Equable Trirectangular Tetrahedron

EQUABLE TRIRECTANGULAR TETRAHEDRON

Balmoral Software

Solutions: 6 (integer dimensions)
Solutions: 3 (non-integer dimensions, conjectured)

A trirectangular tetrahedron is a four-sided polytope created from a corner of a cuboid:

There are exactly 6 equable trirectangular tetrahedra with integer dimensions. It is conjectured that no more than 3 additional equable trirectangular tetrahedra have non-integer dimensions.

Let the lengths of the three orthogonal edges meeting at the corner be A, B and C; these are referred to as the dimensions of the tetrahedron. We begin by showing that the dimensions are rational. The volume of the tetrahedron and the areas of the three non-base faces are integers by convention:

ABC/6, AB/2, AC/2, BC/2 ∈ ℤ

It follows that

is rational, and analogously for B and C. Therefore, we can express A, B and C in terms of their lowest common denominator d as a/d, b/d and c/d, where a, b, c and d are positive integers. Without loss of generality, we may take

a ≥ b ≥ c ≥ 1

Equating volume V and surface area S, we have

[1]

There can be no equable solution when the shortest edge length c/d ≥ 15:

which shows that [1] is not satisfied since the volume exceeds the surface area. Next, assume that c ≤ 6d. Then

so again [1] is not satisfied. Therefore, all equable solutions occur when

6d + 1 ≤ c ≤ 15d - 1

Assume the above inequality holds, and let b > 12d(15d - 1). Then

[a - 6d(15d - 1)][b - 6d(15d - 1)] ≥ [b - 6d(15d - 1)]² > [6d(15d - 1)]²

and it follows that

abc ≥ ab(6d + 1) > 6abd + 6ad(15d - 1) + 6bd(15d - 1) ≥ 6abd + 6acd + 6bcd
= 3d(ab + ac + bc) + 3abd + 3acd + 3bcd

Since

for positive x and y, we have

which shows that [1] is not satisfied. Therefore, all equable solutions occur when

6d + 1 ≤ c ≤ 15d - 1 and c ≤ b ≤ 12d(15d - 1)
[2]

Dividing both sides of [1] by 3abcd, we have

The left side must be positive for an equable solution. Squaring both sides reduces to

[(b - 6d)(c - 6d) - 18d²]a = 6d(b - 3d)(c - 3d) - 54d³ [3]

Since b ≥ c ≥ 6d + 1, the right side is positive. The condition

(b - 6d)(c - 6d) > 18d² [4]

must hold so that division in [3] produces a positive value for the edge length numerator a:

[5]

By convention, the lateral face areas ab/(2d²), ac/(2d²), bc/(2d²) and the volume abc/(6d³) must all be integers. By subtraction, the base area is also an integer since the total surface area is an integer.

For a fixed value of d, it is a simple matter to evaluate the

(9d - 1)(360d² - 45d + 2)/2

feasible values of (b,c) in [2] to find those for which:

[4] is satisfied
an integer value for a is produced by [5]
a ≥ b (to omit duplicates)
ab/(2d²), ac/(2d²), bc/(2d²) and abc/(6d³) are integers
gcd(a,b,c,d) = 1 (to omit duplicates due common factors in fractions)

When d = 1, a search is performed over 1268 values of (b,c), which results in six solutions with all integer edge lengths. When the lowest common denominator is 2, there are 11,492 (b,c) values to search, but none produce additional answers satisfying the above criteria. When d = 3, a search of 40,391 value pairs produces three more solutions, each containing at least one non-integer edge length. It is conjectured that no further solutions exist for d ≥ 4. The 9 known equable trirectangular tetrahedra are as follows, ordered by volume:

d A B C Face AB/2 Face AC/2 Face BC/2 Base V=S

1 18 14 12 126 108 84 186 504

1 24 12 12 144 144 72 216 576

3 36 12 32/3 216 192 64 296 768

1 32 24 8 384 128 96 416 1024

1 54 12 10 324 270 60 426 1080

1 66 18 8 594 264 72 654 1584

3 144 12 28/3 864 672 56 1096 2688

1 168 16 8 1344 672 64 1504 3584

3 336 36 20/3 6048 1120 120 6152 13440

d	A	B	C	Face AB/2	Face AC/2	Face BC/2	Base	V=S
1	18	14	12	126	108	84	186	504
1	24	12	12	144	144	72	216	576
3	36	12	32/3	216	192	64	296	768
1	32	24	8	384	128	96	416	1024
1	54	12	10	324	270	60	426	1080
1	66	18	8	594	264	72	654	1584
3	144	12	28/3	864	672	56	1096	2688
1	168	16	8	1344	672	64	1504	3584
3	336	36	20/3	6048	1120	120	6152	13440

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