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 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

Six snails are situated at the corners of a regular hexagon whose sides are all 10 meters. Each snail is determined to reach its clockwise neighbor, and they all travel at the same speed. But keep in mind that as each snail moves along, slowly but surely, the snail it’s traveling toward is *also* moving toward another snail.

The result is six snails — and six trails of slime — that spiral inward toward the center of the hexagon, as shown in the animation below:

How long is each snail’s slimy trail?

## Riddler Classic

From Julien Beasley, a new spin on the Sultan’s Dowry Problem, a classic problem of matrimony:

The sultan has asked her vizier to present her with 10 candidates for marriage. The vizier has searched the kingdom for the 10 most desirable partners, but he does not know whom the sultan will prefer. If she saw them all at the same time, she would easily be able to rank them from 1 (the best partner) to 10 (the worst partner). But the vizier can only present the candidates one at a time — very hard to sync everybody’s calendars, even back then — and in a random order. Upon seeing each candidate, the sultan must reject or accept him. If a candidate is rejected, the sultan cannot pick him again. But on seeing each new candidate, she knows exactly where he’d stack up relative to the candidates she has rejected. If she strategizes, what’s the highest rank she can expect her chosen candidate to have on average?

For example, if she simply accepted the first candidate presented to her, his rank could be anywhere from 1 to 10 with equal probability, averaging to 5.5. Surely she can do better…

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐJack Altman ÑÑâÐ of Chicago, Illinois, winner of last week’s Riddler Express.

Last week, you were challenged to find the volume of the solid inside the unit cube below, as unwittingly illustrated by artist Sol LeWitt. This time, I’ve included labels for the solid’s eight corners.

This challenge seemed straightforward enough, but there was one slight problem I hadn’t seen coming. As noted by solver Ziyad Gower, the top face (ABCD) of the solid *was not flat*! That meant the top face had to be curved or kinked. Don’t believe me? Check out the animation below, showing how the four corners of that top face don’t lie in the same plane.

Because you can’t tell how the top face is curved from the picture, I accepted a *range* of answers. The two extremes are when the top face is kinked inward (so that corners B and D connected along a straight edge) and when the top face is kinked outward (so that corners A and C are connected along a straight edge). Here’s one more animation, which continuously flips back and forth between the two extremes:

One way to find the volumes of these extremes is to break the solid up into smaller prisms and tetrahedra, find *their* volumes, and then add these values back up. The two extremes turn out to be very close in volume: 38/96 and 39/96, both of which are approximately 0.4. Given the ambiguity of the picture, solver Jack’s answer of 0.396 was as good as any.

Finally, submitter Eli Luberoff also created an interactive three-dimensional graph that calculates the solid’s volume if the top face *were* flat. Had that been the case, this riddle would have been much more fair.

That Sol LeWitt was a pretty tricky artist. The next time his work shows up in this column, I’ll make sure to file it as a Riddler Classic.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐKyle Tauzer ÑÑâÐ of Orlando, Florida, winner of last week’s Riddler Classic.

Last week I wanted to carve the perfect eye for my jack-o’-lantern. The eye had to be a proper triangle, meaning the three corners were connected by “straight” arcs that went directly from one corner to the next via the shortest possible path. I also wanted it to be equilateral (meaning all three sides had the same length) and equiangular (all three corner angles had the same measure). My original carving, seen below, had three 90-degree angles, but resulted in an eye that was too large, taking up an eighth of the pumpkin’s surface area.

Instead, I wanted an eye that took up one-sixteenth of the pumpkin’s surface area. What, then, were the angles between my three cuts?

First off, if you imagine cutting a very, very tiny triangle on the pumpkin’s surface — so tiny that you have to zoom in to see it — it’s almost like cutting a triangle out of a flat surface. It’ll be the typical equilateral triangle you learned about in geometry class, with three equal angles that measure 60 degrees. As the triangle gets larger, and the curvature of the sphere is no longer negligible, the three angles get bigger.

Surprisingly, the relationship between the triangle’s area and the measure of its angles is *linear*. For every degree the angle increases, the resulting triangle covers an additional 5/12 of a percent of the sphere. To see why this is true, check out Hector Pefo’s excellent write-up, which looks at overlapping pieces of the sphere called “lunes.”

To carve an ideal pumpkin eye that’s exactly one-sixteenth of the pumpkin’s surface area, the three angles must each be **75 degrees**. Some solvers were even able to find the solution in their head, noting that a triangle that takes up one-sixteenth of the sphere has an area that’s halfway between zero (when each angle is 60 degrees) and one-eighth (like my original eye, where each angle was 90 degrees). Halfway between 60 degrees and 90 degrees is, sure enough, 75 degrees.

Any which way you carve *your* pumpkin, Happy Halloween from the Riddler!

## 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.