## Making the fastest track

This week’s Riddler Classic is a problem about minimum-time optimization.

While passing the time at home one evening, you decide to set up a marble race course. No Teflon is spared, resulting in a track that is effectively frictionless. The start and end of the track are 1 meter apart, and both positions are 10 centimeters off the floor. It’s up to you to design a speedy track. But the track must always be at floor level or higher — please don’t dig a tunnel through your floorboards. What’s the fastest track you can design, and how long will it take the marble to complete the course?

My solution:
[Show Solution]

## Ellipse packing

You’ve heard of circle packing… Well this week’s Riddler Classic is about ellipse packing!

This week, try packing three ellipses with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio. What is the smallest such larger ellipse you can find? Specifically, how long is its major axis?

Extra credit: Instead of three smaller ellipses, what about other numbers of ellipses?

My solution:
[Show Solution]

## Perfect pizza sharing

This week’s Riddler Classic is about how to cut a pizza to achieve precise area ratios between the slices.

Dean made a pizza to share with his three friends. Among the four of them, they each wanted a different amount of pizza. In particular, the ratio of their appetites was 1:2:3:4. Therefore, Dean wants to make two complete, straight cuts (i.e., chords) across the pizza, resulting in four pieces whose areas have a 1:2:3:4 ratio.

Where should Dean make the two slices?

Extra credit: Suppose Dean splits the pizza with more friends. If six people are sharing the pizza and Dean cuts along three chords that intersect at a single point, how close to a 1:2:3:4:5:6 ratio among the areas can he achieve? What if there are eight people sharing the pizza?

My solution:
[Show Solution]

To jump straight to the results:
[Show Solution]

## Polarization Puzzle

This week’s Riddler Classic is about light polarization.

When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they aren’t perfectly aligned, only the component of the light that’s in the direction of the polarizer passes through. For example, here is what happens if you use two polarizers, the first at 45 degrees, and the second at 90 degrees. The length of the original vector is decreased by a factor of 1/2.

I have tons of polarizers, and each one also reflects 1 percent of any light that hits it — no matter its polarization or orientation — while polarizing the remaining 99 percent of the light. I’m interested in horizontally polarizing as much of the incoming light as possible. How many polarizers should I use?

Here is my solution:
[Show Solution]

## Evil twin

This week’s Riddler Classic is a pursuit problem with a twist. Here is the problem, paraphrased.

You are walking in a straight line (moving forward at all times) near a lamppost. Your evil twin begins opposite you, hidden from view by the lamppost, as illustrated in the figure below.

Assume your evil twin moves precisely twice as fast as you at all times, and they remain obscured from your view by the lamppost at all times. What is the farthest your evil twin can be from the lamppost after you’ve walked the 200 feet as shown?

Here is my solution.
[Show Solution]

## Tetrahedron optimization

This week’s Riddler Classic is a short problem 3D geometry. Here we go! (I paraphrased the question)

A polyhedron has six edges. Five of the edges have length $1$. What is the largest possible volume?

Here is my solution
[Show Solution]

## Outthink the Sphinx

This week’s Riddler Classic is a tricky puzzle that combines logic and game theory.

You will be asked four seemingly arbitrary true-or-false questions by the Sphinx on a topic about which you know absolutely nothing. Before the first question is asked, you have exactly $1. For each question, you can bet any non-negative amount of money that you will answer correctly. That is, you can bet any real number (including fractions of pennies) between zero and the current amount of money you have. After each of your answers, the Sphinx reveals the correct answer. If you are right, you gain the amount of money you bet; if you are wrong, you lose the money you bet. However, there’s a catch. (Isn’t there always, with the Sphinx?) The answer will never be the same for three questions in a row. With this information in hand, what is the maximum amount of money you can be sure that you’ll win, no matter what the answers wind up being? Extra credit: This riddle can be generalized so that the Sphinx asks N questions, such that the answer is never the same for Q questions in a row. What are your maximum guaranteed winnings in terms of N and Q? If you’re just looking for the answer, here it is: [Show Solution] Here is a more detailed write-up of the solution: [Show Solution] ## Baking the biggest pie This week’s Riddler Classic is about baking the biggest pie. Just in time for π day! You have a sheet of crust laid out in front of you. After baking, your pie crust will be a cylinder of uniform thickness (or rather, thinness) with delicious filling inside. To maximize the volume of your pie, what fraction of your crust should you use to make the circular base (i.e., the bottom) of the pie? Here is my solution: [Show Solution] ## Polygons with perimeter and vertex budgets his week’s Riddler Classic involves designing maximum-area polygons with a fixed budget on the length of the perimeter and the number of vertices. The original problem involved designing enclosures for hamsters, but I have paraphrased the problem to make it more concise. You want to build a polygonal enclosure consisting of posts connected by walls. Each post weighs$k$kg. The walls weigh$1$kg per meter. You are allowed a maximum budget of$1$kg for the posts and walls. What’s the greatest value of$k$for which you should use four posts rather than three? Extra credit: For which values of$k\$ should you use five posts, six posts, seven posts, and so on?

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]