Triangles with 60° angle and sides integers

 
Previous blog:   Integer triangles with 120° angle

 

Law of cosines
https://en.wikipedia.org/wiki/Law_of_cosines

c^2 \; = \; a^2 \; + \; b^2 \; - \; 2 \, a \, b \, \cos \, \gamma

If   \gamma \; = \; 60^\circ   —>   \cos 60^\circ \; = \; 0.5

then,     c^2 \; = \; a^2 \; + \; b^2 \; - \; a \, b

 
 
To find triples   (a, b, c)   that satisfy the relation   c^2 \; = \; a^2 \; + \; b^2 \; - \; a \, b

We use the parametric equations

a \; = \; m^2 \; - \; n^2
b \; = \; 2 \,m \,n \; - \; n^2
c \; = \; m^2 \; - \; m \,n \; + \; n^2

where   m   and   n   are integers and   m \; > \; n

(m^2 - n^2)^2 - (m^2 - n^2 ) \,(2 \, m \, n - n^2) + (2 \, m \, n - n^2)^2 \; = \; (m^2 - m \, n + n^2)^2

Perimeter   P :

P \; = \; m^2 \; - \; n^2 \; + \; 2 \, m \, n \; - \; n^2 \; + \; m^2 \; - \; m \, n \; + \; n^2
P \; = \; 2 \, m^2 \; + \; m \, n \; - \; n^2
P \; = \; (2 \, m \; - \; n) \, (m \; + \; n)

 
Let the triple   (x, \; y, \; z)   represents an integer 120° triangle
that is,   x^2 \; + \; x \,y \; + \; y^2 \; = \; z^2

Then the triples   (x+y, \; y, \; z)   and   (x, \; x+y, \; z)   represent two integer 60° triangles :

x^2 \; + \; x \,y \; + \; y^2 \; = \; z^2
x^2 \; - \; x^2 \; + \; x^2 \; + \; x \,y \; - \; x \,y \; + \; x \,y \; + \; y^2 \; = \; z^2
x^2 \; + \; 2 \,x \,y \; + \; y^2 \; - \; x^2 \; - \; x \,y \; + \; x^2 \; = \; z^2
(x+y)^2 \; - \; (x + y) \,x \; + \; x^2 \; = \; z^2

Or

x^2 \; - \; (x+y) \,x \; + \; (x+y)^2 \; = \; z^2

 
Here are the primitives for   a   <   100 :

 
HERON 60-1
HERON 60-2

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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