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 solution to [1] 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]
We will find that the condition
 (b - 6d)(c - 6d) > 18d2 [3]
must also hold in order for a valid solution for the edge length numerator a to be found. Then
 bc - 3(b + c)d > 3(b + c)d - 18d2 [4]
By [2],
b ≥ c ≥ 6d + 1

Applying [4] to the numerator, we have
 [5]
Next, we can solve [1] for a in terms of b, c and d:
Dividing both sides by abc,
 [6]
By [5],
1/(3d) - 1/a - 1/b - 1/c > 0,
and therefore we can square both sides of [6] to obtain
[(b - 6d)(c - 6d) - 18d2]a = 6d(b - 3d)(c - 3d) - 54d3
Since b ≥ c ≥ 6d + 1, the right side is positive.

The condition [3] must hold so that the division produces a positive value for a:

 [7]
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:
• [3] is satisfied
• an integer value for a is produced by [7]
• 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 to 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 values 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:
dABCFace ABFace ACFace BCBaseV=S
118141212610884186504
124121214414472216576
3361232/321619264296768
132248384128964161024
1541210324270604261080
166188594264726541584
31441228/38646725610962688
116816813446726415043584
33363620/360481120120615213440