TITLE: Is the ratio of the side and at least one diagonal of a rhombus always irrational?
QUESTION [15 upvotes]: The ratio between the side of a square $ c = 1 $ and its diagonal is $ \frac 1 { \sqrt 2 } $; a square is a type of rhombus.
The ratio between the side $ c = 1 $ of a rhombus, with angle $ a = \frac \pi 3 $ and its longest diagonal $ AC $ is $ \frac c {AC} = \frac 1 { \sqrt 3 } $, while the other diagonal $ BD = 1 $.
What is the equation for the lengths of the diagonals of a rhombus of side $1$?
Is the ratio of the side and at least one diagonal of a rhombus always irrational? (i.e. not an exact fraction)

REPLY [10 votes]: If $2\alpha$ is the one of the angles in the rhombus and we take the side as the unit of measure, then the ratios you're interested in are $\sin\alpha$ and $\cos\alpha$.
Can they be both rational? Note that
$$
\sin\alpha=\frac{2\tan(\alpha/2)}{1+\tan^2(\alpha/2)},
\qquad
\cos\alpha=\frac{1-\tan^2(\alpha/2)}{1+\tan^2(\alpha/2)},
\qquad
\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}
$$
so that $\sin\alpha$ and $\cos\alpha$ are both rational if and only if $\tan(\alpha/2)$ is rational.
Since $\alpha$ can be any angle satisfying $0<\alpha<\pi/2$, $\tan(\alpha/2)$ can assume any value between $0$ and $1$, among which there are infinitely many rational numbers.
You can try and prove that choices of $\alpha=2\arctan r$, where $0<r<1$ and $r$ is rational, are in one-to-one correspondence with the primitive Pythagorean triples.