Question

Suppose a dodecahedron is centered at the origin and two of its twenty vertices are at (0,0,1) and (0,0,-1). Suppose there is a vertex at (a,0,b) with a, b > 0. There are two different possibilities for a and b. Give the exact values for a and b for each possibility.

Answer 1

From (0,0,1) and (0,0,-1), we have the "height" of the weight-shape-"roof" is h =2
Hence the vertexes of the form (a, 0, b) is calculated by \((\pm (1-h^2),0,(1+h))\)

Answer 2

How do you get that? I havent seen anything like that before for problems like these.

Answer 3

hahaha.. it is from https://en.wikipedia.org/wiki/Dodecahedron
part "Catesian Coordinate" , read it.

Answer 4

I see. I would hope that i wouldnt have to use that, because then Id have to make the argument about why thats true, lol. Unless you see a way to explain why thats true. Otherwise we're using using some sort of triangles along with the spherical trigonometry formulas or something.

Answer 5

ok, hihihi.... I don't know how to prove.

Answer 6

Lol, gotcha. No worries then. Yeah, I assumed there was some way to start incorporating triangles into it all.

Answer 7

Question: is it a regular one??

Answer 8

Not specified, so we can't assume that it is without somehow showin git. If it were regular, I couldnt imagine there being more than one possibility for a and b like the problem states, though
.

Answer 9

If it is a regular one, then its vertices are on the circle center (0,0,0) and radius 1, right?

Answer 10

the sphere, not circle.

Answer 11

(a,0,b) is the vertex on xz plan, we can find it by vector space, right?

Answer 12

oh, forget it. I am crazy.

Answer 13

Yeah, thatd be too eays becuase yeah, itd be on the x-axis, (1,0,0). SO definitely dont think its regular

Answer 14

@zzr0ck3r any idea?? please.

Answer 15

@Empty

Answer 16

I like this picture of the orthogonal projection of the regular dodecahedron:
https://upload.wikimedia.org/wikipedia/commons/3/31/Dodecahedron_t0_H3.png

Now look at the top point and bottom point, these are the points (0,0,1) and (0,0,-1) So to be clear up and down are the z-axis and left and right are the x-axis.

So now where are (a,0,b)? Well it'll be one of the outer most points, since that's going to be along the equator at the plane where y=0.

So there are 10 points equally spaced at y=0 so the angle between them will be \(\frac{2 \pi}{10} = \frac{\pi}{5}\).

Rotating down from the top point we can see we'll have for the points pretty clearly:

\[(a_1,0,b_1) = ( \cos ( \frac{\pi}{2} - \frac{\pi}{5}), 0, \sin ( \frac{\pi}{2} - \frac{\pi}{5}) )\]
\[(a_2,0,b_2) = ( \cos ( \frac{\pi}{2} - 2 \frac{\pi}{5}), 0, \sin ( \frac{\pi}{2} - 2\frac{\pi}{5}) )\]

You can simplify it but I am writing it this way to make it clear how I'm looking at it geometrically.

Answer 17

Actually this doesn't seem right. There is no "equatorial plane at y=0" is there? Hahaha damn.

Answer 18

Can we be certain its regular, though?

Answer 19

I was under the false assumption that both a and b were greater than zero, there's actually no way what I describe can be done.

Also if it's not regular then doesn't this entire question become arbitrary? What's stopping us from placing multiple points anywhere we like?

Answer 20

They are supposed to be greater than 0. That's how its stated in the problem.

Answer 21

And I suppose you're right, lol. I guess you could have mishapen thing you wished.

Answer 22

Hmmm I don't know if it's possible to have a regular dodecahedron with two vertices at (0,0,1) and (0,0,-1) while having 2 other points in the first quadrant of the (x,z) plane formed at y=0.

Answer 23

Just looking at the orthogonal projection you can see where all the points will lie, there are multiple ones but none of them can be along the same meridian from +z to -z while tracing a line from the north to south pole without zig-zagging.

Answer 24

I wasnt sure either. I couldnt find a good image to help me see it. I was supposing you could have it tilted oddly such that it might work, but I had no idea. Not the easiest thing to imagine.

Answer 25

Well they say it's centered at the origin with points at (0,0,1) and (0,0,-1) so we are restricted to just rotating it along this fixed axis and any way we interchange the two vertices here while keeping it centered will be symmerical and unnoticeable so seems kinda funny to me. Hmmm

Answer 26

This is the part of the picture I'm focusing on rotating. We can't rotate around the z-axis to put more than one in the y=0 plane at a time.