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Math Help - [SOLVED] Right Triangle Problem

  1. #1
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    [SOLVED] Right Triangle Problem

    Prompt: List all integer possibilites for the three sides of a right triangle where one side is 2 units shorter than one of the other sides.

    Is there some algebraic method of solving this?


    Also, I found a 3-4-5 triangle and a 6-8-10 triangle. Both of these triangles have the same angles, which are 90 degrees, 36.86989765 degrees, and 53,13010235 degrees. Maybe this could mean something?

    Any suggestions?
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  2. #2
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    From Pithagoras:
    x^2 + (x+2)^2=y^2 where x and y anre ints.
    You can investigate only-int solutions. May be a little complicated be should not be impossible.

    OR (assuming the +2 side can be any side (hypo.) too)
    x^2 + y^2 = (x+2)^2 gives
    y^2 = 4X + 4
    y = 2 * sqrt(x+1)

    x= 8 gives y=6. All are integers.
    x= 15 gives y=8, and x+2=17 all integers and fits the answer.
    So, there should be infinitly many solution in the above format.
    Please note Solution to the first expression may be hard but I suspect if there are infinitely many solutions from: x^2 + (x+2)^2=y^2 probably only 3, 4, and 5. But you got to check that out.

    -O
    Last edited by mr fantastic; February 7th 2009 at 08:02 PM. Reason: Removed link not relevant to question (looks like a signature but isn't)
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  3. #3
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    Hello, KellHS886!

    List all integer possibilites for the three sides of a right triangle
    where one side is 2 units shorter than one of the other sides.

    Is there some algebraic method of solving this?
    All "Pythagorean Triples" can be generated with these parametric equations:

    . . \begin{array}{ccc} a &=& m^2-n^2 \\ b &=& 2mn \\ c &=&m^2+n^2\end{array}\quad\text{ for positive integers }m\text{ and }n,\;m > n.


    It will be found that the required triples are generated for n = 1


    . . \begin{array}{|c|c||c|}\hline<br />
m & n & (a,b,c) \\ \hline \hline<br />
2 & 1 & (3,4,5) \\<br />
3 & 1 & (8,6,10) \\<br />
4 & 1 & (15,8,17) \\<br />
5 & 1 & (24,10,26) \\<br />
6 & 1 & (35,12,37) \\<br />
7 & 1 & (48,14,50) \\<br />
\vdots & \vdots & \vdots<br />
\end{array}

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