Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Special note for this week: We have our first Riddler video! If you recall the story of Planet Xiddler, or even if you don’t and just want to try a new way of solving The Riddler, then watch me walk through the steps to helping the Royal Scientist choose their successor.
Can you solve this puzzle about an alien planet? | FiveThirtyEight
Halloween is just around the corner, which means it’s time for some pumpkin math!
I have a spherical pumpkin. I carefully calculate its volume in cubic inches, as well as its surface area in square inches. Next, I got up to have a piece of Halloween candy (which, naturally, was a Reese’s Peanut Butter Cup).
But when I came back to my calculations, I saw that my units — the square inches and the cubic inches — had mysteriously disappeared from my calculations. But it didn’t matter, because both numerical values were the same!
What is the radius of my spherical pumpkin?
Extra credit: Let’s dispense with 3D thinking. Instead, suppose I have an n-hyperspherical pumpkin. Once again, I calculate its volume (with units inn) and surface area (with units inn−1). Miraculously, the numerical values are once again the same! What is the radius of my n-hyperspherical pumpkin?
The solution to this Riddler Express can be found in the following column.
From Arnaud Quentin comes a question that recently took Netflix subscribers by squid — I mean storm:
Congratulations, you’ve made it to the fifth round of The Squiddler — a competition that takes place on a remote island. In this round, you are one of the 16 remaining competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:
To cross the bridge, you must jump from one pair of squares to the next. However, you must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.
You and your competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.
On average, how many of the 16 competitors will survive and make it to the next round of the competition?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to ÑÑâÐ Jonah Goldstein ÑÑâÐ of Miami, Florida, winner of last week’s Riddler Express.
Last week, Duke Leto Atreides knew for a fact that there were not one, but two traitors within his royal household. The suspects were Lady Jessica, Dr. Wellington Yueh, Gurney Halleck and Duncan Idaho. Leto’s advisor, Thufir Hawat, assisted him in questioning the four suspects. Anyone who was a traitor told a lie, while anyone who was not a traitor told the truth.
Upon interrogation, Jessica said that she was not a traitor, while Wellington similarly said that he was not a traitor. Gurney said that Jessica or Wellington was a traitor. Finally, Duncan said that Jessica or Gurney was a traitor. (Thufir, being the logician that he was, noted that when someone said thing A was true or thing B was true, both A and B could both technically be true.)
After playing back the interrogations in his mind, Thufir was ready to report the name of one of the traitors to the duke. Whose name did he report?
With two traitors among four suspects, there were six possible pairs of traitors for Thufir to consider: Jessica and Wellington, Jessica and Gurney, Jessica and Duncan, Wellington and Gurney, Wellington and Duncan, and Gurney and Duncan. Solver Madeline Argent decided to take a closer look at each of these pairs.
If Jessica and Wellington were the traitors, then both indeed lied about not being traitors. Meanwhile, both Gurney and Duncan told the truth, since Gurney accused both traitors and Dunan accused Jessica. So this was a pair for Thufir to file away.
If Jessica and Gurney were the traitors, then Jessica again lied about not being a traitor. However, Wellington, Gurney and Duncan all would have been telling the truth, which meant none of them could be traitors. With just one traitor (rather than the requisite two), Thufir could discard this scenario. The same could also be said for the case where Jessica and Duncan were the traitors. And if Wellington and Gurney were the traitors, then only Wellington was lying.
Now if Wellington and Duncan were the traitors, then Jessica was telling the truth and Wellington was indeed lying about not being a traitor. Gurney was telling the truth by implicating Wellington, while Duncan was lying by accusing neither traitor. With two liars, this was another pair for Thufir to file away. However, if Gurney and Duncan were the traitors, then only Gurney was lying.
In the end, Thufir knew the traitors either had to be Jessica and Wellington or Wellington and Duncan. No matter what, Dr. Wellington Yueh betrayed his duke.
For the record, this was just a riddle, and not a spoiler.
Solution to last week’s Riddler Classic
Congratulations to ÑÑâÐ Grant Larsen ÑÑâÐ of Elsah, Illinois, winner of last week’s Riddler Classic.
Last week, you had an equilateral triangle. You picked three random points, one along each of its three edges, uniformly along the length of each edge. (In other words, each point along each edge had the same probability of being selected).
With those three randomly selected points, you formed a new triangle inside the original one. What was the probability that the center of the larger triangle could also be found inside the smaller one?
As is often the case for a Classic, a number of solvers started by turning to their computers. Al Shaheen gained some intuition for the problem by re-creating the triangles using a graphing calculator. By running 100 Monte Carlo simulations, one solver found the probability appeared to be a little under 0.5, or 50 percent.
Meanwhile, Peter Ji ran a thousand times as many simulations, finding the solution was closer to 0.46.
To find the answer more precisely, we needed some calculus, along with a helpful few diagrams. Let’s pick any of the three sides of the triangle and suppose the point on that side (which we’ll call A) is a distance p from the median to that side. To keep things relatively simple, suppose each side of the triangle has length 2, so that p is uniformly distributed between 0 and 1. With a little triangular mathemagic (i.e., via coordinate geometry or mass points), you could show that the line connecting A and the triangle’s center passed through another side at point B, a distance (1−p)/(1+3p) from the median to that side. This is illustrated below:
Next, consider the point on the third side, which we’ll call C. (In the above diagram, that would be the top-right side of the triangle.) When C was on the bottom half of its side, the smaller triangle contained the center whenever the point on the bottommost side was to the left of B. Out of a total side length of 2, the fraction that was left of B was (1+p)/(1+3p).
The other half of the time, when C was higher up — say, a distance q past the median to that side — then the bottommost point also had to be to the right of a point D, which was a distance (1−q)/(1+3q) to the left of the median to that side.
Putting this all together, for the 50 percent of the time when C was lower down, the fraction of the time the triangle contained the center was the integral of (1+p)/(1+3p), for p from 0 to 1. The other 50 percent of the time, when C was higher up, the fraction was half a double integral that added (1−p)/(1+3p) and (1−q)/(1+3q). Since p and q were independent, the two parts of the sum were equal, meaning this fraction was just the integral of (1−p)/(1+3p).
Finally, you had to add half the integral of (1+p)/(1+3p) to half the integral of (1−p)/(1+3p), both for p from 0 to 1. With some canceling in the numerators, this became the integral of 1/(1+3p), which was ln(1+3p)/3. Plugging in the bounds of 0 and 1 gave you a final answer of 2ln(2)/3, or about 46.21 percent.
If you were surprised to see a natural logarithm pop up in this solution, then you should check out this write-up by the puzzle’s submitter, Allen Gu. Allen went on to solve a more general version of this problem, where the point that had to lie within both triangles was not the center, but rather any point within the larger triangle. Again, natural logarithms made an appearance, along with a neat heat map:
Among all the points in the larger triangle, the center is most likely to lie within the smaller, random triangle. I, for one, say we chalk that up to symmetry.
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at email@example.com.