various methods for finding the edges of a convex hull. Dmcq (talk) 17:19, 17 May 2017 (UTC)[reply]

Whoops, that's two edit conflicts in a row! If you impose that the polyhedron should be convex, Dmcq's answer is right (though someone should take a look at that article -- numerous external links in the body is not good). Otherwise, the answer is that it is not uniquely determined. --JBL (talk) 17:22, 17 May 2017 (UTC)[reply]