Pythagorean pyramid, find minimal volume?

This question was asked on this website about a year ago. At that time I gave a detailed explanation of how to find the answer so I’m not going to repeat it here.

(It was to some extent a trial and error method but directed at a restricted range of possible solutions. It is in the middle of the second page of my starred questions.)

One answer is for the perpendicular sides to be 44, 117, 240.

Then you have

44^2 + 117^2 = 125^2

44^2 + 240^2 = 244^2

117^2 + 240^2 = 267^2

The pyramid thus has six integer edges of 44, 117, 125, 240, 244, 267. I believe that (1/3)*44*117*240 = 411840 is the smallest possible volume but I’m not sure.

EDIT. Sorry, volume should be (1/6)*44*117*240 = 205920.

Mathsmanretired is right – the answer is (44, 117, 240) indeed and is known long ago.

Your question is related to the so called ‘Euler brick’ – a cuboid with integer edges and face diagonals (follow the link in Sources below for a Wiki Article on the subject). In some sources I have met also the requirement the inner diagonal to have integer length too (described as ‘Perfect Cuboid’ in the article – one of still unsolved famous problems), but if we accept the definition in the article, the questions about the cuboid and the right tetrahedron are equivalent.

This is an example when an elementary computer search (no need one to be a wizard as my friend Mr. Daftary has supposed) does all the job with minimal efforts. A code like the following, written in less than a minute:

for x = 1 to 12 do for y = x to 12 do for z = y to 12 do

if frac(sqrt(x*x + y*y) = 0 and frac(sqrt(y*y + z*z) = 0 and frac(sqrt(z*z + x*x) = 0 then Output(x, y, z),

produced for less than 5 seconds on my computer all solutions, less than 12: (44, 117, 240), (85, 132, 720), (88, 234, 480)*, (132, 351, 720)*, (140, 480, 693), (160, 231, 792), (176, 468, 960)*, (220, 585, 12)*, (240, 252, 275), (480, 504, 550)*, (720, 756, 825)*, (960, 18, 11), (18, 11, 1155)

/non-primitive solutions with asterisks/

Such example of computer superiority over a human mind makes me feel bad, but that is the life!

Final remarks:

1) Probably more information (about analytical methods to treat the problem etc.) is available googling ‘Euler Brick’

2) Another problem (I think interesting also) just came to my mind: consider a tetrahedron, whose faces are 4 right triangles (3 of its edges are 2 by 2 perpendicular – like a Greek ‘Π’, one of whose legs is twisted at a right angle, its longest edge is a diameter of the circumscribed sphere), unlike the right tetrahedron in question (it has 3 right and 1 acute triangle as faces, the three 2 by 2 perpendicular edges meet in a vertex). Can anybody find such tetrahedron with all 6 edges having integer length? Hint: read carefully the Wiki article.

Source(s): http://en.wikipedia.org/wiki/Euler_brick

Assuming the pyramid could desire to be a familiar sq. pyramid, the main important sq. pyramid interior a cube with area length d, could have top f = d and base area e = d (such that it shares a cube’s area as its base), and its apex on the middle of the choice cube area, such that its quantity V[pyramid] = (a million/3) e^2 f = d^3 / 3 = one-0.33 the quantity V[cube] of the cube. hence, V[pyramid] is maximized as V[cube] is maximized. A cube interior a cylinder with radius b and top c, could have quantity V[cube] such that: V[cube] = min(2b, c)^3 *** Eq. a million considering that c is inversely proportional Assuming the two caps of the cylinder are such that their around perimeters are parallel “small circles” of the sector, (that’s, for the cylinders optimum radius b[max] = a million, c = 0; and additionally, for the cylinder’s optimum top c[max] = 2, b = 0). enable factor o be the middle of the sector, and enable factor p be on the edge of a around cap. enable oq be a radius of the cylinder such that pq is perpendicular to the around cap. Then, op = the around radius a = a million, pq = a million/2 the cylinder’s top = c/2, and oq = the cylinder’s radius b opq is an exact triangle with hypotenuse op. with the help of the Pythagorean Theorem, c^2 / 4 + b^2 = a million => c = 2 sqrt(a million – b^2) *** Eq. 2 considering that 0 < b < a million and 0 < c < 2, we pick purely evaluate the valuable branches of the sqrts. Now, b is inversely proportional to c, and the two are monotonic, meaning that (with the help of Eq. a million), V[cube] is maximized the place b = c. So our Pythagorean equation could properly be simplified to: b^2 / 4 + b^2 = a million => b = 2 sqrt(5) / 5 And, so V[pyramid, max] = V[cube, max] / 3 = b^3 / 3 = (2 sqrt(5) / 5)^3 / 3 = 8 sqrt(5) / seventy 5 *** answer ? 5.7% of the quantity of the sector — because of the fact the inradius of a unit cube is a million/2 and a circumradius is sqrt(3)/2, i might think of packing the sector interior the cube might maximize area utilization. the main awkward quantity to fill seems to be the pyramid. So i might wager a maximal order might bypass like this (from innermost to outermost): pyramid, cylinder, cube, sphere i might wager that a minimum order may be cylinder, cube, sphere, pyramid

Let x^2 + y^2 = l^2

y^2 + z^2 = m^2 and

z^2 + x^2 =n^2

=>

x^2 = (1/2)(m^2 + n^2 – l^2)

y^2 = (1/2)(l^2 – m^2 + n^2) and

z^2 = (1/2)(l^2 + m^2 – n^2)

Then we have to try to fit the integer lengths of hypotenuses of Pythagorean triples so that Each RHS of the above equation is a perfect square.

I could not figure out any analytical method to do it, but some computer wiz can develop a software programme to find the least values of xyz satisfying above conditions.