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April 11th, 2019, 03:17 AM   #11
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My response was incorrect. I now agree with Skipjack. I forgot to add the side of the square when calculating the hypotenuse.
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April 11th, 2019, 05:51 AM   #12
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Damn! I erroneously used the longer leg instead of the hypotenuse
when doing one of my "similar triangle a-go-go!".
Agree with Skip.
Turns out to be quite an easy problem.
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April 11th, 2019, 11:01 AM   #13
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Piece of cake.
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April 11th, 2019, 11:45 AM   #14
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Thanks. Got the result from the concept of similarity of triangles.
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April 11th, 2019, 01:16 PM   #15
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Quote:
Originally Posted by happy21 View Post
Thanks. Got the result from the concept of similarity of triangles.
This works as a formula:
length of sides of square = abc / (a^2 +b^2 + a*b)
(a = shorter leg, b = other leg, c = hypotenuse)

Only primitive case that has integer results (up to a = 999):
a = 111, b = 148, c = 185 : square sides = 60
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April 11th, 2019, 11:27 PM   #16
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Quote:
Originally Posted by Denis View Post
This works as a formula:
length of sides of square = abc / (a^2 +b^2 + a*b)
(a = shorter leg, b = other leg, c = hypotenuse)

Only primitive case that has integer results (up to a = 999):
a = 111, b = 148, c = 185 : square sides = 60
The above formula was derived from that concept of similarity only...right?
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April 12th, 2019, 04:58 AM   #17
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Quote:
Originally Posted by happy21 View Post
The above formula was derived from that concept of similarity only...right?
Of course...
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