Does a rectangle's diagonal to its perimeter have a ratio, like Pi? The ratio of a circle's diameter to circumference merited its very own greek letter, Pi.
I have been wondering if there is such a similar, or important, ratio in a rectangle, i.e. its diagonal to perimeter?
Would I be right in guessing that it's because it's not a single value, but a more complex formula involving trigonometry? Or perhaps not a widely used one?
Many thanks in advance.
 A: You can see with a couple examples that there is no such "rectangle $\pi$" - or more precisely, that different rectangles can have different perimeter/diagonal ratios.
First, consider a square. By the Pythagorean theorem we know that the perimeter/diagonal ratio of a square is just ${4\over \sqrt{2}}=2\sqrt{2}.$
Now consider a "really really thin" rectangle - e.g. a rectangle of length $1$ and height $\epsilon$ for some very small positive $\epsilon$. The length of a diagonal is "basically" $1$ and the perimeter is "basically" $2$, so we get a perimeter/diagonal ratio of "basically" $2$. Of course that's not immediately rigorous, but it should still be pretty convincing - and you can now verify it by e.g. seeing what happens if we set $\epsilon={1\over 2}$.
That said, any two similar rectangles will share the same perimeter/diagonal ratio. For example, "$\pi$ for squares" makes sense and is $2\sqrt{2}$, per the above.
Note that any two circles are automatically similar to each other. The picture becomes nicer if we think of the analogy $$\mbox{circles:ellipses::squares:rectangles.}$$ Just as with rectangles, we can find ellipses with different perimeter/"maximal-diameter" ratios - but any two similar ellipses (e.g. two circles) will have the same ratio.
A: For any regular n-gon define its radius as the radius of its incircle and $\pi_n$ as the ratio of its perimeter and its diameter.  Then, of course, by definition, is $2\pi_nr$ and it's easy to show that its area equals $\pi_nr^2$.
For example, $\pi_3=3\sqrt3$ and $\pi_4=4$; in general $\pi_n=n\cdot\tan(\pi/n)$.
A: Trigonometric solution

$a=d\cos x \qquad b=d\sin x$
perimeter : $2(a+b)=2d\:(\cos x+\sin x)$
$k$ = perimeter/diagonal ratio
$k = \dfrac{2(a+b)}{d}=2(\cos x+\sin x)$
$\cos x = \sin \Big( x + \dfrac{\pi}{2} \Big)$
$\sin u + \sin v = 2\sin\Big( \dfrac{u+v}{2} \Big)\cos\Big( \dfrac{u-v}{2} \Big)$
$u = x+\dfrac{\pi}{2} \; \& \; v = x \quad \Rightarrow \quad k = 2\sqrt{2}\sin\Big(x+\dfrac{\pi}{4}\Big)$

$\color{green}{y_1=2\sin x}\;,\; \color{red}{y_2=2\cos x}\;,\; \color{blue}{y_3=k=2(\sin x+\cos x)}$
$x\in \Big[ 0, \dfrac{\pi}{2} \Big] \quad \Rightarrow \quad k\in \big[ 2, 2\sqrt{2} \big]$
$\pi \approx 3.14 > 2\sqrt{2} \approx 2.83$
Conclusion
The perimeter/diagonal ratio of a rectangle cannot be equal to $\pi$.
