Feb. 7, 2020,

at

8:00 AM

# How Many More Palindrome Dates Will You See?

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

From James Anderson comes a palindromic puzzle of calendars:

This past Sunday was Groundhog Day. Also, there was a football game. But to top it all off, the date, 02/02/2020, was palindromic, meaning it reads the same forwards and backwards (if you ignore the slashes).

If we write out dates in the American format of MM/DD/YYYY (i.e., the two digits of the month, followed by the two digits of the day, followed by the four digits of the year), how many more palindromic dates will there be this century?

## Riddler Classic

Also on Super Bowl Sunday, math professor Jim Propp made a rather interesting observation:

At first glance, this might look like one of those annoying memes about order of operations that goes viral every few years — but it’s not.

When you write lengthy mathematical expressions using parentheses, it’s always clear which “open” parenthesis corresponds to which “close” parenthesis. For example, in the expression (1+2(3−4)+5), the closing parenthesis after the 4 pairs with the opening parenthesis before the 3, and *not* with the opening parenthesis before the 1.

But pairings of other mathematical symbols can be more ambiguous. Take the absolute value symbols in Jim’s example, which are vertical bars, regardless of whether they mark the opening or closing of the absolute value. As Jim points out, |−1|−2|−3| has *two* possible interpretations:

- The two left bars are a pair and the two right bars are a pair. In this case, we have 1−2·3 = 1−6 = −5.
- The two outer bars are a pair and the two inner bars are a pair. In this case, we have |−1·2−3| = |−2−3| = |−5| = 5.

Of course, if we gave each pair of bars a different height (as is done in mathematical typesetting), this wouldn’t be an issue. But for the purposes of this problem, assume the bars are indistinguishable.

How many different values can the expression |−1|−2|−3|−4|−5|−6|−7|−8|−9| have?

## Solution to last week’s Riddler Express

Congratulations to 👏 Stephen Kloder 👏 of Berkeley, California, winner of last week’s recent Riddler Express.

Last week’s Riddler Express was inspired by Nick Brett’s roll at the recent World Indoor Bowls Championships in Great Yarmouth, England:

You were asked to imagine yourself in Nick’s shoes, trying to split two of your opponent’s bowls. Each bowl was a sphere with a radius of 1, and your opponent’s two red bowls were separated by a distance of 3 — that is, their centers were separated by a distance of 5. If the angle between the path your bowl was on and the line connecting your opponent’s bowls was 75 degrees, then you could indeed split your opponent’s bowls, as shown in the animation below.

What was the *minimum* such angle that would have allowed your bowl to split your opponents’ bowls without hitting them?

Stephen reasoned that, when the path of the green bowl made the minimum angle, it would be tangent to the two red bowls as it passed by them, meaning it just barely grazes each of them at a single point.

Solver Thomas Stone sketched it out and applied some trigonometry.

Triangle ABC in Thomas’s sketch is a right triangle, since a circle’s radius is always perpendicular to the tangent it intersects. That means the sine of the angle in question was equal to 2 (twice the radius of each bowl) divided by 2.5 (half the distance between the centers of your opponent’s two bowls). The angle itself turns out to be approximately **53.13 degrees**.

And if you’re still not convinced, here’s an animation of that perfect roll:

Nick Brett may be a great professional indoor bowler, but no one can match the precision of trigonometry.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Daniel Hennessy 👏 of West Chester, Ohio, winner of last week’s Riddler Classic.

Last week you looked at Magna-Tiles that were isosceles triangles with one 30 degree angle and two 75 degree angles. If you arranged 12 of these tiles with their 30 degree angles in the center, they would lay flat and form a regular dodecagon. If you put fewer (between three and 11) of those tiles together in a similar way, they would form a pyramid whose base was a regular polygon.

To maximize the volume contained within the resulting pyramid, how many tiles should you use?

Solver Sam Koehn used design software to mock up the different pyramids and measure their volumes directly. Here are his results for pyramids of between four and 11 tiles:

Plugging in a side length of 3 inches for each tile’s shortest side, Sam found that the volume was maximized when **10 tiles** were used, for a total volume of about 73 cubic inches.

It was, of course, possible to calculate this without visually rendering the pyramids (but now that you’ve seen them, that’s definitely the cooler way to solve the problem). Many solvers set pencil to paper and slogged through the messy trigonometry, which was apparently last week’s theme.

Solver David Zimmerman found the volume as a precise function of the number of tiles. For any given number of tiles, he first measured the dimensions of the polygonal base they would form, and then used the Pythagorean theorem to find the resulting pyramid’s height. At this point, the volume can be found by multiplying the area of the base by the height and dividing by three. I’ll spare you the exact formula, which has cotangents and cosecants to spare.

The puzzle’s author, Rob Berger, took a similar approach, and graphed the area of the base, the height, and the volume for each of the possible pyramids. As you increase the number of tiles, the area of the base increases and the height decreases. Sure enough, the product of the base and height is maximized when there are 10 tiles.

In this problem, the greatest number of tiles you could put together was 12, which resulted in a flat pancake shape. Several solvers were interested in what would happen if this number of tiles, 12, was increased. In other words, if you have some large number *N* of congruent isosceles triangles that fit together to form a pancake, what number of triangles are needed to form a pyramid with maximal area?

By taking limits and setting derivatives equal to zero, solver Hector Pefo found that the volume is maximized when you use slightly more than 80 percent of the tiles, or, more precisely, a fraction of them that’s close to √(2/3).

So if you ever find yourself in a position where you have lots of skinny triangles arranged in a circular formation, and you need to build a container out of them, be sure to get rid of a fifth of them before you break out the glue.

## 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 protected].

## Footnotes

Important small print: Please wait until Monday to publicly share your answers. In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor. @xaqwg