Is the ratio of the side and at least one diagonal of a rhombus always irrational?

Question

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)

$ 2 \cos (\alpha / 2 ) $ and $ 2 \sin (\alpha / 2 ) $ with ratio $ \cot (\alpha / 2 ) $ – is that what you wanted? — PJTraill, Oct 04 '17 at 19:38
Imagine a smooth transition from a square to a nearly flat rhombus. You can achieve any ratio you want between 1/2 and infinity. — Brady Gilg, Oct 09 '17 at 22:58
@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 — adrienlucca.net, Oct 12 '17 at 08:01

score 61 · Answer 1 · edited Oct 05 '17 at 18:23

61

No, you can make a rhombus out of four identical Pythagorean right triangles, such as (3, 4, 5).

edited Oct 05 '17 at 18:23

templatetypedef

10,206

answered Oct 04 '17 at 15:23

Will Jagy

139,541

1

This doesn't answer the question; the long diagonal of that rhombus is sqrt(97) units long. – Robyn Oct 05 '17 at 03:05
12

@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 $6/5$. – ShreevatsaR Oct 05 '17 at 03:47
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. – Robyn Oct 05 '17 at 04:19
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@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. – Level River St Oct 05 '17 at 04:55
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Thank you, this is 1/2 of the answer I needed – adrienlucca.net Oct 05 '17 at 10:12
1

Short solution. – Feb 28 '21 at 04:02

egreg · Accepted Answer · 2017-10-16T10:37:18.717

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.

Did you not mean $ \frac 1 { 2 \sin \alpha } $ rather than $ \frac 1 2 \sin \alpha $ (etc.)? — PJTraill, Oct 16 '17 at 09:59
@PJTraill No, they are actually without the $1/2$ factor (removed now) — egreg, Oct 16 '17 at 10:36

Is the ratio of the side and at least one diagonal of a rhombus always irrational?

2 Answers2