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 Pythagorean Chain (Posted on 2006-03-20)
Take a right triangle with integer sides A, B, & C.
(C need not be the hypotenuse.)

To side C attach another right triangle with integer sides C, D & E.

On this new triangle attach another right triangle to either side D or E. Continue the process of attaching a new right triangle to the previous; creating a chain of right triangles.

Three further rules:
1. No side length may be repeated.
2. No triangles may overlap.
3. No side may have length over 10000.

How many triangles can you make in this chain?

 See The Solution Submitted by Jer Rating: 3.5000 (4 votes)

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 Ideas... | Comment 5 of 18 |

OK, here's an idea:

Starting with the smallest possible trangle, the triangles that get added are going to get larger and larger.  To get the most triangles in the chain, we want to minimize this effect.

We can't build a unique integer right triangle off of the two smaller sides, so the next one we could build would be 5, 12, 13.

You could then build off triangles of size:

12, 35, 37   and   35, 612, 613.

If you continue by making 612 or 613 the smallest side of the new triangle, the triangles get too large to satisfy rule #3.

So that's one approach that maxes out at just four triangles before they get too large.  With this approach, rule #2 is irrelevant, since you can always orient the new triangles away from the preceding ones.

This of course is no proof, just an example.  I have a hunch that there probably exists a solution with more than four triangles, but I haven't found it yet, or a general proof of the limit.

I found these nifty formulas for generating these integer triplets - maybe someone can put them to use in finding a more optimal solution to this problem:

Given two integers p and q (where p > q), a right triangle can be formed by:

side a = 2pq
side b = p^2 - q^2
side c = p^2 + q^2

 Posted by tomarken on 2006-03-20 15:37:29

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