Right Square & Right Rhombic Pyramid

EQUABLE RIGHT SQUARE & RIGHT RHOMBIC PYRAMID

Balmoral Software

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

T > B/4
1 ≤ B ≤ 4T - 1
[1]

If h represents the height of the pyramid, then

V = (1/3)Bh

By equability, the total surface area is

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

so the height

[3]

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

p² + q² = 4a² [4]

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

B = pq/2 [5]

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

T² = (1/4)[(p/2)²h² + (q/2)²h² + (p/2)²(q/2)²)] [6]

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

T² = a²h²/4 + B²/16

Using [3], this reduces to

[7]

and so a² is rational. Collecting terms in T, we have

4(B² - 36a²)T = B³ + 36a²B > 0

Therefore, an equable solution must have

B > 6a

Since a non-square rhombus is non-cyclic, its area does not exceed that of the corresponding square:

B ≤ a², [8]

from which it follows that

B ≥ 37 [9]

Substituting [7] in [8],

[10]
[11]

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

B² - (4T - 36)B + 144T ≤ 0 [12]

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

q = 2B/p [13]

Substituting into [4], we have

p² + (2B/p)² = 4a²
[14]

The 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:

[15]

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

T ≥ 53
37 ≤ B ≤ 4T - 1 by [9] and [1]
If inequality [12] holds:
Calculate V by [2], h by [3] and a² by [7]
Calculate p and q by [14] and [13]
Calculate e_p and e_q by [15]

For example, when T = 53, there are 21 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

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

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

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 a = B/12.

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

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

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

h = 3 + 12T/B [16]

and therefore,

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

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

[18]

Since T ∈ ℤ and 12 is relatively prime to 5, it follows that 12 | B, or B/12 ∈ ℤ. Substituting [18] in [7],

a² = (B/12)²,

so a ∈ ℤ.

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 = mx²

and

h = my²

for some positive integers m, x, y. It follows that

my² - mx² = 6
m(y² - x²) = 6

Since the left side is a product of two integers, we have the following possibilities:

m y² - x²

1 6 Difference of two squares cannot be 6

2 3 Only possibility is 2² - 1²

3 2 Difference of two squares cannot be 2

6 1 Difference of two squares cannot be 1

The only solution is h = my² = (2)2² = 8 and

so a = B/12, QED.

Right Square Pyramids

For a square base, there is equality in [11]:

[19]

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

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)
From [2], V = S = 2B²/(B - 36)
From [7], a² = B
a⁴ - B² = 0
p² = 2a² = 2B by [14]
p²/4 = B/2

The equable right square pyramids are listed below, ordered by volume:

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

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 the same area for all five pyramid faces is B = T = 60, h = 15 and V = S = 300.
The only solution with an integer base side is a = 12, h = 8 and V = S = 384.

Top Page

Home

m	y² - x²
1	6	Difference of two squares cannot be 6
2	3	Only possibility is 2² - 1²
3	2	Difference of two squares cannot be 2
6	1	Difference of two squares cannot be 1