Radish pie

I recently came across a rather peculiar recipe for something called Babylonian radish pie. Intrigued, I began to follow the directions, which said I could start with any number of cups of flour.

Any number? I mean, I had to start with some flour, so zero cups wasn’t an option. But according to the recipe, any positive value was fair game. Next, I needed a second amount of flour that was 3 divided by my original number. For example, if I had started with two cups of flour, then the recipe told me I now needed 3 divided by 2, or 1.5, cups at this point.

I was then instructed to combine these amounts of flour and discard half. Apparently, this was my new starting amount of flour. I was to repeat the process, combining this amount with 3 divided by it and then discarding half. The recipe told me to keep doing this, over and over. Eventually, I’d have the proper number of cups of flour for my radish pie.

How many cups of flour does the recipe ultimately call for?

So if $x_0 \gt 0,$ with $$x_n = \frac{1}{2} \left( x_{n-1} + \frac{3}{x_{n-1}} \right),$$ for $n \geq 1.$ Let's assume that $x_n \to x^*,$ then the limit would have to satisfy $$x^* = \frac{1}{2} \left( x^* + \frac{3}{x^*} \right)$$ or equivalently $x^* = \frac{3}{x^*}$ or $x^* = \sqrt{3}.$

Strips by a million cuts

One morning, Phil was playing with my daughter, who loves to cut paper with her safety scissors. She especially likes cutting paper into “strips,” which are rectangular pieces of paper whose shorter sides are at most $1$-inch long.

Whenever Phil gives her a piece of standard printer paper ($8.5$ inches by $11$ inches), she picks one of the four sides at random and then cuts a $1$-inch wide strip parallel to that side. Next, she discards the strip and repeats the process, picking another side at random and cutting the strip. Eventually, she is left with nothing but strips.

On average, how many cuts will she make before she is left only with strips?

Let's define $S(m,n)$ as the expected number of strips for an $m$ by $n$ sheet of paper. By definition, we have $S(m,n) = 0$ for all $m , n \gt 0$ such that $\min \{ m, n \} \leq 1.$ Since there is a $0.5$ probability that the length side is reduced and $0.5$ probability that the width side is reduced by the next cut, we have the following recursion formula: $$S(m,n) = \frac{1}{2} S(m-1, n) + \frac{1}{2} S(m, n-1) + 1.$$

After using the fact that $S(m,n) = S(n,m)$ for all $m,n \gt 0$ and seeding the cache with $S(1,k) = 0$ for $k = 1, \dots, 11$, a recursive function needs $52$ calls to arrive at $$S(8.5,11) = 14.29058837890625.$$

We can muck about with the recursive definition to note a few things about this function $S(m,n).$ Firstly, we get $$S(m,n) = 2 - \frac{1}{2^{\lceil n \rceil -2}}, \,\, \forall m \in (1,2], n > 0.$$ Let's focus on $n \in \mathbb{N}$ as the extension to all $n > 0$ is trivial. The base case holds by the boundary condition since $S(m,1) = 2 - \frac{1}{2^{-1}} = 0.$ If the equation holds for some $n \in \mathbb{N}$ then since $m \in (1,2],$ $S(m-1,\cdot) \equiv 0,$ so \begin{align*} S(m,n+1) &= \frac{1}{2} S(m-1, n+1) + \frac{1}{2} S(m, n) + 1 \\ &= \frac{1}{2} \left( 2 - \frac{1}{2^{n-2}} \right) + 1 \\ &= 2 - \frac{1}{2^{n-1}}. \end{align*}

We can similarly use the recursion formula and induction to build up $$S(m,n) = 4 - \frac{\lceil n\rceil +3}{2^{\lceil n\rceil-1}}, \,\, \forall m \in (2,3], n > 0$$ and $$S(m,n) = 6 - \frac{\lceil n \rceil^2 + 7\lceil n \rceil + 16}{2^{\lceil n \rceil+1}}, \,\, \forall m \in (3,4], n > 0,$$ and so on. We can then also get the general asymptotic behavior again by induction, that $$S(m,n) = 2(\lceil m \rceil -1) - \frac{O\left(\lceil n\rceil^{\lceil m \rceil -2}\right)}{2^{\lceil n \rceil}}$$

Dakota Jones and the Slightly Smaller Set of Constraints

Earlier this year, Dakota Jones used a crystal key to gain access to a hidden temple, deep in the Riddlerian Jungle. According to an ancient text, the crystal had exactly six edges, five of which were 1 inch long. Also, the key was the largest such polyhedron (by volume) with these edge lengths.

However, after consulting an expert, Jones realized she had the wrong translation. Instead of definitively having five edges that were 1 inch long, the crystal only needed to have four edges that were 1 inch long. In other words, five edges could have been 1 inch (or all six for that matter), but the crystal definitely had at least four edges that were 1 inch long.

The translator confirmed that the key was indeed the largest such polyhedron (by volume) with these edge lengths. Once again, Jones needs your help. Now what is the volume of the crystal key?

Using the same setup of as the earlier post, without loss of generality, Dakota will assume that the equilateral triangle is situated along the $xy$-plane with vertices as $(0,0,0),$ $(1,0,0)$ and $(1/2, \sqrt{3}/2,0).$ Let $v = (x,y,h)$ be the remaining vertex. Again, without loss of generality, let's assume that the distance from $(0,0,0)$ to $v$ is $1,$ that is, $$x^2 + y^2 + h^2 = 1.$$ Furthermore, the volume of such a triangular pyramid would be $$V(x,y,h) = \frac{1}{3} B h = \frac{\sqrt{3}}{12} h.$$

Therefore, Dakota now needs to find \begin{align*} \max \,\, & \frac{\sqrt{3}}{12} h \\ \text{s.t.} \,\, & x^2 + y^2 + h^2 = 1,\end{align*} which is solved by the optimal vertex of $\hat{v} = (0,0,1),$ for a maximal volume of $$V_\max = V(1) = \frac{\sqrt{3}}{12}.$$