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Prove this is a right triangle?

  1. Nov 19, 2004 #1
    How can I prove that a right triangle whose sides are n^2+1 , n^2-1, and 2n is indeed a right triangle? The hint is to use the pythagorean theorem which is a^2 + b^2 = C^2 . I dont know what to do? i tried squaring the 3 terms but I cant get the first 2 to equal the last term (term c which is probably 2n^2) Please help me someone I dont get this at all :cry:
     
  2. jcsd
  3. Nov 19, 2004 #2
    It just says it is a right triangle. It doesn't say which side is the hypotenuse. Why don't you try making (n^2+1) as your c value.
     
  4. Nov 20, 2004 #3
    [MATH] n^2+1, n^2-1 , 2n[/MATH]
    The proof of these three sides form a right-angled triangle is quite easy by Pyth. theorem.
    Three sides, just get the sum of the square of two sides, if it is equal to the third side, it is indeed a right ... by converse of Pyth.theorem.
    There are 3 combination (3C2) but since n must be a postive number, n^2 is the longest side(unless n=1, which is rejected by n >1), therefore, just test for the sum of the squares of the other two sides.
     
    Last edited: Nov 20, 2004
  5. Nov 20, 2004 #4

    Pyrrhus

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    You could use the idea that

    [tex] a^2 + b^2 + c^2 = a^2 + b^2 +c^2 [/tex]

    [tex] a^2 + b^2 + c^2 = c^2 +c^2 [/tex]

    [tex] a^2 + b^2 + c^2 = 2c^2 [/tex]

    so

    [tex] \sqrt{\frac{a^2 + b^2 + c^2}{2}} = c [/tex]

    If get one of the sides of your triangle when your work the left part of the equation then it is a right triangle, and you know that's the hypotenuse.

    Part you might get stuck:

    [tex] \sqrt{n^4 + 2n^2 +1} = c [/tex]

    [tex] \sqrt{(n^2 + 1)^2} = c [/tex]

    [tex] n^2 + 1 = c [/tex]
     
    Last edited: Nov 20, 2004
  6. Nov 20, 2004 #5
    That's a slick one cyclovenom, but why not just determine which of those three numbers is largest and set it equal to c? n^2+1 is obviously >= 2n and n^2-1 for all n>=1 which are the only valid values of n that would give you non-negative sides.

    a^2 + b^2 = c^2

    (2n)^2 + (n^2-1)^2 = (n^2+1)^2

    4n^2 + n^4 - 2n^2 + 1 = n^4 + 2n^2 + 1

    n^4 + (4n^2 - 2n^2) + 1 = n^4 + 2n^2 + 1
     
  7. Nov 20, 2004 #6
    n cannot be 1.
     
  8. Nov 20, 2004 #7
    Thanks sooo Much !!

    Thanks sooo much everyone :smile: I'm so happy I got this question, you guys helped soo much. I understand the question so much better now thank you once again YAY!!!! Wow you guys are sooo Smart, I really appreciate it..
     
  9. Nov 20, 2004 #8
    You are welcome.
    My method is learnt from my textbook last year.
    It is the conversion of Pyth. theorem. 3,4,5 Determine the hypotheuse ( i.e. 5)
    3^2+4^2=25
    5^2=25
    So it is a right-angled triangle(converse of Pyth. theorem)
    In this case, it is obvious that n>1, and n^2+1 bigger than others ( since n^2+1-2n = (n-1)^2, which the smallest value is 0 which in other words, which n attains 1 , there are two equal side[ But n >1 ])
     
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