# Can You Split The States?

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.

## Riddler Express

You’ve been hired to design a new logo for Riddler City. The mayor is a little eccentric and has requested that the logo have at least two lines of symmetry that intersect at an angle of precisely 1 radian, or 180/𝜋 (approximately 57.3) degrees.

What sorts of logos meet the mayor’s requirements? (You can give one example or describe what all possible logos have in common.)

Submit your answer

## Riddler Classic

Earlier this month, Adam Kotsko asked the Twitterverse to choose four contiguous U.S. states for a new breakaway nation:

This got Philip Bump wondering about states you could pick such that the remaining (previously) contiguous states were no longer contiguous, but rather broken up into two near-halves by area. Treating the Upper Peninsula of Michigan as distinct from the Lower Peninsula, Philip hypothesized that removing Illinois, Missouri, Oklahoma and New Mexico would create two near-equal halves.

Suppose you remove a set of states (not necessarily four) so that you have two distinct contiguous regions among the lower 48 states, where the larger region has area *A* and the smaller region has area *B*. What states should you remove to maximize area *B*? (Was Philip’s intuition correct?) And what percentage of the lower 48 states’ combined area does B represent?

For reference, here are the areas of the states according to the U.S. Census Bureau. Like Philip, you can treat Michigan’s peninsulas as distinct “states” here. Michigan’s Upper Peninsula has an area of 16,377 square miles. Also, don’t worry about islands, Minnesota’s Northwest Angle, etc. — you can assume they are contiguous with the rest of their respective states.

Submit your answer

## Solution to last week’s Riddler Express

Congratulations to 👏 Max G. 👏 of Chicago, winner of last week’s Riddler Express.

Last week, Max the Mathemagician (no relation to this week’s winner, as far as I know!) was calling for volunteers. He had a magic wand of length 10 that could be broken anywhere along its length (fractional and decimal lengths were allowed). After the volunteer chose these breakpoints, Max would multiply the lengths of the resulting pieces. For example, if they broke the wand near its midpoint and nowhere else, the resulting product would have been 5×5, or 25. If the product was the largest possible, they’d win a free backstage pass to his next show. (Amazing, right?)

You raised your hand to volunteer, and you and Max briefly made eye contact. As he called you up to the stage, you knew you had this in the bag. What was the maximum product you could have achieved?

First off, for any given number of pieces, the product was always maximized when they were all the same length. For example, had you broken the wand into two pieces, if the length of the first piece was x, then the length of the other was (10−*x*). You could experiment with different values, but the maximum of *x*·(10−*x*) occurred when *x* was 5, giving you the optimal product of 25. To see why this was true, you could graph the function *f*(*x*) = *x*·(10−*x*) and spot the peak of its parabolic arch, or you could take the derivative of the function and set it equal to zero.

When you had three pieces, again the product was maximized when all three were the same. Dividing a length of 10 among three pieces meant each piece had a length of 10/3. Multiplying these values together gave you (10/3)^{3}, or about 37.04. And with four pieces, each segment had a length of 10/4, or 2.5. Their product was 2.5^{4}, or 39.0625.

So far, as the number of pieces increased, the product increased as well. Would this pattern continue?

No, it would not. With five pieces, the product was 2^{5}, or 32. Sure enough, the maximum product occurred when there were exactly four pieces, which meant the optimal product was indeed **39.0625**.

But what was so special about the number 2.5? Why did *that* length piece result in a larger product than the other values mentioned so far?

If you generalized to the case of *N* pieces, the product became (10/*N*)* ^{N}*. Let’s rewrite this product as a function of the length

*L*of each resulting piece (which equals 10/

*N*), so that it becomes

*L*

^{10/}

*, or (*

^{L}*L*

^{1/}

*)*

^{L}^{10}. This shows that regardless of the original length of the unbroken magic wand, your best strategy was to pick a piece length

*L*that maximized the function

*L*

^{1/}

*— an expression that is maximized by*

^{L}*e*, or about 2.71828.

However, it didn’t make sense to have lengths that were *precisely e*, since you needed a whole number of pieces that were all equal in length. And that’s why the optimal length was 2.5 — because 2.5 was fairly close to *e*.

For extra credit, Zax the Mathemagician (no relation to either Max) had the same routine in his show, only his wand has a length of 100. What was the maximum product now?

Again, your best bet was to look for piece lengths close to *e*. Splitting Zax’s wand into 37 equal pieces meant that each piece was about 2.703 units long, which was as close to *e* as you could get. And so the answer to the extra credit was (100/37)^{37}, or about **9.47 quadrillion**.

Looks like everyone is winning free backstage passes to some mathemagic shows!

## Solution to last week’s Riddler Classic

Congratulations to 👏 Brian Schoep 👏 of Highlands Ranch, Colorado, winner of last week’s Riddler Classic.

Last week’s Classic came courtesy of Alexander Zhang of Lynbrook High School, California. Alexander won first place in the mathematics category at this year’s International Science and Engineering Fair, and has long had an interest in topology — which *just might *have been related to the puzzle.

You were asked to consider the following image, which showed a particular uppercase sans serif font:

Alexander thought many of these letters were equivalent, but he left it to you to figure out how and why. He also had a message for you:

It might not have looked like much, but Alexander assured me that it was equivalent to exactly one word in the English language.

What was Alexander’s message?

There wasn’t a lot to go on here, and the message provided — “YIRTHA” — was woefully insufficient for reverse engineering a cipher.

The clues were Alexander’s interest in topology, and the fact that he thought many of the letters were somehow equivalent. From there, many solvers recognized that some letters in the alphabet are homeomorphic, which is how a topologist might classify objects as being equivalent. More specifically, two letters are homeomorphic if one can be deformed into the other without cutting or gluing (and vice versa). So while certain features like holes and three-way intersections must be preserved, curves and lines make no difference.

Almost half of the letters in the typeface shown above can be made with a single open (i.e., there are no holes) stroke: {C, G, I, J, L, M, N, S, U, V, W, Z}. These 12 letters are all homeomorphic and make up the largest equivalence class.

Meanwhile, four letters have three “tails” that all meet up at a single point, with no holes: {E, F, T, Y}. Two letters have a single hole: {D, O}. Two letters have one hole and one tail: {P, Q}. (Q’s topology is particularly dependent on the choice of font, which is why an image of a specific font had to be used in this puzzle.) Two more letters have one hole and two tails: {A, R}. One letter has two holes: {B}. Another letter has four tails: {X}. Finally, two letters have a “bar” with two tails on either end: {H, K}. (Like Q, K’s topology depends on the font. In the image above, K’s bar is rather short.)

So what did these equivalence classes have to do with Alexander’s message? If you took each letter in “YIRTHA” and replaced it with another letter in its equivalence class (or left the letter untouched), there were 1,536 possible six-letter combinations. Many solvers worked out the solution by hand, letter by letter. Others, like Elaine H. of Tampa, Florida and Stefano Perfetti of Zurich, Switzerland, wrote code to search through all the letter combinations and check them against a dictionary. Either way, only *one* equivalent combination was an English word:

Yes, Alexander’s message was “EUREKA,” the famous interjection uttered by Archimedes and the motto of Alexander’s home state of California.

Solver Laurent Lessard got in on the fun with his own encoded message describing topological equivalence.

I will say, I found it rather surprising how many words in the English language have unique topologies. Of the 15,000 or so six-letter words (courtesy of Peter Norvig’s word list), almost 25 percent of them are unique. That was definitely my YIRTHA! moment last week.

## 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]