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Pythagorean triples in problem concerning matches

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data
    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 any one 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. Relevant 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)


    3. The attempt at a solution
    ive found an answer but im not quite sure whether it's the correct answer. my answer was 60 as the minimum. im 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 couldnt be acquired through the pythag. triples rule.
     
  2. jcsd
  3. Mar 23, 2012 #2
    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?
     
  4. Mar 23, 2012 #3
    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
     
  5. Mar 23, 2012 #4
    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.
     
  6. Mar 24, 2012 #5
    Thanks for that, I now know how to explain my answer.
     
  7. Apr 27, 2012 #6

    berkeman

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    Staff: Mentor

    Is this question from the Math Challenge Stage competition?
     
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