# Pythagorean triples in problem concerning matches

• diana.hole
In summary, the conversation discusses a problem where Steve creates two right-angled triangles using matches of equal length, without being able to split the matches. The smallest number of matches that can be used for this is 60, found using the pythagorean theorem and the formula for generating pythagorean triples. There is also a discussion about the limitations of the formula and how it can account for multiples of previous triples.
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.

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?

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

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.

zooxanthellae said:
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.

diana.hole said:

## 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?

## 1. What are Pythagorean triples?

Pythagorean triples are sets of three positive integers (a, b, c) that satisfy the Pythagorean theorem (a² + b² = c²). In other words, the sum of the squares of the two shorter sides of a right triangle is equal to the square of the longest side.

## 2. How are Pythagorean triples relevant to problems concerning matches?

In problems concerning matches, Pythagorean triples can be used to determine the number of matches needed to create a specific shape or design. By using the Pythagorean theorem, we can calculate the length of the hypotenuse (longest side) which represents the number of matches needed.

## 3. Can Pythagorean triples be used for any shape or design?

Yes, Pythagorean triples can be used for any shape or design that can be broken down into right triangles. This includes squares, rectangles, and even more complex shapes.

## 4. How can I find Pythagorean triples for a given problem concerning matches?

One method is to use the Pythagorean triple formula (a = m² - n², b = 2mn, c = m² + n²) where m and n are any positive integers. Another method is to use a table of Pythagorean triples to find a set that fits the given problem.

## 5. Are Pythagorean triples the only way to solve problems concerning matches?

No, there are other methods such as using geometry or algebra to find a solution. However, Pythagorean triples are a useful tool and can often simplify the problem and provide a quick solution.

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