April 11th, 2019, 03:17 AM  #11 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
My response was incorrect. I now agree with Skipjack. I forgot to add the side of the square when calculating the hypotenuse.

April 11th, 2019, 05:51 AM  #12 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Damn! I erroneously used the longer leg instead of the hypotenuse when doing one of my "similar triangle agogo!". Agree with Skip. Turns out to be quite an easy problem. 
April 11th, 2019, 11:01 AM  #13 
Senior Member Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 
Piece of cake. 
April 11th, 2019, 11:45 AM  #14 
Senior Member Joined: Jan 2012 Posts: 140 Thanks: 2 
Thanks. Got the result from the concept of similarity of triangles.

April 11th, 2019, 01:16 PM  #15  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
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  
April 11th, 2019, 11:27 PM  #16 
Senior Member Joined: Jan 2012 Posts: 140 Thanks: 2  The above formula was derived from that concept of similarity only...right?

April 12th, 2019, 04:58 AM  #17 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  

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