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Topic: semi perimeter a square (Read 2553 times) |
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Christine
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semi perimeter a square
« on: Mar 30th, 2014, 1:26pm » |
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Can you find a triangle whose area, sides are integers and the semi perimeter is a square?
Is there an algorithm to generate them?
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pex
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Re: semi perimeter a square
« Reply #1 on: Mar 30th, 2014, 5:27pm » |
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on Mar 30th, 2014, 1:26pm, Christine wrote:| Can you find a triangle whose area, sides are integers and the semi perimeter is a square? |
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Yes. The simplest triangle with integer sides and area that I can think of has sides 3,4,5. Its semiperimeter is 6, so scaling the whole thing up by a factor of 6 works: the 18,24,30 triangle fits the requirements.
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Christine
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Posts: 159
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Re: semi perimeter a square
« Reply #2 on: Apr 10th, 2014, 3:18pm » |
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on Mar 30th, 2014, 5:27pm, pex wrote:
Yes. The simplest triangle with integer sides and area that I can think of has sides 3,4,5. Its semiperimeter is 6, so scaling the whole thing up by a factor of 6 works: the 18,24,30 triangle fits the requirements. |
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Thanks. I'm looking for a primitive Heronian triangle
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towr
wu::riddles Moderator
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Re: semi perimeter a square
« Reply #3 on: Apr 10th, 2014, 11:13pm » |
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The smallest primitive with those properties is 8,5,5 (which is just 2 (5,4,3)'s put together)
You can generate a list based on http://en.wikipedia.org/wiki/Heronian_triangle#Exact_formula_for_Heronia n_triangles
e.g.
Python Code:from fractions import gcd
for m in range(1000):
for n in range(1,m+1):
for k in range(int(((m**2*n)/(2*m+n))**0.5-1), int((m*n)**0.5)+1):
if gcd(m,gcd(n,k))==1 and m*n > k**2 >= (m**2*n)/(2*m+n) and m>=n>=1:
a = n*(m**2+k**2)
b = m*(n**2+k**2)
c = (n+m)*(m*n-k**2)
g = gcd(a,gcd(b,c))
a /= g
b /= g
c /= g
s = (a+b+c)/2
if int(s**0.5)**2==s:
print (m,n,k, a,b,c, (a+b+c)/2) |
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(697, 657, 104) is the first I've found to also have a square area.
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| « Last Edit: Apr 10th, 2014, 11:18pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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