# One for you, WonderBoy

• Jan 10th 2010, 08:54 AM
Wilmer
One for you, WonderBoy
Right triangle: sides 135-352-377
Isosceles triangle: sides 132-366-366
Both have perimeter = 864 and area = 23760

Find another case where a right triangle and an isosceles triangle have the
same area and perimeter; these rules apply, of course:
- all sides are integers
- new case is primitive (not a multiple of above)
• Jan 10th 2010, 09:17 AM
wonderboy1953
Working on it Wilmer
Without a computer, just a calculator. Give me a couple of days.
• Jan 10th 2010, 09:28 AM
Wilmer
Quote:

Originally Posted by wonderboy1953
Without a computer, just a calculator. Give me a couple of days.

Use computer if you wish.
• Jan 11th 2010, 06:31 AM
wonderboy1953
This was a toughie
Tried applying Goldbach's Conjecture. No success.

Next I flipped Riemann's Conjecture. No luck here.

I next tried a backward FLT with Selmer Groups and a little Galois's Group mixed in. Nothing doing here either.

I guess we'll have to settle for Right Triangle: 270, 704, 754 and
Isosceles Triangle: 264, 732, 732 with...

perimeter of 1728 and area equaling 95,040.
• Jan 11th 2010, 09:35 AM
Wilmer
Quote:

Originally Posted by wonderboy1953
I guess we'll have to settle for Right Triangle: 270, 704, 754 and
Isosceles Triangle: 264, 732, 732 with...

• Jan 11th 2010, 02:30 PM
wonderboy1953
Back to the drawing board
I'll keep trying.
• Jan 12th 2010, 07:14 AM
wonderboy1953
If this really requires a computer
Then I will decline this challenge until I think of a shortcut or can run a program on a computer to solve this problem.
• Jan 15th 2010, 10:08 AM
Wilmer
Fyi
I don't believe there is another primitive case.

I discovered this in 2003, and sent it to MathWorld; they show it here:
Heronian Triangle -- from Wolfram MathWorld
• Jan 16th 2010, 06:24 AM
rainer
Very nice. So, can you prove that no other primitive case exists?

In my naivete I was trying to take a "constrained optimization" approach to this problem. It of course didn't work, but I still am not totally clear as to why it didn't work.
• Jan 16th 2010, 06:36 AM
Jester
Quote:

Originally Posted by Wilmer
I don't believe there is another primitive case.

I discovered this in 2003, and sent it to MathWorld; they show it here:
Heronian Triangle -- from Wolfram MathWorld

Did you run the lengths up to 400,000 on a computer (and how long did that take)?
• Jan 16th 2010, 09:25 AM
wonderboy1953
Wilmer
Did you know that you can only post problems you already know the answer to in this section? "I don't believe there is another primitive case."
• Jan 16th 2010, 09:37 AM
Wilmer
YES, I know!
However, this one HAS an answer: "it is not known..."
A bit like the Euler Brick:
Euler Brick -- from Wolfram MathWorld

Perhaps Mr Fantastic can issue a statement as to
"it is not known" being a valid solution HERE. (Cool)
• Jan 16th 2010, 10:03 AM
Wilmer
Quote:

Originally Posted by Danny
Did you run the lengths up to 400,000 on a computer (and how long did that take)?

Yes I did. How long? Forgot!
I just did a run getting ALL right triangles where short leg < 100000:
there are 1,521,629 triangles; took 12 minutes 7 seconds (included
was the checking of each to see if it met the criteria).

I hit on this quite accidentally; like, tried it "just to kill time"!
I was surprised that MathWorld "accepted" it.

My computer "search" is simple enough:
get right triangle sides a,b,c (c being hypotenuse, of course)

I use this triangle as half the isosceles triangle.

p = 2(a + c) ; k = 2ab

I then check if there is a right triangle d,e,f (d<e, f the hypotenuse)
that meets the "same perimeter-area" and integer condition:
d,e = [p^2 + 2k +- SQRT(p^4 - 12p^2 k + 4k^2)] / (4p)
(d being the -, e being the +)

So ONLY 2 variables are looped: a and b.