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estudier

  • 3 years ago

All Pythagorean triples where area = perimeter (numerically)?

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  1. estudier
    • 3 years ago
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    More algebra...:-)

  2. RadEn
    • 3 years ago
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    5,12,13 using trial and error method ^^

  3. estudier
    • 3 years ago
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    That's one.....

  4. CheetahBooks1
    • 3 years ago
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    what do you mean?

  5. estudier
    • 3 years ago
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    By what?

  6. RadEn
    • 3 years ago
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    one more: 6,8,10

  7. estudier
    • 3 years ago
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    That's two....

  8. RadEn
    • 3 years ago
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    hehe...

  9. RadEn
    • 3 years ago
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    ok, general form tripel pythagorean number is : a=x b=(x^2-1)/2 c=(x^2+1)/2 with a<b<c (a,b,c are sides of a right-triangle) perimeter = area x + (x^2-1)/2 + (x^2+1)/2 = x(x^2-1)/2, simplify it we get : x^3 - 4x^2 - 5x =0 x(x^2 - 4x - 5) = 0 x(x-5)(x+1)=0 x=5 satifisfy it (so, a=5, b=12, c=13) with same way, for a=2x b=x^2-1 c=x^2+1 perimeter = area 2x+x^2-1+x^2+1 = 2x(x^2-1)/2, simplify it we get : x^3 - 2x^2 - 3x = 0 x(x^2 - 2x - 3) =0 x(x-3)(x+1)=0 x=3 satifisfy it (so, a=6, b=8, c=10) i have done for a=kx b=k(x^2-1) c=k(x^2+1) with k=3,4,5,6... there are x satisfy it but the result just reflication of on... and its rest not satisfy for x which make for a,b,c integer number

  10. RadEn
    • 3 years ago
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    so, just 2 triple only...

  11. CheetahBooks1
    • 3 years ago
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    3,4,5 is a triple, too!

  12. RadEn
    • 3 years ago
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    but, its area not equal its perimeter...

  13. estudier
    • 3 years ago
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    Yes, good enough:-)

  14. RadEn
    • 3 years ago
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    thanks...

  15. estudier
    • 3 years ago
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    you can have an easier algebra by eliminating z from x^2+y^2=z^2 and xy/2 = x + y +z leads to xy -4x -4y +8 = 0 -> (x-4)(y-4) = 8 etc.

  16. RadEn
    • 3 years ago
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    just combine of (x-4)(y-4) = 8, and we will get 2 triple only are : (x,y,z) = 5,12,13 and 6,8,10 but yours easier than me :)

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