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August 31

Slices of regular polyhedra?

) Are all convex quadrilaterals a cross section of a regular polyhedron? If not, which aren't?
) Which Regular polygons are the cross section of a regular polyhedron? I'm pretty sure 3,4,5 (slice off an edge), 6 the center of a cube on point and 10, the equator of a dodecagon. Any others?Naraht (talk) 02:00, 31 August 2019 (UTC)[reply]

1: if even some triangles cannot be produced this way—such as triangles having an (obtuse) angle greater than the maximal dihedral angle in the Platonic solids—then it’s pretty clear that “nearby” quadrilaterals also cannot. 2: it is called “dodecahedron”. Incnis Mrsi (talk) 06:52, 31 August 2019 (UTC)[reply]

1. Hmm. So a 179-.5-.5 degree triangle can't be created at all...

2. Whoops. You are correct.Naraht (talk) 22:45, 31 August 2019 (UTC)[reply]

Multiplying conjugate polynomials

We know that (a-b)(a+b) is a^2 - b^2.

However, the product of conjugate polynomials in general gets complicated easily. The rule, for clarification, is that they have the same terms, with the difference is that in one polynomial the signs alternate and in the other they're all positive. Let's try these conjugates:

(a-b+c)(a+b+c)

Is the product a^2 - b^2 + c^2?? No, there's one additional term, 2ac.

Now let's try these conjugates:

(a-b+c-d)(a+b+c+d)

Their product is a^2 - b^2 + c^2 - d^2 + 2ac - 2bd.

What's the general rule for multiplying conjugates with any number of terms?? Georgia guy (talk) 21:04, 31 August 2019 (UTC)[reply]

Regroup the terms to make it fit the form of the basic one. (a-b+c-d)(a+b+c+d) = ((a+c)-(b+d))((a+c)+(b+d)) = (a+c)^2 - (b+d)^2

--76.69.116.4 (talk) 22:27, 31 August 2019 (UTC)[reply]

That's multiplying 2 conjugate polynomials with 4 terms; I already did that above. Can anyone extend the sequence and find the product of 2 conjugates with 5 terms?? This means:

(a-b+c-d+e)(a+b+c+d+e)

Georgia guy (talk) 20:43, 1 September 2019 (UTC)[reply]

For the general case, you can put the positive terms in the first set of brackets, and the negative terms in the 2nd set, so for n=5, this would give you (a-b+c-d+e)(a+b+c+d+e) = ((a+c+e)-(b+d))((a+c+e)+(b+d)) = (a+c+e)^2 - (b+d)^2 Iffy★Chat -- 21:13, 1 September 2019 (UTC)[reply]