FiveThirtyEight Riddler

Can You Bake The Biggest π?

This week’s Riddler Classic asks:

This Sunday, March 14, is Pi Day! To celebrate, you are planning to bake a pie. You have a sheet of crust laid out in front of you. After baking, your pie crust will be a cylinder of uniform thickness (or rather, thinness) with delicious filling inside.

To maximize the volume of your pie, what fraction of your crust should you use to make the circular base (i.e., the bottom) of the pie?

Note: If you solve this riddle by baking an optimal pie, you automatically win.


First, happy Pi Day! Now on to new business. Setting up this problem, we are trying to maximize the volume of pie filling that can fit inside the cylindrical pie crust. The volume of filling will be:
\[V=\pi r^2\cdot h\]
with \(V\) being the volume of pie filling, \(pi r^2\) is the surface area of the base of the pie crust, and \(h\) the height of the walls. We are going to maximize \(V\) with respect to \(r\), but first we have to constrain \(h\) in terms of \(r\). The sheet of pie crust, before being shaped into a pie, will have some area \(A\), and from it we must extract two circular sections with area \(A_b=\pi r^2\) and also the rectangular wall which will have area \(A_h=2\pi r\cdot h\). This gives
\[A=2\pi r^2+2\pi r\cdot h.\]
From this, we can extract that
\[h=\frac{A-2\pi r^2}{2\pi r}\]
which will be substituted into our volume equation:
\[V=\pi r^2\cdot\left(\frac{A-2\pi r^2}{2\pi r}\right).\]
We need to maximize \(V\) with respect to \(r\), which we will do by taking the partial derivative and setting it equal to 0, to find the maximum
\[\frac{\partial V}{\partial r}=\frac{A}{2}-3\pi r^2=0.\]
Now, we just need to find the value of \(\pi r^2\) (the area of the circle) in terms of \(A\) to find the fraction of \(A\) used to make the base. With some simple arithmetic, we get that
\[\pi r^2=\frac{A}{6},\]
and that the fraction is 1/6.

Can You Hunt More Mystery Numbers?

This week’s Riddler Express asks:

By all accounts, Riddler Nation had a lot of fun hunting for the mysterious numbers a few weeks back. So here’s what we’re going to do: For the next four weeks, the Riddler Express will feature a similar puzzle that combines multiplication and logic. We’ll be calling these CrossProducts.

For your first weekly CrossProduct, there are five three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

Can you find all five three-digit numbers and complete the table?


Like the Riddler Classic from several weeks ago, this question can be solved by finding the factors of numbers given and deducing (through logic or software) what the three-digit numbers are. Rather than write it up again, I’ll direct you to my writeup from before.

This week’s answers are 359, 591, 818, 578, and 572.

Fun With Unit Cubes

This week’s Riddler Express asks:

If you have young children (or if you’re still a child at heart), you probably have small blocks somewhere in your home.

I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:

If you too have blocks at home (I mean, of course you do), see if you can make the same arrangement.

Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?


The first step I took in solving this problem was to figure out what to measure to arrive at the answer. Consider this isometric diagram of the arrangement:

In this perspective, we’re looking directly through the nearest base unit cube (which appears as a solid-lined square in the middle). The portion of the suspended cube that sits within the recess created by the other three cubes is a pyramid. The base of this pyramid is an equilateral triangle of length 1 (inherited from the sides of the supporting unit cubes). For the next step, I’ll focus on this pyramid alone:

In this view of the pyramid (looking directly down on it), we see the base is a unit equilateral triangle, and the walls are isoceles right triangles with hypotenuse 1 and edges of \(\frac{1}{\sqrt{2}}\). To figure out the total height of the pyramid, we’ll need to inspect the triangle which represents the horizontal projection of this pyramid.

This triangle, depicted below, has a base of \(\frac{\sqrt{3}}{2}\), and one of the sides is the \(\frac{1}{\sqrt{2}}\) vertex of the pyramid.

The height of this triangle, we can figure out using trigonometry, is \(\frac{1}{\sqrt{6}}\). This number tells us how deep into the recess the suspended cube falls. From here, we can simply subtract this from 1 to get our answer: \(1-\frac{1}{\sqrt{6}}\approx\)0.592.

Can You Survey Riddler City?

This week’s Riddler Express asks:

You’re reviewing some of the survey data that was randomly collected from the residents of Riddler City. As you’ll recall, the city is quite large.

Ten randomly selected residents were asked how many people (including them) lived in their household. As it so happened, their answers were 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

It’s your job to use this (admittedly limited) data to estimate the average household size in Riddler City. Your co-worker suggests averaging the 10 numbers, which would give you an answer of about 5.5 people. But you’re not so sure.

Would your best estimate be exactly 5.5, less than 5.5 or greater than 5.5?


This is a sneaky question. It takes a careful reading to realize the issue at hand. At first glance, it seems like the estimator, 5.5, should be unbiased. Only after realizing that using a sample of residents to estimate the average household size is not unbiased. Let’s walk through a small example to demonstrate.

Imagine that instead of many, many citizens, Riddler City has exactly 10 residents. These 10 residents are split between two households: the town hermit, who lives alone, and a large family of nine, who all live in the second household. In this example, the average household size is 5 (check for yourself!) If we were to survey all 10 residents about their household size and use the proposed estimator, we would miscalculate the estimated mean household size as 8.2.

Returning to the question at hand, a household with 10 residents is 10-times as likely to be sampled as a household with one resident. While it’s dangerous to make any assumption about the underlying distribution from a sample of one, I suspect few would object to the assertion that it’s quite unlikely that there are exactly as many one-person households in Riddler City as 1o-person ones, considering that each appeared once in the sample. I would wager that there are many more one-person households than 10-person ones, and that this method overestimates the true average household size. The actual average household size is less than 5.5.