Let $s$ be the edge length of a regular dodecahedron. As a function of $s$, what is the dodecahedron's minimum and maximum diagonal (i.e. cross-section)?
Minimum and maximum diagonal of a dodecahedron
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0How do you mean "diagonal (i.e. cross-section)"? – 2011-07-16
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0Do you allow diagonals within a face? – 2011-07-16
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0@Joseph Malkevitch, yes, I do allow diagonals within a face - this should allow the shortest possible diagonal across the dodecahedron. – 2011-07-17
3 Answers
For unit edge length, the longest diagonal is 2.802517076888147. If this is not what you
mean (as joriki's comment question indicates), my apologies.
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0Américo and Ross and I (fortunately!) all agree on the longest diagonal, which is the same as the diameter of the circumscribed sphere. (Sorry to disappear as the question was being clarified.) – 2011-07-17
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0That's right, we all agree on that. – 2011-07-17
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0@Joseph, I apologize for disappearing! I was on a trip and without internet access for awhile. Yes, your interpretation of the longest diagonal is correct! And very nice graphic! – 2011-07-17
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0@Joseph, for fun, plugging in your value of "2.802517076888147" into the inverse symbolic calculator generates:"(5+5^(1/2))*15^(1/2)" which equal to exactly 1/10th the exact expression for the longest diagonal. =) – 2011-07-17
In this post of mine I posted this link to a small notebook (see section 15.3. Dodecaedro, in Portuguese), where I compute the relation
$$\frac{l}{d}=\frac{1}{2\sqrt{3}\cos \left( \frac{\pi }{5}\right) }=\frac{2}{% \sqrt{3}\left( 1+\sqrt{5}\right) }\approx 0.356\,82,$$
where $l$ is the edge length (denoted $s$ by you) and $d$ is the diagonal length connecting two opposite vertices.
So
$$\frac{d}{l}=2\sqrt{3}\cos \left( \frac{\pi }{5}\right) =\frac{\sqrt{3}% \left( 1+\sqrt{5}\right) }{2}\approx 2.802\,5.$$
Top figure: edges (black), diagonal (red), axes of faces (magenta). Bottom figure: cross-section rotated with respect to the top figure.
It you want the cross-section area, then I think the bottom figure may help. One would have to compute the height (brown), the radius of the circumscribed circle to the faces, and its apothem.
As far as the other "diagonals" I have not computed them.
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0Can you also provide the length of the vertical line, i.e. the minimum distance through the centre? – 2011-07-16
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0@Henri: I think it is $2r_{i}$, where $r_{i}$ is the radius on the incribed circle (see figure p. 98 or the bottom figure above). In p.100 I found $$r_{i}=l% \sqrt{\left( \frac{\sqrt{3}}{4}\left( 1+\sqrt{5}\right) \right) ^{2}-\left( \frac{1}{2}\right) ^{2}}.$$ So $$2\frac{r_{i}}{l}=2\sqrt{\left( \frac{\sqrt{3}}{% 4}\left( 1+\sqrt{5}\right) \right) ^{2}-\left( \frac{1}{2}\right) ^{2}}% \approx 2.618.$$ – 2011-07-16
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0@Américo Tavares, very nice, thank you! – 2011-07-17
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0@Roger Harris: You are welcome! – 2011-07-17
Wikipedia gives the radius of the circumscribed sphere as $\frac{\sqrt{15}+\sqrt{3}}{4}s\approx 1.401s$. Your maximum diameter is twice this. It also gives the radius of the inscribed sphere as $\frac{\sqrt{250+110\sqrt{5}}}{20}s\approx 1.1135s$. Is this whatyou menant by minimum diagonal-it is between two face centers.
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0@Ross, perfect, this is exactly what I was looking for. I can't believe I missed the wikipedia article... – 2011-07-17