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 
If h represents the height of the pyramid, then
V = (1/3)Bh,
and therefore h = 3V/B is rational. By equability, the total surface area is
 S = B + 4T = V, 
so the height is rational. The side a of the rhombus base can be calculated from the preceding parameters. The relationship between the diagonals p and q of the rhombus base is
 p2 + q2 = 4a2 
The area of the rhombus can be expressed in terms of its diagonals as
 B = pq/2 
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)] 
Substituting  and  into , we have
T2 = a2h2/4 + B2/16
Using , this reduces to and 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, 
from which it follows that
 B ≥ 37 
From , we then have
4T - 1 ≥ B ≥ 37

 T ≥ 10 

Substituting  in , B2 - (4T - 36)B + 144T ≤ 0 
The left side is a parabola in B opening upwards, so its negative portion is between the two real roots of the quadratic: By , the lower bound for B is positive. For an equable solution, the interval for B must contain at least one integer: which reduces to
T ≥ 52.4571
Therefore, for T ≥ 53, there is at least one integer solution for B in .

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

 q = 2B/p 
Substituting into , we have
p2 + (2B/p)2 = 4a2 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: 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  and 
• If inequality  holds:
• Calculate V by , h by  and a2 by 
• Calculate p and q by  and 
• Calculate ep and eq by 
For example, when T = 53, there are 21 equable right rhombic pyramids:
BThV=Sapqepeqp/q
7853145/13 = 11.15382908.836812.793012.194212.857812.71151.05
7953873/79 = 11.05062918.901313.062612.095612.836412.59731.08
8053219/20 = 10.95002928.964713.267912.059112.802812.50031.10
8153293/27 = 10.85192939.026813.442512.051412.764712.41261.12
8253441/41 = 10.75612949.087813.596612.061912.724412.33151.13
8353885/83 = 10.66272959.147713.735012.085912.682912.25601.14
8453 74/7 = 10.57142969.206313.860612.120712.640612.18541.14
8553891/85 = 10.48242979.263813.974712.164912.597712.11921.15
8653447/43 = 10.39532989.320214.078312.217412.554412.05731.15
8753299/29 = 10.31032999.375314.172112.277712.510611.99951.15
8853225/22 = 10.22733009.429314.256112.345612.466211.94571.15
8953903/89 = 10.14613019.482214.330612.421012.421111.89591.15
9053151/15 = 10.06673029.533814.395212.504112.375111.85021.15
9153909/91 = 9.98903039.584314.449512.595612.327911.80861.15
9253228/23 = 9.91303049.633614.492712.696012.279211.77141.14
9353305/31 = 9.83873059.681814.523612.806812.228411.73901.13
9453459/47 = 9.76603069.728714.540212.929612.174911.71191.12
9553921/95 = 9.69473079.774514.539613.067812.117711.69101.11
9653 77/8 = 9.62503089.819114.516013.226712.054911.67811.10
9753927/97 = 9.55673099.862614.458013.418211.982911.67661.08
9853465/49 = 9.48983109.904814.328413.679111.890411.69771.05
For T ≤ 1000, there are 1,889,244 equable right rhombic pyramids.

It's easy to see from  and  that all five pyramid faces have the same area if and only if h = 15, which is equivalent to . Also, if h = 8, then a is an integer.

### Right Square Pyramids

For a square base, there is equality in  and . Solving  for T, we have 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, and the corresponding values of T determined by . Of these, ten satisfy equality in . The equable right square pyramids are listed below, ordered by volume:
Base area BTriangle face area THeight hV = SBase side aLateral edge
7254122888.485313.4164
6060153007.746015.9687
10854932410.392311.6189
4884243846.928224.4949
1446083841211.6619
44110334846.633233.3317
25284758815.874513.2288
40190608006.324660.1664
36011020/380018.973714.9815
68419019/3144426.153419.5477
The only solution with integer base sides is a = 12, with h = 8 and V = S = 384. The only solution with the same area for all five pyramid faces is B = T = 60, h = 15 and V = S = 300.