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Posers and Puzzles

Posers and Puzzles

  1. Standard member ark13
    Enola Straight
    06 Mar '08 00:08
    Find, and prove that you have found every right triangle that satisfies the following conditions:

    -Its side lengths are integers
    -Its area is numerically equivalent to its perimeter

    You can use Euclid's formula for calculating Pythagorean triples for this problem, but if you're really clever, you won't need it. I just want to see what people come up with.
  2. 06 Mar '08 00:38 / 1 edit
    Lets see:

    Area = ab/2
    Perimeter = a+b+sqrt(a^2+b^2)

    ab/2 = a+b+sqrt(a^2+b^2)
    ab/2-a-b = sqrt(a^2+b^2)
    ab-2a-2b=2sqrt(a^2+b^2)
    squaring:
    a^2*b^2+4a^2+4b^2-4a^2*b-4ab^2+8ab = 4a^2+4b^2
    a^2*b^2-4a^2*b-4ab^2+8ab = 0
    Divide by ab:
    ab-4a-4b+8=0
    ab-4a-4b+16=8
    (a-4)(b-4)=8

    a-4 and b-4 are integers, so the possibilities are
    {a-4,b-4} = {-1,-8}, {-2,-4}, {1,8}, {2,4}
    {a,b} = {3,-4}, {2,0}, {5,12}, {6,8}

    Of course only the last two are OK so the solutions are:

    The triangles' sides are 5,12,13 or 6,8,10.

    Am I really clever?
  3. Standard member ark13
    Enola Straight
    06 Mar '08 22:43
    Originally posted by David113
    Lets see:

    Area = ab/2
    Perimeter = a+b+sqrt(a^2+b^2)

    ab/2 = a+b+sqrt(a^2+b^2)
    ab/2-a-b = sqrt(a^2+b^2)
    ab-2a-2b=2sqrt(a^2+b^2)
    squaring:
    a^2*b^2+4a^2+4b^2-4a^2*b-4ab^2+8ab = 4a^2+4b^2
    a^2*b^2-4a^2*b-4ab^2+8ab = 0
    Divide by ab:
    ab-4a-4b+8=0
    ab-4a-4b+16=8
    (a-4)(b-4)=8

    a-4 and b-4 are integers, so the possibilities are
    {a-4,b-4} = {-1,-8}, { ...[text shortened]... so the solutions are:

    The triangles' sides are 5,12,13 or 6,8,10.

    Am I really clever?
    Oh, you are really clever. Exactly what I was looking for. Nice work.