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Pythagorean chain converse (Posted on 2006-06-16) Difficulty: 3 of 5
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 integer right triangles.

No side length may be repeated.

If n is the number of triangles in the chain, what is the minimum largest side for n=2, 3, 4, 5, 6, 7, 8, 9, 10.

No Solution Yet Submitted by Jer    
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Some Thoughts n = 6 (spoiler) | Comment 2 of 5 |
Why am I still talking to myself?  When does Charlie weigh in?

Well, I'll start.

n   min
--  ---
1    5
2   13
3   15
4   17
5   17
6   25

The smallest 8 triangles are:
3     4     5
    6     8    10
    5    12    13
    9    12    15
    8    15    17
12 16 20
7 24 25
15 20 25
And they can be chained:

n = 1
------
3-4-5

n= 2
------
3-4-5
5-13-12

n = 3
-------
3-4-5
5-13-12
12-9-15

n = 4
-------
3-4-5
5-13-12
12-9-15
15-17-8

n = 5
-------
3-4-5
5-13-12
12-9-15
15-17-8
8-6-10

n = 6
-------
3-4-5
5-13-12
12-16-20
20-25-15
15-17-8
8-10-6

(note that 20 is not an achievable minimum if n = 6, because the a side of 12 comes up in 3 of the first 6 triangles)

(note that I formed chains 2,3,4,5 by adding one more to the previous chain.  for n = 6, I needed to rearrange)







  Posted by Steve Herman on 2006-06-17 08:01:43
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