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Topic: Heron triangle - Ratio (Read 649 times) |
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Christine
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Heron triangle - Ratio
« on: Jul 28th, 2015, 6:35pm » |
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Let A/P be ratio of area to perimeter
To find triangles such that A/P is a prime number.
Here are 3 examples:
(15, 15, 24) ---> A/P = 108/54 = 2
(20, 20, 24) ---> A/P = 108/54 = 3
(30, 39, 39) ---> A/P = 540/108 = 5
Can you prove that there are no others?
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towr
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Re: Heron triangle - Ratio
« Reply #1 on: Jul 28th, 2015, 10:49pm » |
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So, for a triangle a,b,c with 0 < a <= b <= c < a+b
We have semiperimeter s = 1/2 (a+b+c)
perimeter P = 2s
area A = sqrt(s(s-a)(s-b)(s-c))
And then we want 1/2 sqrt((s-a)(s-b)(s-c)/s) to be prime.
With a quick script, the first 10 I find are
(9, 40, 41) => 2
(9, 75, 78) => 2
(10, 24, 26) => 2
(10, 35, 39) => 2
(11, 25, 30) => 2
(11, 90, 97) => 2
(12, 16, 20) => 2
(12, 50, 58) => 2
(13, 14, 15) => 2
(13, 84, 85) => 3
So there are others.
We can also get other primes than 2, 3 and 5:
(29, 420, 421) => 7
(46, 528, 530) => 11
(54, 728, 730) => 13
(72, 646, 650) => 17
(80, 798, 802) => 19
(100, 621, 629) => 23
(124, 957, 965) => 29
(143, 732, 775) => 31
(185, 444, 481) => 37
(195, 687, 738) => 41
(215, 516, 559) => 43
(221, 879, 940) => 47
So at this point it might be interesting to find a way to generate them.
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| « Last Edit: Jul 28th, 2015, 10:49pm by towr » |
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towr
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Re: Heron triangle - Ratio
« Reply #2 on: Jul 28th, 2015, 11:01pm » |
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Any triangle (6k, 8k, 10k) gives area/perimeter = k
So we can trivially get any prime.
It might also be interesting to consider only primitive triangles.
[edit]
(k2+8, k2+16, 2k2+8) also gives A/P = k
and for odd numbers that's always a primitive triangle.
[/edit]
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| « Last Edit: Jul 28th, 2015, 11:19pm by towr » |
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Christine
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Re: Heron triangle - Ratio
« Reply #3 on: Jul 29th, 2015, 12:01am » |
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Awesome!
I posted 3 isosceles Heron triangles.
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Christine
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Re: Heron triangle - Ratio
« Reply #4 on: Jul 29th, 2015, 12:10am » |
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In the case of isosceles
I used:
b/4 sqrt(4*a^2 - b^2) = p*(2*a + b)
p is a prime number.
I tried some values for p (2,3,5,..). The weird thing is I get negative values for b and positive values for a
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towr
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Re: Heron triangle - Ratio
« Reply #5 on: Jul 29th, 2015, 10:44am » |
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on Jul 29th, 2015, 12:10am, Christine wrote:In the case of isosceles
I used:
b/4 sqrt(4*a^2 - b^2) = p*(2*a + b) |
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You can square both sides, and simplify to
b^2 * (2*a - b) = 16 * p (2*a + b)
then solve for a to get
a = [16*p^2*b + b^3] / [2*b^2 - 32*p^2]
This make a slightly bigger than 1/2*b as b grows very large, so (given a p) if a is ever an integer for large b, then a triangle should be possible. But for these types of equations the solution might be very large if there is one.
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| « Last Edit: Jul 29th, 2015, 10:44am by towr » |
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