Infinite cake

This week’s Riddler Express is a short problem about infinite series. Let’s dig in! (I paraphrased the question)

You and your infinitely many friends are sharing a cake, and you come up with several different methods of splitting it.

  1. Friend 1 takes half of the cake, Friend 2 takes a third of what remains, Friend 3 takes a quarter of what remains after Friend 2, Friend 4 takes a fifth of what remains after Friend 3, and so on.
  2. Friend 1 takes $1/2^2$ (or one-quarter) of the cake, Friend 2 takes $1/3^2$ (or one-ninth) of what remains, Friend 3 takes $1/4^2$ of what remains after Friend 3, and so on.
  3. Same as previous, with even denominators only: Friend 1 takes $1/2^2$ of the cake, Friend 2 takes $1/4^2$ of what remains, Friend 3 takes $1/6^2$ of what remains after Friend 2, and so on.

For each of these methods, after your infinitely many friends take their respective pieces, how much cake is left?

Here is my solution
[Show Solution]

Baking the biggest pie

This week’s Riddler Classic is about baking the biggest pie. Just in time for π day!

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?

Here is my solution:
[Show Solution]

Flip to freedom

This week’s Riddler Classic is a problem about coin flipping. The text of the original problem is quite long, so I will paraphrase it here:

There are $n$ prisoners, each with access to a random number generator (generates uniform random numbers in $[0,1]$) and a fair coin. Each prisoner is given the opportunity to flip their coin once if they so choose. If all of the flipped coins come up Heads, all prisoners are released. But if any of the flipped coins come up Tails, or if no coins are flipped at all, the prisoners are not released. If the prisoners cannot communicate in any way and do not know who is flipping their coin or not, how can they maximize their chances of being released?

Here is my solution:
[Show Solution]

When did the snow start?

This week’s Riddler Classic is a neat calculus problem:

One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.

At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s speed is inversely proportional to the depth of the snow — if you were to double the amount of snow on the ground, the plow would move half as fast.

In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.

When did it start snowing?

Here is my solution:
[Show Solution]

Settlers in a circle

In this Riddler problem, the goal is to spread out settlements in a circle so that they are as far apart as possible:

Antisocial settlers are building houses on a prairie that’s a perfect circle with a radius of 1 mile. Each settler wants to live as far apart from his or her nearest neighbor as possible. To accomplish that, the settlers will overcome their antisocial behavior and work together so that the average distance between each settler and his or her nearest neighbor is as large as possible.

At first, there were slated to be seven settlers. Arranging that was easy enough: One will build his house in the center of the circle, while the other six will form a regular hexagon along its circumference. Every settler will be exactly 1 mile from his nearest neighbor, so the average distance is 1 mile.

However, at the last minute, one settler cancels his move to the prairie altogether (he’s really antisocial). That leaves six settlers. Does that mean the settlers can live further away from each other than they would have if there were seven settlers? Where will the six settlers ultimately build their houses, and what’s the maximum average distance between nearest neighbors?

Here is my solution:
[Show Solution]

Alice and Bob fall in love

In this interesting Riddler problem, we’re dealing with a possibly unbounded sequence of… children? Here it goes:

As you may know, having one child, let alone many, is a lot of work. But Alice and Bob realized children require less of their parents’ time as they grow older. They figured out that the work involved in having a child equals one divided by the age of the child in years. (Yes, that means the work is infinite for a child right after they are born. That may be true.)

Anyhow, since having a new child is a lot of work, Alice and Bob don’t want to have another child until the total work required by all their other children is 1 or less. Suppose they have their first child at time T=0. When T=1, their only child is turns 1, so the work involved is 1, and so they have their second child. After roughly another 1.61 years, their children are roughly 1.61 and 2.61, the work required has dropped back down to 1, and so they have their third child. And so on.

(Feel free to ignore twins, deaths, the real-world inability to decide exactly when you have a child, and so on.)

Five questions: Does it make sense for Alice and Bob to have an infinite number of children? Does the time between kids increase as they have more and more kids? What can we say about when they have their Nth child — can we predict it with a formula? Does the size of their brood over time show asymptotic behavior? If so, what are its bounds?

Here is an explanation of my derivation:
[Show Solution]

If you’re just interested in the answers to the questions, here they are:
[Show Solution]

Tether your goat!

A geometry problem from the Riddler blog. Here it goes:

A farmer owns a circular field with radius R. If he ties up his goat to the fence that runs along the edge of the field, how long does the goat’s tether need to be so that the goat can graze on exactly half of the field, by area?

Here is my solution:
[Show Solution]

The war game

This Riddler puzzle is about game theory… War or peace?

Two countries are eyeing each other’s gold. At the beginning of the game, the “strength” of each country’s army is drawn from a continuous uniform distribution and lies somewhere between 0 (very weak) and 1 (very strong). Each country knows its own strength but not that of its opponent. The countries observe their own strength and then simultaneously announce “peace” or “war.”

If both announce “peace,” then they each stay quietly in their own territory, with their own gold, which is worth \$1 trillion (so each “wins” \$1 trillion). If at least one announces “war,” then they go to war, and the country with the stronger army wins the other’s gold. (That is, the stronger country wins \$2 trillion, and the other wins \$0.)

What is the optimal strategy of each country (declaring “peace” or “war”) given its strength?

Extra credit: What if the countries don’t announce at the same time and instead one announces first and the other second? What if the value of winning the war were \$5 trillion rather than \$2 trillion?

Here is my solution for the first part, where both countries declare their intentions simultaneously.
[Show Solution]

Here is my solution for the second part, where the countries declare their intentions sequentially.
[Show Solution]

Baking the optimal cake

This Riddler puzzle asks about finding the maximum-volume shape subject to constraints.

A mathematician who has a birthday coming up asks his students to make him a cake. He is very particular and asks his students to make him a three-layer cake that fits under a hollow glass cone he has on his desk. He requires that the cake fill up the maximum amount of volume under the cone as possible and that the layers of the cake have straight vertical sides. What are the thicknesses of the three layers of the cake in terms of the height of the glass cone? What percentage of the cone’s volume does the cake fill?

Here is my solution.
[Show Solution]

Here, I go into more detail about bounding the optimal cake volume as the number of layers becomes large.
[Show Solution]

Can you outrun the angry ram?

The Riddler puzzle this week appears simple at first glance, but I promise you it’s not!

You, a hard-driving sheep farmer, are tucked into the southeast corner of your square, fenced-in sheep paddock. There are two gates equidistant from you: one at the southwest corner and one at the northeast corner. An angry, recalcitrant ram enters the paddock from the southwest gate and charges directly at you at a constant speed. You run — obviously! — at a constant speed along the eastern fence toward the northeast gate in an attempt to escape. The ram keeps charging, always directly at you.

How much faster than you does the ram have to run so that he catches you just as you reach the gate?

Here is a very simple solution by Hector Pefo. Minimal calculus required!
[Show Solution]

And here is my solution, which finds an equation for the path of the ram but requires knowledge of calculus and differential equations.
[Show Solution]