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]

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]

Riddler Football Playoffs

This week’s Riddler Classic is a probability question inspired by the ongoing World Cup.

The Riddler Football Playoff (RFP) consists of four teams. Each team is assigned a random real number between 0 and 1, representing the “quality” of the team. If team $A$ has quality $a$ and team $B$ has quality $b$, then the probability that team $A$ will defeat team $B$ in a game is $\frac{a}{a+b}$.

In the semifinal games of the playoff, the team with the highest quality (the “1 seed”) plays the team with the lowest quality (the “4 seed”), while the other two teams play each other as well. The two teams that win their respective semifinal games then play each other in the final.

On average, what is the quality of the RFP champion?

My solution:
[Show Solution]

Randomly cutting a sandwich

This week’s Riddler Classic is geometry puzzle about randomly slicing a square sandwich.

I have made a square sandwich, and now it’s time to slice it. But rather than making a standard horizontal or diagonal cut, I instead pick two random points along the perimeter of the sandwich and make a straight cut from one point to the other. (These points can be on the same side.)

What is the probability that the smaller resulting piece has an area that is at least one-quarter of the whole area?

My solution:
[Show Solution]

Fall color peak

This week’s Riddler Classic is a seasonal puzzle about leaves changing color.

The trees change color in a rather particular way. Each tree independently begins changing color at a random time between the autumnal equinox and the winter solstice. Then, at a random later time for each tree — between when that tree’s leaves began changing color and the winter solstice — the leaves of that tree will all fall off at once. At a certain time of year, the fraction of trees with changing leaves will peak. What is this maximal fraction?

My solution:
[Show Solution]

Another way to solve the problem, courtesy of Matthew Wallace:
[Show Solution]

Catch the grasshopper

This week’s Riddler classic is a probability problem about a grasshopper!

You are trying to catch a grasshopper on a balance beam that is 1 meter long. Every time you try to catch it, it jumps to a random point along the interval between 20 centimeters left of its current position and 20 centimeters right of its current position. If the grasshopper is within 20 centimeters of one of the edges, it will not jump off the edge. For example, if it is 10 centimeters from the left edge of the beam, then it will randomly jump to anywhere within 30 centimeters of that edge with equal probability (meaning it will be twice as likely to jump right as it is to jump left). After many, many failed attempts to catch the grasshopper, where is it most likely to be on the beam? Where is it least likely? And what is the ratio between these respective probabilities?

My solution:
[Show Solution]

Cone crawling

This week’s Riddler Classic is a geometry problem about traversing the surface of a cone

The circular base of the cone has a radius of 2 meters and a slant height of 4 meters. We start on the base, a distance of 1 meter away from the center. The goal is to reach the point half-way up the cone, 90 degrees around the cone’s central axis from the start, as shown. What is the shortest path?

Here is my solution:
[Show Solution]

The luckiest coin

This week’s Riddler Classic is about finding the “luckiest” coin!

I have in my possession 1 million fair coins. I first flip all 1 million coins simultaneously, discarding any coins that come up tails. I flip all the coins that come up heads a second time, and I again discard any of these coins that come up tails. I repeat this process, over and over again. If at any point I am left with one coin, I declare that to be the “luckiest” coin.

But getting to one coin is no sure thing. For example, I might find myself with two coins, flip both of them and have both come up tails. Then I would have zero coins, never having had exactly one coin.

What is the probability that I will at some point have exactly one “luckiest” coin?

Here is my solution:
[Show Solution]

Triangle Trek

This week’s Riddler Classic is a problem involving traversing a triangle.

Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures 15 degrees, and sides AB and AC both have length 1.

Amare must:

  • Start at point B.
  • Second, touch a point — any point — on side AC.
  • Third, touch a point — any point — back on side AB.
  • Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).

What is the shortest distance Amare can travel to complete the desired path?

I solved the problem in two different ways. The elegant solution:
[Show Solution]

And the more complicated 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]