Tiling squares

This week’s Fiddler is about tiling a square with smaller squares.

Suppose you have infinitely many 3-by-3 cm tiles and infinitely many 5-by-5 cm tiles. You want to use some of these tiles to precisely cover a square whose side length is a whole number of centimeters. Tiles may not overlap, and they must completely cover the larger square, without jutting beyond its borders. What is the smallest side length this larger square can have, such that it can be precisely covered using at least one 3-by-3 tile and at least one 5-by-5 tile?

Extra credit:
This time, you have an infinite supply of square tiles for each odd whole number side length (as measured in centimeters) greater than 1 cm. In other words, you have infinitely many 3-by-3 cm tiles, infinitely many 5-by-5 cm tiles, infinitely many 7-by-7 cm tiles, and so on. You want to use one or more of these tiles to precisely cover a square whose side length is $N$ cm, where $N$ is an integer. Once again, tiles may not overlap, and they must completely cover the larger square without jutting beyond its borders. What is the largest integer N for which this task is not possible?

My solution:
[Show Solution]

Showcase Showdown

This week’s Fiddler is based on “Showcase Showdown” on the game show “The Price is Right”.

Suppose we have some number of players. Player A is the first to spin a giant wheel, which spits out a real number chosen randomly and uniformly between 0 and 1. All spins are independent of each other. After spinning, A can either stick with the number they just got or spin the wheel one more time. If they spin again, their assigned number is the sum of the two spins, as long as that sum is less than or equal to 1. If the sum exceeds 1, A is immediately declared a loser.

After A is done spinning (whether once or twice), B steps up to the wheel. Like A, they can choose to spin once or twice. If they spin twice and the sum exceeds 1, they are similarly declared the loser. This continues until all players are done. Whoever has the greater value (that does not exceed 1) is declared the winner.

Assuming all players play the game optimally, what are player A’s chances of winning?

My solution:
[Show Solution]

The weaving loom problem

This week’s Fiddler is a classic problem.

A weaving loom consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to a picture that looks like this:

As the number of hooks goes to infinity, what does the shape trace out?

Extra credit: If four looms are rotated and superimposed as shown below, what is the area of the white region in the middle?

My solution:
[Show Solution]

Betting on football with future knowledge

This week’s Fiddler is a football-themed puzzle with a twist: you can see the future! Sort of.

You know ahead of time that your football team will win 8 of their 12 remaining games, but you don’t know which ones. You can place bets on every game, placing bets either for or against your team. You can bet any amount up to how much you currently have. You want to implement a betting strategy that guarantees you’ll have as much money as possible after the 12 games are complete. If you did so, then after the 12 games how much money would you be guaranteed to have if you started with $100?

My solution:
[Show Solution]

A more elegant alternative solution, due in large part to a clever observation by Vince Vatter.
[Show Solution]

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]