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, and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

During a break at a music festival, the crew is launching T-shirts into the audience using a T-shirt cannon. And you’re in luck — your seat happens to be in the line of flight for one of the T-shirts! In other words, if the cannon is strong enough and the shirt is launched at the right angle, it will land in your arms.

The rows of seats in the audience are all on the same level (i.e., there is no incline), they are numbered 1, 2, 3, etc., and the T-shirts are being launched from directly in front of Row 1. Assume also that there is no air resistance (yes, I know, that’s a big assumption). You also happen to know quite a bit about the particular model of T-shirt cannon being used — with no air resistance, it can launch T-shirts to the very back of Row 100 in the audience, but no farther.

The crew member aiming in your direction is still figuring out the angle for the launch, which you figure will be a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up). Which row should you be sitting in to maximize your chances of nabbing the T-shirt?

Submit your answer

Riddler Classic

Last week’s Riddler column garnered six comments on Facebook. However, every single one of those comments was spam. Sometimes, spammers even reply to other spammers’ comments with yet more spam. This got me thinking.

Over the course of three days, suppose the probability of any spammer making a new comment on this week’s Riddler column over a very short time interval is proportional to the length of that time interval. (For those in the know, I’m saying that spammers follow a Poisson process.) On average, the column gets one brand-new comment of spam per day that is not a reply to any previous comments. Each spam comment or reply also gets its own spam reply at an average rate of one per day.

For example, after three days, I might have four comments that were not replies to any previous comments, and each of them might have a few replies (and their replies might have replies, which might have further replies, etc.).

After the three days are up, how many total spam posts (comments plus replies) can I expect to have?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Pat Walsh 👏 of Prospect Park, Pennsylvania, winner of last week’s Riddler Express.

Last week, you were walking along the middle of a wide sidewalk when you saw someone walking toward you from the other direction, also down the middle of the sidewalk, 12 feet away. Being responsible citizens, you passed each other while maintaining a distance of at least 6 feet at all times. By the time you reached each other’s original positions, you were back in the middle of the sidewalk again. You were asked to assume that the other person followed the same path you did, but flipped around (since they were walking in the opposite direction).

You wanted to know the shortest distance you and the other person could walk so that you could switch positions, all while staying at least 6 feet apart at all times. What was this distance? (Note: Although the problem didn’t explicitly say it, the following solutions all assume that you and your counterpart were always walking at the same speed.)

Many solvers thought the optimal path was a rhombus whose diagonals had lengths of 12 feet and 6 feet, as shown below:

While this path clocked in at 6√5, or about 13.42 feet, it was not the correct answer. In the above animation, the two circles each had radii of 3 feet. In other words, any time the circles overlapped, you and your counterpart had come within 6 feet of each other. Sure enough, this rhomboidal solution had a fair bit of overlap, and so it did not meet the puzzle’s social distancing requirement.

Another popular solution was a “do-si-do,” where you and your counterpart walked directly toward each other until you were exactly 6 feet apart. From there, you walked in opposite directions around a circle whose radius was 3 feet, before finally continuing down the middle of the sidewalk to your respective destinations:

As the animation shows, this path, which has length 6 + 3𝜋, or about 15.42 feet, respected social distancing. But was it the shortest path?

As it turned out, it was not. The optimal path had three parts: two tangent lines to that central circle whose radius was 3 feet, and a shorter arc of the circle itself. To show this was faster than the do-si-do, here they are in a race against each other:

With a little geometry, solver Len Chyall found the exact length of this path was 6√3 + 𝜋, or approximately 13.53 feet.

@xaqwg To be safe and efficient it's a distance of 13.534 ft. That's a little less than 13' 6-7/16" as measured with a tape measure. pic.twitter.com/DdAUxVuz0q — Len Chyall (@lchyall) April 6, 2020

Last week’s extra credit asked what would happen if the person walking toward you had no intention of straying from the center of the sidewalk (sigh), and it was entirely up to you to maintain a distance of at least 6 feet.

In this case, the shortest path again had three parts: two tangent lines and an arc. But this time, because of the asymmetry in the problem (i.e., you and your counterpart were no longer following the similarly shaped paths), the lengths of the tangent lines had to be calculated separately, and the “arc” between them wasn’t a true arc — the circle it wrapped around was a moving target.

Solver Angela Zhou animated the solution, finding the shortest distance was about 17.45 feet.

Steve Schaefer arrived at the same answer analytically, but there was a fair bit of calculus and at least one transcendental equation involved — beyond the scope of this column.

In the end, if the person walking toward you had any sense of decency and mirrored your path, you could have shaved about 4 feet off your journey. It pays to stay safe and keep your distance.

Solution to last week’s Riddler Classic

Congratulations to 👏 Greg Couillard 👏 of Ithaca, New York, winner of last week’s Riddler Classic.

Last week you were asked about a mysterious snowplow. From the moment it began snowing one morning, the snow fell at a constant rate, and it continued the rest of the day.

At noon, a snowplow began to clear the road. The more snow there was on the ground, the slower the plow moved. In fact, the plow’s speed was inversely proportional to the depth of the snow — if you doubled the amount of snow on the ground, the plow moved half as fast.

During its first hour on the road, the plow traveled 2 miles. During the second hour, the plow traveled only 1 mile.

When did it start snowing?

Several solvers recognized this problem — perhaps they were in the same calculus class as the puzzle’s submitter, Phil Imming. It’s more likely they saw it in their own calculus class. Indeed, this riddle has been attributed to Ralph Palmer Agnew’s differential equations textbook, published in 1942. There’s even a video with more than 300,000 views that walks through the solution.

Suppose it started snowing at time t = 0. For any time t, the height of the snow was proportional to t, which meant the speed of the plow was proportional to 1/t. We’re looking for two consecutive hours where the plow moved twice as much during the first hour as it did in the second. And we can find the distance the plow traveled in a given hour by taking the integral of its speed — again, proportional to 1/t — over time. (Some solvers tried a shortcut where they looked at the average amount of snow on the ground during the hour, but this led to incorrect solutions.) In other words, if the plow started at time t 0 , then the integral of 1/t from t 0 to t 0 +1 was twice the integral of 1/t from t 0 +1 to t 0 +2.

The integral of 1/t is ln(t), the natural log of t. That meant for the plow to have gone twice as far in the first hour as the second, t 0 had to satisfy the following equation: ln(t 0 /(t 0 +1)) = 2ln((t 0 +1)/(t 0 +2)). With the help of some logarithmic identities, this equation became t 0 2 + t 0 − 1 = 0 — a quadratic — whose positive solution was t 0 = (−1 + √5)/2, or one divided by the golden ratio.

In other words, the plow started about 0.618 hours, or about 37 minutes, after the snow started falling. Since you were told the plow started at noon, that meant it started snowing 37 minutes before noon, at approximately 11:23 a.m.

Looking back on this problem, what would have happened if the snowplow had started just before 11:23 a.m.? There would have been no snow on the ground, and since the plow’s speed was inversely proportional to the amount of snow … it would have momentarily been zipping around at an infinite speed. So I’m just glad it waited a few minutes before starting!

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 riddlercolumn@gmail.com.