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Let's say we have a square with side length $a$. Let the center of the first circle be at a corner of the square and let $r = a$. If we want the circle to be the same as the diagonal of the square we need $r = \infty$.

Now I want to create more circles with $r = [a, \infty)$. Let's call the circle with $r = \infty$ fully bent and the circle with $r = a$ not bent at all. I want the circle to bend uniformly. For example: How do I calculate $r$ of a new circle if I want it to be in the middle between fully bent and not bent at all?

Gabriellalala
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    I assume that the centers for circles with $r \not = a$ are not on the corner of the square? And that you want every circle to intersect with two corners of the square? – healynr Dec 30 '21 at 16:53
  • Yes that is correct – Gabriellalala Dec 30 '21 at 16:56
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    Consider the diagonal that goes through the center of the first circle. We know that every circle's center must be on this line, and every circle must intersect the other diagonal of the square at some point. For instance, on the infinite circle, it intersects it on its midpoint. Is your notion of "halfway-bent" that this intersection is halfway between where the first and infinite circles intersect the second diagonal? – healynr Dec 30 '21 at 17:03
  • Yes that is also true. I know how to calculate the length of each circle between the $2$ corners of the square that the circle goes through but I don't know if that helps. $l_p = r \alpha$ and $l_c = 2r \sin (\alpha / 2)$. $l_p$ is the length of the part of the circle and $l_c$ is the chord. – Gabriellalala Dec 30 '21 at 17:07

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Let $AB$ be the diagonal of the square that is regarded as a arc of an infinite circle, and let $C$ be the corner of the square that is the centre of the first circle:

Bending a quarter circle into a line

In the general case, if the height of the circular segment $ADBF$ cut off by the chord $AB$ is $h,$ then $$ r - \sqrt{r^2 - \frac{a^2}2} = h, $$ therefore $$ r + \sqrt{r^2 - \frac{a^2}2} = \frac{a^2}{2h}, $$ therefore $$ r = \frac12\left(h + \frac{a^2}{2h}\right) = \frac{2h^2 + a^2}{4h}. $$ Check: if $$ h = a - \frac{a}{\sqrt2}, $$ then $2h^2 = (3 - 2\sqrt2)a^2,$ therefore $$ r = \frac{(4 - 2\sqrt2)\sqrt2a^2}{4(\sqrt2 - 1)a} = a. \ \checkmark $$ In the case of interest, $$ h = \frac12\left(a - \frac{a}{\sqrt2}\right) = \frac{(\sqrt2 - 1)a}{2\sqrt2}, $$ therefore $$ 2h^2 = \frac{(3 - 2\sqrt2)a^2}4, $$ therefore $$ r = \frac{(7 - 2\sqrt2)2\sqrt2a^2}{4\cdot4(\sqrt2 - 1)a} = \frac{7 - 2\sqrt2}{4(2 - \sqrt2)}a = \frac{10 + 3\sqrt2}8a \bumpeq 1.7803a. $$

More neatly: $$ \frac{r}{a} = \frac12\left(\frac{h}{a} + \frac{a}{2h}\right). $$

Check: if $$ \frac{h}{a} = 1 - \frac1{\sqrt2}, $$ then $$ \frac{a}{h} = 2\left(1 + \frac1{\sqrt2}\right), $$ therefore $$ \frac{r}{a} = \frac12\left(1 - \frac1{\sqrt2}\right) + \frac12\left(1 + \frac1{\sqrt2}\right) = 1.\ \checkmark $$

In the case of interest, $$ \frac{h}{a} = \frac12\left(1 - \frac1{\sqrt2}\right), $$ therefore $$ \frac{a}{h} = 4\left(1 + \frac1{\sqrt2}\right), $$ therefore $$ \frac{r}{a} = \frac14\left(1 - \frac1{\sqrt2}\right) + \left(1 + \frac1{\sqrt2}\right) = \frac14\left(5 + \frac3{\sqrt2}\right) = \frac{10 + 3\sqrt2}8. $$

Calum Gilhooley
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