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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)

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    $\begingroup$ $ 2 \cos (\alpha / 2 ) $ and $ 2 \sin (\alpha / 2 ) $ with ratio $ \cot (\alpha / 2 ) $ – is that what you wanted? $\endgroup$ – PJTraill Oct 4 '17 at 19:38
  • $\begingroup$ Thank you, this is 1/2 of the answer I needed $\endgroup$ – adrienlucca.wordpress.com Oct 5 '17 at 10:11
  • $\begingroup$ Imagine a smooth transition from a square to a nearly flat rhombus. You can achieve any ratio you want between 1/2 and infinity. $\endgroup$ – Brady Gilg Oct 9 '17 at 22:58
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    $\begingroup$ @BradyGilg yes you are right, but this does not answer my question: take the two diagonals AC and BD that show up for every case, is at least one of the ratio to the side AB/AC or AB/BD always irrational? This was the question. Answered by egreg $\endgroup$ – adrienlucca.wordpress.com Oct 12 '17 at 8:01
  • $\begingroup$ I'm aware of that. $\endgroup$ – Brady Gilg Oct 12 '17 at 17:41
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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.

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  • $\begingroup$ Did you not mean $ \frac 1 { 2 \sin \alpha } $ rather than $ \frac 1 2 \sin \alpha $ (etc.)? $\endgroup$ – PJTraill Oct 16 '17 at 9:59
  • $\begingroup$ @PJTraill No, they are actually without the $1/2$ factor (removed now) $\endgroup$ – egreg Oct 16 '17 at 10:36
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No, you can make a rhombus out of four identical Pythagorean right triangles, such as (3, 4, 5).

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  • $\begingroup$ This doesn't answer the question; the long diagonal of that rhombus is sqrt(97) units long. $\endgroup$ – Robyn Oct 5 '17 at 3:05
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    $\begingroup$ @Robyn What are you talking about? The rhombus has sides $(5, 5, 5, 5)$ and its two diagonals have lengths $8$ and $6$ respectively. Of course you scale everything to a fifth of those values, so that the rhombus has side $1$ and the diagonals have sides $8/5$ and $6/5$. $\endgroup$ – ShreevatsaR Oct 5 '17 at 3:47
  • $\begingroup$ Oh, I see! You're right. I was thinking of the rhombus with sides (5,6,5,6) that could be made by arranging the same 4 triangles in a different way. $\endgroup$ – Robyn Oct 5 '17 at 4:19
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    $\begingroup$ @Robyn A rhombus is a special kind of parallelogram that has all sides of equal length. the shape you were thinking of was a parallelogram, but not a rhombus. $\endgroup$ – Level River St Oct 5 '17 at 4:55
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    $\begingroup$ Thank you, this is 1/2 of the answer I needed $\endgroup$ – adrienlucca.wordpress.com Oct 5 '17 at 10:12

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