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

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 ∈ ℤIt 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 ≥ 1Equating volume V and surface area S, we have

There can be no solution to [1] when the shortest edge length c/d ≥ 15:

[1]

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

[a - 6d(15d - 1)][b - 6d(15d - 1)] ≥ [b - 6d(15d - 1)]and it follows that^{2}> [6d(15d - 1)]^{2}

abc ≥ ab(6d + 1) > 6abd + 6ad(15d - 1) + 6bd(15d - 1) ≥ 6abd + 6acd + 6bcdSince for positive x and y, we have= 3d(ab + ac + bc) + 3abd + 3acd + 3bcd

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

We will find that the condition

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

must also hold in order for a valid solution for the edge length numerator a to be found. Then

(b - 6d)(c - 6d) > 18d ^{2}[3]

By [2],

bc - 3(b + c)d > 3(b + c)d - 18d ^{2}[4]

b ≥ c ≥ 6d + 1Applying [4] to the numerator, we have

Next, we can solve [1] for a in terms of b, c and d:

[5]

Dividing both sides by abc,

By [5],

[6]

1/(3d) - 1/a - 1/b - 1/c > 0,and therefore we can square both sides of [6] to obtain

[(b - 6d)(c - 6d) - 18dSince b ≥ c ≥ 6d + 1, the right side is positive.^{2}]a = 6d(b - 3d)(c - 3d) - 54d^{3}

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

By convention, the lateral face areas ab/(2d

[7]

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

(9d - 1)(360dfeasible values of (b,c) in [2] to find those for which:^{2}- 45d + 2)/2

- [3] is satisfied

- an integer value for a is produced by [7]

- a ≥ b (to omit duplicates)

- ab/(2d
^{2}), ac/(2d^{2}), bc/(2d^{2}) and abc/(6d^{3}) are integers

- gcd(a,b,c,d) = 1 (to omit duplicates due to common factors in fractions)

d A B C Face AB Face AC Face BC 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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