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        riddles >> medium >> Heron triangle - Ratio
        
(Message started by: Christine on Jul 28^th, 2015, 6:35pm)

Title: Heron triangle - Ratio
Post by Christine on Jul 28^th, 2015, 6:35pm

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?

Title: Re: Heron triangle - Ratio
Post by towr on Jul 28^th, 2015, 10:49pm

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.

Title: Re: Heron triangle - Ratio
Post by towr on Jul 28^th, 2015, 11:01pm

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]

Title: Re: Heron triangle - Ratio
Post by Christine on Jul 29^th, 2015, 12:01am

Awesome!

I posted 3 isosceles Heron triangles.

Title: Re: Heron triangle - Ratio
Post by Christine on Jul 29^th, 2015, 12:10am

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???

Title: Re: Heron triangle - Ratio
Post by towr on Jul 29^th, 2015, 10:44am

on 07/29/15 at 00:10:53, Christine wrote:

In the case of isosceles

I used:

b/4 sqrt(4*a^2 - b^2) = p*(2*a + b)

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.