$a \le b \le c$ and $\gcd(a,b,c) = 1$ such that they are side lengths of a triangle one of whose angle is $60^\circ$.

Asked Feb 03 '23 at 23:46

Active Feb 03 '23 at 23:46

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Determine all triples $(a,b,c)$ of positive integers with $a \le b \le c$ and $\gcd(a,b,c) = 1$ such that they are side lengths of a triangle one of whose angle is $60^\circ$.

Here if we assume that the angle between sides with lengths $b$ and $c$ is $60^\circ$, then we have

$$a^2 = b^2+c^2-2bc \cos(60^\circ)$$ and hence $a^2 = b^2+c^2-bc$.

I am not able to proceed after that.

asked Feb 03 '23 at 23:46

user5210

5

Does this answer your question? Solving for $z^2 = x^2 -xy + y^2$ - found using an Approach0 search. Note the search results also include the more general Quadratic Diophantine Equation $x^2 + axy + y^2 = z^2$. – John Omielan Feb 03 '23 at 23:56
3

If $a\le b\le c$, then the 60$^\circ$ angle has to be between sides with length $a$ and $c$, actually. – Daniel Schepler Feb 03 '23 at 23:59

$a \le b \le c$ and $\gcd(a,b,c) = 1$ such that they are side lengths of a triangle one of whose angle is $60^\circ$.

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