A pseudo-square has the following properties:
- It is a simple, closed curve.
- It has four sides, all the same length.
- Each side is either a straight line segment or the arc of a circle.
- The four sides are joined at four corners, with each corner having an internal angle of 90 degrees or 270 degrees.
The pseudo-square pictured above has two straight sides, which run radially between arcs of two concentric circles. Assuming this is a unit pseudo-square (i.e., each side has length 1), what is its area?
As you can see, I've pre-doctored the image from the prompt with some parameters. Let $r$ be the radius of the smaller circle and let $\theta$ be the angle inscribed between the two straight line edges. From here, if the shape is a unit pseudo-square then the larger radius is $1+r,$ so we can find the formula for the area of the shape in terms of $r$ and $\theta,$ as $$A(r,\theta) = \pi r^2 + \frac{1}{2} \theta \left((1+r)^2 - r^2\right) = \pi r^2 + \frac{1}{2} \theta (1 + 2r).$$ That is all well and good, but we now have to find the particular value of $r$ and theta that allow for this shape to be unit pseudo-square. In particular, the length of the arc on the larger of the two concentric circles is $$\ell = \theta (1 + r).$$ The length of the other arc is $$\tilde{\ell} = (2\pi - \theta) r.$$ Since the shape is a unit pseudo-square we have the nonlinear system of equations \begin{align*} \theta ( 1 + r) &= 1 \\ (2\pi - \theta) r &= 1.\end{align*} By setting $\theta = (1 + r)^{-1}$ and plugging into the second equation we get $$\left(2\pi - \frac{1}{1+r}\right) r = 1,$$ which is equivalent to the quadratic equation $$2\pi r^2 + 2(\pi - 1) r - 1 = 0.$$ Thus, if the shape is a unit pseudo-square then the smaller radius $r$ is equal to $$r^* = \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi}.$$
By plugging $\theta = (1+r)^{-1}$ into the area formula we get $A(r) = \pi r^2 + \frac{1+2r}{2(1+r)},$ so plugging in $r^*$ from above we get the area of the unit psuedo-square with two straight lines is \begin{align*}A^* = A(r^*) &= \pi \left( \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} \right)^2 + \frac{ 1 + 2 \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} }{2 \left( 1 + \frac{1-\pi + \sqrt{1+\pi^2}}{2\pi} \right)}\\ &= \frac{1+\sqrt{1+\pi^2}}{2\pi} \approx 0.683874197466......\end{align*}
Can you find a unit pseudo-square that has three curved sides and just one straight side? What is the area of your new unit pseudo-square?
Without loss of generality, we can assume that the one flat side is positioned along the bottom. We would need to have two curved sides connected to this flat bottom that curve towards each other, with there being a third circle that is tangent to the circular arcs. The resulting shape is sort of like the shape of some pawns in chess. See the figure below. We again insert some parameters. Let the flat side be the line segment from $(-\frac{1}{2},0)$ to $(\frac{1}{2},0).$ Let the two symmetric sides attached to flat bottom be arcs of the circles centered at $(-a,0)$ and $(a,0),$ respectively, each with radius $a+\frac{1}{2},$ with subtended angle $\theta$. Let the final arc be from the circle centered $(0,b)$ with radius $r$ and subtended angle $2(\pi - \theta).$
Let's first try to figure out how to quantify the area of the pseudo-square with three curved sides. The region denoted by $B$ has area given by $$A_B = \frac{1}{2} (\pi - \theta) r^2.$$ The region denoted by $C$ has area given by $$A_C = \frac{1}{2} r^2 \tan \theta.$$ Finally, the region denoted by $D$ has aread given by $$A_D = \frac{1}{2} \theta \left(a + \frac{1}{2}\right)^2 - \frac{1}{2} a^2 \tan \theta.$$ The area of the entire pseudo-square with three curved sides is thus \begin{align*}A = A(a,r,\theta) &= 2 \left(A_B + A_C + A_D\right) \\ &= (\pi - \theta + \tan \theta) r^2 + \theta \left(a + \frac{1}{2}\right)^2 - a^2 \tan \theta.\end{align*}
With some creative trigonometry we can obtain $r = r(a,\theta),$ thus reducing the dimensions of the problem a smidge. The region denoted by $C$ is a triangle whose height is $r$ and base $\beta,$ which is some portion of line segment of total length $a + \frac{1}{2},$ since that line segment is a radius of the circle centered at $(-a,0)$ of radius $a+\frac{1}{2}.$ Since the line segment, negative $x$-axis and positive $y$-axis form a triangle, we can compute $\beta$ as $$\beta = a + \frac{1}{2} - a\sec \theta.$$ Since $r = \beta \cot \theta,$ we have $$r = r(a,\theta) = \left( \left(a+\frac{1}{2}\right) - a \sec \theta \right) \cot \theta = \frac{ (a + \frac{1}{2}) \cos \theta - a }{ \sin \theta }.$$ Therefore, we have $$A = A(a,\theta) = (\pi - \theta + \tan \theta) \left( \frac{ \left(a + \frac{1}{2}\right) \cos \theta - a }{\sin \theta} \right)^2 + \theta \left(a + \frac{1}{2} \right)^2 - a^2 \tan \theta.$$
OK, now that that significantly more intricate foundational work is out of the way, we need to ensure that each of these curves sides is unit length. Since the more vertical arcs are symmetric, we only need to quantify $\ell=(a+\frac{1}{2}) \theta$ and $\tilde{\ell} = 2(\pi - \theta) r,$ and set them both equal to $1$ to obtain the nonlinear system of equations \begin{align*} (a + \frac{1}{2}) \theta &= 1 \\ 2(\pi - \theta) r = (\pi - \theta) \frac{(a+1/2)\cos \theta - a}{\sin \theta} &= 1\end{align*} Solving for $a$ in the second equation and then plugging back into the first we get $$\frac{ \sin \theta - \pi + \theta }{ 2(\pi - \theta) ( \cos \theta - 1) } \theta = 1.$$ This non-linear equation is solved with $$\theta^* \approx 0.74960359...$$ which leads to $$a^* = \frac{2-\theta^*}{2\theta^*} \approx 0.8340377...$$ which ultimately leads to the area of a unit pseudo-square with three curved sides of $$A^* = A(a^*, \theta^*) \approx 0.8317044...$$
