Equable Trirectangular Tetrahedron


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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
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)
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:
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: 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:
dABCFace AB/2Face AC/2Face BC/2BaseV=S

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