Without a computer, just a calculator. Give me a couple of days.

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- Jan 10th 2010, 09:54 AM #1

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## 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, 10:17 AM #2

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- Jan 10th 2010, 10:28 AM #3

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- Jan 11th 2010, 07:31 AM #4

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## 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, 10:35 AM #5

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- Jan 11th 2010, 03:30 PM #6

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- Jan 12th 2010, 08:14 AM #7

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- Jan 15th 2010, 11:08 AM #8

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## 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, 07:24 AM #9

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- Jan 16th 2010, 07:36 AM #10

- Jan 16th 2010, 10:25 AM #11

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- Jan 16th 2010, 10:37 AM #12

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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.

- Jan 16th 2010, 11:03 AM #13

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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.