p**2 + b**2 = 60**2 or 60**2 + b**2 = h**2 ( same as p**2 + 60**2 = h**2 )

where p, b, and h are the lengths of perpendicular, base, and height.

I am not going to solve them but eq 1 will probably have a limited number of solutions (or may be even none) but eq 2 will have unlimited solutions unless there is some mathematical trick involving some property of squares. I guess, one could write a simple program that can use brute force algorithm to start putting values for b as 1 to N and seeing whether sq root of the sum of 60**2 + b**2 is an integer or not.

I) Assume hypotenuse = 60 a^2 + b^2 = 60^2 a^2 = 60^2 - b^2 = (60+b)(60-b) a = sqrt[(60+b)(60-b)] where 0 < b <60

So, for a to be an integer 60 + b and 60 - b should be squares of integers

So, let 60 + b = m^2, where m is an integer, and 60< (60+b)<120 Possible values of m are 8, 9 and 10

FOr all possible values of m, find 60 - b such that 60 - b is a square For m = 8, b= 4, 60 - b = 56, not a square, won't work For m = 9, b= 21, 60 - b = 39, not a square, won't work For m = 10, b= 40, 60 - b = 20, not a square, won't work

So, there is no right angle triangle with all integer side lengths have a hypotenuse that is 60 units long

II) Assume side = 60 a ^2 + 60^2 = b^2 a ^2 = b^2 - 60^2 = (b+60)(b-60)where b>60 a = sqrt[(b+60)(b-60)] where b>60 let b+60 = m^2 where m is an integer > 10 b - 60 = n^2 where n is an integer > 0

m^2 - n^2 = 120 (m+n)(m-n) = 120

Let m + n = 2q, m-n = 2r where q and r are integers Solving we get m = q + r, and n = q-r

so, 4.q.r = 120 q.r = 30 = 2.3.5

So, q = 15, r = 2 OR q = 6, r = 5 OR q = 10, r = 3

So, m = 17, n = 13 OR m = 11, n = 1 OR m= 13, n = 7

So, a = 221 OR a = 11 OR a = 91 b = 229 OR b = 61 OR b = 109

So, there are 3 rt triangles with integer side lengths that have a side that is 60 units long (221, 60, 229); (11, 60, 61); (91, 60, 109)

Originally posted by Jayesh Lalwani: I) Assume hypotenuse = 60 ... So, there is no right angle triangle with all integer side lengths have a hypotenuse that is 60 units long

The proof on this assumption can't be right:

The famous 3 - 4 - 5 triangle can be multiplied by 12: 36 - 48 - 60 -> 1296 + 2304 = 3600, so there is at least one triangle with hypotenuse 60.

Each number system has exactly 10 different digits.

Ryan,last year you posed this puzzle.and we solved it.

Namma Suvarna Karnataka

Ryan McGuire
Ranch Hand

Joined: Feb 18, 2005
Posts: 1013

3

posted

0

Originally posted by Arjunkumar Shastry: Ryan,last year you posed this puzzle.and we solved it.

Damn!

I just gave someone else grief for posting variations on a previous theme, and then I go and post an exact duplicate of one of my own. However the discussion we've gotten this time around does prove that it's still new to someone.

Yeah that's it... I knew I'd posted it before and I just wanted to see if people would realize it's a previously-solved problem.

You could at least replace the number 60 with something much bigger, like 60^4. That way you'd weed out the O(N^2) solutions, as well as invite some additional possible numeric errors if the programmer isn't careful. [ March 28, 2006: Message edited by: Jim Yingst ]

"I'm not back." - Bill Harding, Twister

Ryan McGuire
Ranch Hand

Joined: Feb 18, 2005
Posts: 1013

3

posted

0

Originally posted by Jim Yingst: You could at least replace the number 60 with something much bigger, like 60^4. That way you'd weed out the O(N^2) solutions, as well as invite some additional possible numeric errors if the programmer isn't careful.

[ March 28, 2006: Message edited by: Jim Yingst ]

Ok, how about this:

What number less than or equal to 10,000 is in the most (all-integer) pythagorean triples?