Pythagorean triples in problem concerning matches

[diana.hole]

Homework Statement

I've encountered a problem in which i need help with to explain my answer:
problem: Steve puts matches of equal length end-to-end to create three sides of a triangle. He possesses an unlimited supply of matches and cannot split anyone of the matches in half, thirds etc. Steve made a right-angled triangle with a certain number of matches. He then used all those matches to make a different shaped right-angled triangle. What is the smallest number of matches Steve could have used?

2. Homework Equations
i used pythagoras' theorem and the formula used to find out the pythag. triples (a=n^2- m^2, b=2nm, c=n^2+ m^2, where n is larger than m)

The Attempt at a Solution

ive found an answer but I am not quite sure whether it's the correct answer. my answer was 60 as the minimum. I am also not quite sure how to justify my answer or explain the process i went through. also, the three sides of both triangles created are considered to be pythag. triples, but they couldn't be acquired through the pythag. triples rule.

 

[zooxanthellae]

I'm not exactly sure what you mean by not being allowed to use the "pythagorean triples rule". Do you mean you can't use a formula to generate triples?

 

[diana.hole]

I need to find 2 pythagorean triples which the sum of all sides must be equal to one another.
I found 2 pythagorean triples (26, 24, 10) & (25, 20, 15), but I couldn't generate those triples from the pythagorean triples formula I used.
I'm quite terrible at explaining things so I apologise

 

[zooxanthellae]

If you're using the a = n^2 - m^2, b = 2nm, c = n^2 + m^2 method of generating triples, notice that it doesn't account for multiples of previous triples. So, for example, (3,4,5) is generated but 5(3,4,5) = (15,20,25) is not. But (15,20,25) is nonetheless a triple. You just have to generate it as a multiple of (3,4,5). You explained it fine.

 

[diana.hole]

If you're using the a = n^2 - m^2, b = 2nm, c = n^2 + m^2 method of generating triples, notice that it doesn't account for multiples of previous triples. So, for example, (3,4,5) is generated but 5(3,4,5) = (15,20,25) is not. But (15,20,25) is nonetheless a triple. You just have to generate it as a multiple of (3,4,5). You explained it fine.

Thanks for that, I now know how to explain my answer.

 

[berkeman]

Homework Statement

I've encountered a problem in which i need help with to explain my answer:
problem: Steve puts matches of equal length end-to-end to create three sides of a triangle. He possesses an unlimited supply of matches and cannot split anyone of the matches in half, thirds etc. Steve made a right-angled triangle with a certain number of matches. He then used all those matches to make a different shaped right-angled triangle. What is the smallest number of matches Steve could have used?

2. Homework Equations
i used pythagoras' theorem and the formula used to find out the pythag. triples (a=n^2- m^2, b=2nm, c=n^2+ m^2, where n is larger than m)

The Attempt at a Solution

ive found an answer but I am not quite sure whether it's the correct answer. my answer was 60 as the minimum. I am also not quite sure how to justify my answer or explain the process i went through. also, the three sides of both triangles created are considered to be pythag. triples, but they couldn't be acquired through the pythag. triples rule.

Is this question from the Math Challenge Stage competition?