*Spoiler for New Scientist Enigma 1779 “Four triangles” *(Follow the link to see the puzzle.)

*Edited to correct two omissions, one pointed out by Jim Olson.*

Is there an easier way to do this than to find or build a list of all Pythagorean triples with hypotenuse < 100 and start looking through it?

I find four chains of four if you allow the hypotenuse of each smaller triangle to be *either* leg of the next:

- 9-12-15 / 15-20-25 / 25-60-65 / 65-72-97
- 9-12-15 / 15-36-39 / 39-52-65 / 65-72-97
- 12-16-20 / 15-20-25 / 25-60-65 / 65-72-97
- 12-16-20 / 20-48-52 / 39-52-65 / 65-72-97

and the first two satisfy the requirement that the hypotenuse of the smaller triangle be the *smaller* leg of the next. In either case, the smallest and largest sides are 9 and 97.

Here’s a diagram of the second of these:

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*Related*

You missed a series that satisfies the conditions:

9,12,15 ; 15,20,25 ; 25,60,65 and 65,72,97.

Both series give the same solution: 9mm and 97mm.

Right! Post corrected.