How many different Pythagorean triples exist in which 60 is the length of one of the two smaller integers? Show your work.?

this is math that I don't know

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  • Dixon
    Lv 7
    7 months ago

    If either of the first two numbers of a primitive triple is a factor of 60, that triple can be scaled up to give a unique Pythagorean triple.

    So just work methodically through a primitive Pythagorean triples table and count the factors of 60 in the short sides until you are out of range.

    If you don't have access to a table you would have to generate one by trial and elimination but in the real world we can just get a list with Google.

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  • 7 months ago

    60^2 + x^2 = y^2

    3600 = y^2 - x^2

    3600 = (y + x)(y - x)

    y - x = 3600 / (y + x)

    Remember that x and y are positive integers.

    As x gets bigger, y gets bigger, so y + x gets bigger, so y - x gets smaller.

    But y is the hypotenuse so it must be bigger than x, so it y - x must be positive.

    Since x and y are integers, y - x is an integers, so it must be at least 1.

    Can y - x = 1? In that case:

    y + x = 3600 and y = x + 1

    So x + x + 1 = 3600, which means x = (3600 - 1) / 2 = 1799.5, which is not an integer, so that's impossible.

    Therefore, y - x can't be less than 2.

    I wrote a simple computer program to try x as every positive integer starting with 1, which yields:

    11^2 + 60^2 = 61^2

    25^2 + 60^2 = 65^2

    32^2 + 60^2 = 68^2

    45^2 + 60^2 = 75^2

    63^2 + 60^2 = 87^2

    80^2 + 60^2 = 100^2

    91^2 + 60^2 = 109^2

    144^2 + 60^2 = 156^2

    175^2 + 60^2 = 185^2

    221^2 + 60^2 = 229^2

    297^2 + 60^2 = 303^2

    448^2 + 60^2 = 452^2

    899^2 + 60^2 = 901^2

    Now that x is 899 and y is 901, their difference is 2, and we know from above that this difference can't go lower than 2, so we're done at this point.

    Answer: 13

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  • Let's look at some primitive triples. These are all primitive triples with a hypotenuse less than 100

    (3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)

    (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53)

    (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73)

    (13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)

    A primitive triple is just a pythagorean triple that is the most reduced form of similar triangles. 3-4-5 and 6-8-10 are similar, but 3-4-5 is the primitive form. Additionally, we have these, too:

    (20, 99, 101) (60, 91, 109) (15, 112, 113) (60, 221, 229)

    https://en.wikipedia.org/wiki/Pythagorean_triple

    So the question here is how many of these can have 60 as one of the leg lengths. So we can just eliminate any of them where one of the leg lengths isn't a divisor of 60

    (3, 4, 5) (5, 12, 13) (8, 15, 17) (20, 21, 29) (12, 35, 37) (11, 60, 61) (20, 99, 101) (60, 91, 109) (15, 112, 113) (60, 221, 229)

    And then we can scale up accordingly:

    3-4-5 =>> 60-80-100

    3-4-5 =>> 45-60-75

    5-12-13 =>> 60-144-156

    5-12-13 =>> 25-60-65

    8-15-17 =>> 32-60-68

    20-21-29 =>> 60-63-87

    12-35-37 =>> 60-175-185

    11-60-61 =>> 11-60-61

    20-99-101 =>> 60-297-303

    60-91-109 =>> 60-91-109

    15-112-113 =>> 60-448-452

    60-221-229 =>> 60-221-229

    I'm counting 12.

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  • 7 months ago

    How many different Pythagorean triples exist in which

    60 is the length of one of the two smaller integers?

    Only one: (11, 60, 61)

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