# Pythagorean triangles

1. Nov 10, 2005

### ceptimus

Pythagorean triangles are right angled triangles where each of the three sides is an integer length. The 3,4,5 triangle is the best known.

Find three Pythagorean triangles that each have the same area.

2. Nov 11, 2005

### LarrrSDonald

15 112 113
42 40 58
70 24 74
(leg leg hypotenuse)
All three have area 840.

Gotten by:
I essentially used the old formula for generating primitive (i.e. non-reducable by division)pythagorian triangles:
$$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$
...where m>n and m and n are relativily prime and have different parities. I have no idea how to prove this, but appearently this spans all primitives and only primitives. I'd seen it a few times but had to look it up. Anyway, this means that:
$$2A=(m^2-n^2)(2mn)$$
$$A=m^3n-mn^3$$
I generated the areas for all m,n in the 1-10 span, replacing those not fitting the bill by 0:
Code (Text):
0,   0,    0,    0,    0,    0,    0,    0,    0,    0
6,   0,    0,    0,    0,    0,    0,    0,    0,    0
0,   30,   0,    0,    0,    0,    0,    0,    0,    0
60,  0,    84,   0,    0,    0,    0,    0,    0,    0
0,   210,  0,    180,  0,    0,    0,    0,    0,    0
210, 0,    0,    0,    330,  0,    0,    0,    0,    0
0,   630,  0,    924,  0,    546,  0,    0,    0,    0
504, 0,    1320, 0,    1560, 0,    840,  0,    0,    0
0,   1386, 0,    2340, 0,    0,    0,    1224, 0,    0
990, 0,    2730, 0,    0,    0,    3570, 0,    1710, 0

Right away, the two 210s pop out at me. These are (m,n) 5,2 and 6,1. Feeding them into the original formula gives 21,20 and 35,12 as their respective legs (areas indeed 210). Now, it's obviously possible to scale them up by a set factor and still have them remain pythagorian (though no longer primitive). Scaling the legs up scales the area up by it's square, i.e. if multiply the legs by two and the area increases by 4. Hence, if there is another area that can be obtained by multiplying 210 with a square number, there's a third one. As I start thinking about how one might find intersection between the up-scalings of these, I notice that 840 (210*4) is already on the board at 8,7. This gives 15,112 (hyp 113, area 840). I scale up the others by two (21,20 -> 42,40 and 35,12 -> 70,24). Re-crunching just to make sure, they are indeed still pythagorian and all three now equal in area.

[EDIT]I went ahead and reduced the primitives to a list, multiplied each by all square numbers small enough to hit the largest and sorted/analyzed. Hence, I can now also say with some confidence that this is the smallest answer possible (figured it was, but I didn't exactly know). There also seems to be no others at least up to A=~10k that aren't further upscales of these. I checked for four or more up to A=~100k, but no such luck.
[EDIT]Just for the heck of it, here's the smallest four with equal areas (felt a little odd about the three case since using a computer as I admittedly did one could have exhaustive searched the integer triangles up to 200 or so - the algo used here is way overkill):
(leg leg hyp)
13230 8360 15650
10395 10640 14875
14950 5544 20706
5985 18480 19425
Areas are 55301400

Last edited: Nov 12, 2005