This Riddler Classic puzzle is about cutting circles out of rectangles!

You’re on a DIY kick and want to build a circular dining table which can be split in half so leaves can be added when entertaining guests. As luck would have it, on your last trip to the lumber yard, you came across the most pristine piece of exotic wood that would be perfect for the circular table top. Trouble is, the piece is rectangular. You are happy to have the leaves fashioned from one of the slightly-less-than-perfect pieces underneath it, but there’s still the issue of the main circle. You devise a plan: cut two congruent semicircles from the perfect 4-by-8-foot piece and reassemble them to form the circular top of your table. What is the radius of the largest possible circular table you can make?

Here is my solution to the case of a general rectangular table. The result may surprise you!

[Show Solution]

We’ll solve the more general problem in which we vary the dimensions of the rectangle. Specifically, we’ll suppose the rectangle has a height of $1$ and a width of $x \ge 1$. After some thought, you can convince yourself that there are two arrangements that lead to large areas. The first arrangement is to use semicircles with flat sides aligned with the top and bottom sides of the table. Here is a diagram:

In this case, when the semicircle is as large as possible, it will either touch the other semicircle, or it will touch the opposite side of the rectangle, whichever comes first. The top side enforces the restriction $r \le 1$. Meanwhile, being tangent to the other semicircle can be characterized by calculating the width and height in terms of $r$. Specifically,

\begin{align*}

2r \sin \theta &= 1 \\

2r + 2r \cos \theta &= x

\end{align*}Rearranging and using the fact that $\sin^2\theta + \cos^2\theta = 1$ for any $\theta$, we conclude that:

\[

\left(\frac{1}{2r}\right)^2 + \left(\frac{x}{2r}-1\right)^2 = 1

\]Solving this equation for $x$ yields, and together with $r\le 1$ found before, we conclude that:

\[

r = \textrm{min}\left( \frac{x^2+1}{4x}, 1 \right)

\]

The other competitive arrangement of semicircles involves a slanted arrangement depicted in the figure below. Here, both semicircles are tangent to two sides, touch a third side, and also touch each other along the flat edge.

The algebra in this case is more complicated, but it can be resolved in a similar fashion. Once again, we calculate the height and width as a function of $r$ and $x$, and we obtain:

\begin{align*}

r + r\sin\theta &= 1 \\

2r + r\cot\theta-r\cos\theta &= x

\end{align*}We’ll work to eliminate $\theta$ as we did in the simpler case. Solving for $\sin\theta$ in the first equation and substituting into the second, we obtain:

\begin{align*}

r + r\sin\theta &= 1 \\

2r + r\cos\theta\left(\frac{2r-1}{1-r}\right) &= x

\end{align*}Once again, use the fact that $\sin^2\theta + \cos^2\theta = 1$ for any $\theta$, and we find:

\[

\left(\frac{1-r}{r}\right)^2 + \left(\frac{1-r}{2r-1}\right)^2\left(\frac{x}{r}-2\right)^2 = 1

\]Unfortunately, this equation isn’t as easy to solve as the previous one. Clearing all fractions, we are left with a fourth degree polynomial in $r$:

\[

4 r^4 -4(x+4)r^3+(x+4)^2r^2-(6+4x+2x^2)r+(1 + x^2) = 0

\]For any fixed $r$, this can be solved numerically. There is a closed-form expression for $r$ in terms of $x$, but it’s quite messy and not particularly instructive.

The interesting result here is that different schemes are optimal for different $x$ values. In the animation below, I plot $\frac{\pi r^2}{x}$ as a function of $x$. This is the area ratio of the circular area to the rectangular area. The transition occurs at $x\approx 2.47243$.

For a moment I was hoping the crossover point was 4/phi, but it looks like it’s really (7 – 17^0.5 + (62 – 14*17^0.5)^0.5)/2

What is meant by two congruent semicircles? Look at the flag of the Republic of Korea. There you will see a circle divided into two congruent pieces. Each piece is bounded by a semicircle and a wavy line joining the two ends of the semicircle. Are these congruent pieces congruent semicircles? I argue that they are, because the word semicircle does not define an area, but only an arc. Congruent pieces dividing a circular object into N pieces must each include a continuous circular arc that divides the circular perimeter in N equal parts. They must also each include, on an edge, the center of the circular arc. But there is no reason to suppose that the pieces have to include straight radial lines joining the center to the ends of the arc.

In the context of this Riddler, we can find a solution that gives a bigger circular table top than the one described in the posted solution. Only two congruent pieces are involved, and they can be made by one zigzag cut in the rectangle, a displacement that converts the shape into a square with two opposite corners snipped off, and then a circular cut with diameter 2^1.5 feet. This is larger that the given solution.

You’re right — if you allow for congruent shapes that don’t necessarily include the radial lines, then you can fit a larger circular area inside the rectangle. I took “semicircle” to mean half of a circle as formed by a straight cut through the center, but your interpretation could lead to some very interesting solutions as well!

This seems to be a maxima, minima problem as well. Is there a calculus solution as well?

It’s definitely a constrained optimization problem, e.g. you can parameterize the positions of the segments and their orientations and sizes, and ask yourself what choice of parameters maximizes the area. However, the problem is made very challenging because we must include constraints that prevent the segments from overlapping the boundary or each other. This renders the problem non-convex and probably hopeless to solve in general.