*Balmoral Software*

Solutions: 3

(6, 8 or 12 faces)

A dipyramid is created from two identical right pyramids, each of which has a base that is a regular polygon of n sides of length a, joined base-to-base. The height of the entire dipyramid is h. The area of the regular polygon is

BThe volume of the dipyramid is twice that of the pyramid:_{n}= (n/4)a^{2}cot(180°/n)

and is assumed to be an integer by convention. The 2n faces of the dipyramid are identical isosceles triangles, each of which has a base of length a and two sides of length e, where

V = 2(1/3)(h/2)B _{n}= (n/12)a^{2}hcot(180°/n)[1]

eand R is the circumradius of the regular polygon:^{2}= (h/2)^{2}+ R^{2}

The area of each triangle then is

R = (a/2)csc(180°/n) [2]

which is also assumed to be an integer by convention. By equability,

which reduces to

V = 2nT = S, [3]

Substituting [4] in [1] and [3], we have

[4]

where the real number c

[5]

Some values of c

c _{n}= 3tan(180°/n)[6]

For h > 6, T has a minimum value of at h = , and two real solutions for h exist for each integer value of T above this minimum. Therefore, there are infinitely-many equable isosceles dipyramids. We henceforth restrict the analysis to the more distinctive cases where c

__Case n = 4__

In the first case, we have

so the fraction on the right is a positive integer, and therefore its numerator is at least as large as its denominator:

[8]

108h ≥ hWe can iterate [8] over the range 7 ≤ h ≤ 108 to find all solutions for which 2T is an even integer; the single result is^{2}- 36

h = 12, T = 24

__Case n = 3__

Now let h = for some positive integer k. For n = 3, we have

As before,

108k ≥ kIf h > 6, then k > = 3.464, so we iterate over the range 4 ≤ k ≤ 108 to find the solution^{2}- 12

k = 4, h = , T = 72

__Case n = 6__

Similarly, we have for n = 6:

We can then iterate over the range 4 ≤ k ≤ 36 to find the solution36k ≥ k

^{2}- 12

k = 4, h = , T = 24It follows from [4] and [6] that

and from [2] that

R^{2}= (a^{2}/4)(1 + 9/c_{n}^{2})

__Summary__

The three equable solutions are:

Both the equable octahedron and the equable dodecahedron have identically-sized faces.

n c _{n}h a R e T V = S 3 12 72 432 4 3 12 24 192 6 24 288

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