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Solutions: 10 (square) Solutions: ∞ (rhombic) |

There are exactly 10 equable right square pyramids and infinitely-many equable right rhombic pyramids. Denote the area of the rhombus base by the positive integer B, the area of one of its four congruent lateral triangular faces by the positive integer T, and its volume by the integer V. Since each triangular face is a vertical projection above one-quarter of the base, it is evident that

If h represents the height of the pyramid, then

T > B/4 1 ≤ B ≤ 4T - 1

[1]

V = (1/3)BhBy equability, the total surface area is

so the height

S = B + 4T = V, [2]

is rational. The relationship between the side a and the diagonals p and q of the rhombus base is

[3]

The area of the rhombus can be expressed in terms of its diagonals as

p ^{2}+ q^{2}= 4a^{2}[4]

Since every rhombus is orthodiagonal, each triangular face of the pyramid is the base of a trirectangular tetrahedron having edge lengths p/2, q/2 and h bounding the trihedral angle. The square of the area of the triangular face then is

B = pq/2 [5]

Substituting [4] and [5] into [6], we have

T ^{2}= (1/4)[(p/2)^{2}h^{2}+ (q/2)^{2}h^{2}+ (p/2)^{2}(q/2)^{2})][6]

TUsing [3], this reduces to^{2}= a^{2}h^{2}/4 + B^{2}/16

and so a

[7]

4(BTherefore, an equable solution must have^{2}- 36a^{2})T = B^{3}+ 36a^{2}B > 0

B > 6aSince a non-square rhombus is non-cyclic, its area does not exceed that of the corresponding square:

from which it follows that

B ≤ a ^{2},[8]

Substituting [7] in [8],

B ≥ 37 [9]

For B ≥ 37, the right side of [11] is minimized at so T must be at least 53. The inequality in [10] can also be written

[10] [11]

Without loss of generality, we can take p ≥ q and write [5] as

B ^{2}- (4T - 36)B + 144T ≤ 0[12]

Substituting into [4], we have

q = 2B/p [13]

pThe lateral edge lengths of the pyramid are the hypotenuses associated with the mutually-orthogonal half-diagonals p/2 and q/2 and the height h:^{2}+ (2B/p)^{2}= 4a^{2}

[14]

Since T is unbounded, there are infinitely-many equable right rhombic pyramids. They can be generated over the following domain:

[15]

- T ≥ 53
- 37 ≤ B ≤ 4T - 1 by [9] and [1]
- If inequality [12] holds:
- Calculate V by [2], h by [3] and a
^{2}by [7] - Calculate p and q by [14] and [13]
- Calculate e
_{p}and e_{q}by [15]

For T ≤ 1000, there are 1,889,244 equable right rhombic pyramids.

B T h V=S a p q e _{p}e _{q}p/q 78 53 145/13 = 11.1538 290 8.8368 12.7930 12.1942 12.8578 12.7115 1.05 79 53 873/79 = 11.0506 291 8.9013 13.0626 12.0956 12.8364 12.5973 1.08 80 53 219/20 = 10.9500 292 8.9647 13.2679 12.0591 12.8028 12.5003 1.10 81 53 293/27 = 10.8519 293 9.0268 13.4425 12.0514 12.7647 12.4126 1.12 82 53 441/41 = 10.7561 294 9.0878 13.5966 12.0619 12.7244 12.3315 1.13 83 53 885/83 = 10.6627 295 9.1477 13.7350 12.0859 12.6829 12.2560 1.14 84 53 74/7 = 10.5714 296 9.2063 13.8606 12.1207 12.6406 12.1854 1.14 85 53 891/85 = 10.4824 297 9.2638 13.9747 12.1649 12.5977 12.1192 1.15 86 53 447/43 = 10.3953 298 9.3202 14.0783 12.2174 12.5544 12.0573 1.15 87 53 299/29 = 10.3103 299 9.3753 14.1721 12.2777 12.5106 11.9995 1.15 88 53 225/22 = 10.2273 300 9.4293 14.2561 12.3456 12.4662 11.9457 1.15 89 53 903/89 = 10.1461 301 9.4822 14.3306 12.4210 12.4211 11.8959 1.15 90 53 151/15 = 10.0667 302 9.5338 14.3952 12.5041 12.3751 11.8502 1.15 91 53 909/91 = 9.9890 303 9.5843 14.4495 12.5956 12.3279 11.8086 1.15 92 53 228/23 = 9.9130 304 9.6336 14.4927 12.6960 12.2792 11.7714 1.14 93 53 305/31 = 9.8387 305 9.6818 14.5236 12.8068 12.2284 11.7390 1.13 94 53 459/47 = 9.7660 306 9.7287 14.5402 12.9296 12.1749 11.7119 1.12 95 53 921/95 = 9.6947 307 9.7745 14.5396 13.0678 12.1177 11.6910 1.11 96 53 77/8 = 9.6250 308 9.8191 14.5160 13.2267 12.0549 11.6781 1.10 97 53 927/97 = 9.5567 309 9.8626 14.4580 13.4182 11.9829 11.6766 1.08 98 53 465/49 = 9.4898 310 9.9048 14.3284 13.6791 11.8904 11.6977 1.05

All five pyramid faces have the same area if and only if h = 15, which is
equivalent to .

Also, the base side a is an integer if and only if h = 8, which is equivalent to

Proof. For the first part, note that in [3],

is equivalent to T = B. Then from [7],

so .

For the second part, note that [3] is equivalent to

and therefore,

h = 3 + 12T/B [16]

If h = 8 in [16], then 12T/B = 5, or

4T = B(h/3 - 1) [17]

Since T ∈ ℤ and 12 is relatively prime to 5, it follows that

[18]

aso a ∈ ℤ.^{2}= (B/12)^{2},

Conversely, note that by substituting [17] in [7], we have

If a ∈ ℤ, then

is the square of a rational, so we can write

h - 6 = mxand^{2}

h = myfor some positive integers m, x, y. It follows that^{2}

mySince the left side is a product of two integers, we have the following possibilities:^{2}- mx^{2}= 6m(y

^{2}- x^{2}) = 6

The only solution is h = my

m y ^{2}- x^{2}1 6 Difference of two squares cannot be 6 2 3 Only possibility is 2 ^{2}- 1^{2}3 2 Difference of two squares cannot be 2 6 1 Difference of two squares cannot be 1

so a = B/12, QED.

Since T is an integer, B - 36 must divide the numerator. The integer quotient is

[19]

so B - 36 must divide 2592. Thirty candidate values for B can then be generated from the divisors of 2592. Of these, ten produce integer values of T in [19]. The remaining parameters can all be determined in terms of B:

From [3], h = 6B/(B - 36)The equable right square pyramids are listed below, ordered by volume:From [2], V = S = 2B

^{2}/(B - 36)From [7], a

^{2}= Ba

^{4}- B^{2}= 0p

^{2}= 2a^{2}= 2B by [14]p

^{2}/4 = B/2

The only solution with the same area for all five pyramid faces is

Base area B Triangle face area T Height h V = S Base side a Lateral edge 72 54 12 288 8.4853 13.4164 60 60 15 300 7.7460 15.9687 108 54 9 324 10.3923 11.6189 48 84 24 384 6.9282 24.4949 144 60 8 384 12 11.6619 44 110 33 484 6.6332 33.3317 252 84 7 588 15.8745 13.2288 40 190 60 800 6.3246 60.1664 360 110 20/3 800 18.9737 14.9815 684 190 19/3 1444 26.1534 19.5477

The only solution with an integer base side is a = 12, h = 8 and

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