Friday 24 January 2020

Looking for Perfectly Amicable Boxes

It started with a Twitter post by Mario Livio, who is head of public outreach at the Space Telescope Science Institute (the folks who operate the Hubble Telescope) but most of us know him better as a writer of popular math and science. I especially liked his book on the so-called Golden Ratio.

Mario's tweet was, "For buffs:118=14+50+54=18+30+70=15+40+63=21+25+72 and product of each triple is 37,800."

I got to wondering what was the smallest number that would have more than one partition into three parts so that the product of the three was the same. Turns out after some pencil scribbles, that 13 appears to be the smallest: 13= 1+6+6 = 2+2+9 and both have a product of 36.
Fourteen had two pairs. One with a product of 40 and the other with a product of 72 (I leave the actual values to the interested reader to find.)

Twenty-one seemed to be the smallest with three different pairs. Twenty-two and twenty three were interesting because they had the same product: 5+8+9 = 22, 4+9+10 = 23, but both have a product of 360. Twenty-five also produced a pair with a product of 360.

By thirty-nine I found what I believe is the smallest value with a three different partitions with the same product: 6 + 8 + 25 = 5 + 10+ 24 = 4 + 15 + 20; all three with a product 10 1200.
Forty-nine had two different triples.

I finally resorted to a computer search (it seemed quicker even with my poor programming skills than continuing to try to see a pattern in theses. Did I miss something easy?)to see if there was a quadruple before the 118 that Mario had found. By the time I reached the mid 70's I was repeatedly finding triple matches that had products of 5400, but no quadruples, and finally I concluded that 118 was indeed the smallest number that would have four partitions into three parts which had the same product.

So why is two partitions into three parts interesting? Well, for one thing, I now know that there are two different box shapes (rectangular Parallelepipeds) which have a volume of 36 for which the total sum of the edge lengths is 52 (4x13); and that for integer sides and volume, there is no smaller total side length for which that is true. So the L,W,H relation of 2,4,9 and 1,6,6 are special boxes.

The whole thing did get me off on another search. I wondered if there were distinct integer sided boxes for which the edge lengths had the same total and the surface areas are also equal. The short form would be, are there two partitions of a number n into x,y,z and a,b,c such that xy+xz+yz = ab+ac+bc?

Turns out they are pretty common. The smallest is for n = 9 (a total edge length of 36) with sides of 1,4,4 and 2,2,5. Each would have a surface area of 48 sq units.

With multiple examples of boxes that had the same total side lengths and either the same surface area or the same volume, as was left to wonder about the trifecta.... are there pairs, or triples of Boxes which share the same total edge lengths, surface area, and volume.  I already had decided they were obviously to be called Amicable Box Pairs (or triples, etc) ; but did they exist.

Certainly there were  boxes which had the surface area equal to the volume, such as the 6x6x6 cube.  In fact an infinite variety of them if we don't restrict ourselves to the integers.  Any box which has the volume more, or less than equal can be scaled up or down by some dimension to make them equal.  But could we get more than one of them, and could any two of them have the same sum of the edges.  We couldn't start with a cube, like 6,6,6 because no other three numbers which would have the same product with the sum of the edges equal to 72.

At this point I suspended my search due to the late hour, but wondered.  If I ask it, will they answer.... so it's on you know deep mathematical minds and computer programmers.  Do Amicable Box Pairs or Triples exist?

Hans Havermann answered my question:
Chris Thompson writes on the Sequence Fanatics discussion list:

I take it that *Perfectly* Amicable Boxes refers to the question "are there pairs, or triples of Boxes which share the same total edge lengths, surface area, and volume". In which case, no there aren't any.

If the box is a X b X c, specifying the edge lengths fixes a+b+c, the surface ab+bc+ca, and the volume abc. Therefore you have fixed the cubic (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc, and therefore its three roots, up to a permutation.

Or to put it another way, if you have fixed all the elementary symmetric polynomials of a set of numbers, you have fixed the numbers themselves, up to a permutation. 
Thanks To both Hans and Chris. 

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