# Right triangles

Ryan McGuire

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Ryan McGuire

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Posts: 1069

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Ram Bhakt

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posted 10 years ago

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.

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.

Jayesh Lalwani

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posted 10 years ago

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

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)(221, 60, 229); (11, 60, 61); (91, 60, 109)

posted 10 years ago

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.

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.

posted 10 years ago

Beside 36-48-60, all other possibilities are:

60-11-61

60-25-65

60-32-68

60-45-75

60-63-87

60-80-100

60-91-109

60-144-156

60-175-185

60-221-229

60-297-303

60-448-452

60-899-901

The answer is 14.

60-11-61

60-25-65

60-32-68

60-45-75

60-63-87

60-80-100

60-91-109

60-144-156

60-175-185

60-221-229

60-297-303

60-448-452

60-899-901

Originally posted by Ryan McGuire:

How may right triangles with all integer side lengths have a side that is 60 units long?

The answer is 14.

Each number system has exactly 10 different digits.

Arjunkumar Shastry

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Ryan McGuire

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posted 10 years ago

Damn!

I just gave someone else grief for posting variations on a previous theme, and then I go and post an

Yeah that's it... I

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.

Jim Yingst

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posted 10 years ago

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 ]

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

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

Ryan McGuire

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posted 10 years ago

Ok, how about this:

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

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?

Jim Yingst

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Garrett Rowe

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Jim Yingst

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