# Can You 'Waffle' Your Way To A Proof?

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 or send me an email.

## Riddler Express

This week’s Express is an oldie but a goodie:

Consider the infinite points in the coordinate plane, and suppose that each point is one of two colors: red or blue. It turns out there must be at least two points of the same color that are a distance 1 apart.

Why? Draw any equilateral triangle with side length 1. All three vertices are a distance 1 from each other, and at least two of them must be the same color, whether red or blue.

Now suppose every point in the plane is one of *three* colors: red, green or blue. Once again, it turns out there must be at least two points of the same color that are a distance 1 apart.

How can you show this is true using just seven points in the plane?

## Riddler Classic

As you may have heard, two high school students from St. Mary’s Academy in New Orleans, Louisiana — Ne’Kiya Jackson and Calcea Johnson — recently discovered a novel proof of the Pythagorean Theorem.

Their proof applied the law of sines (which itself can be derived from equivalent expressions for a triangle’s area and has *no dependency* on the Pythagorean theorem, thereby avoiding any circular logic) to the construction below:

Atop the figure are two reflected right triangles with legs *a* and *b *(with *a* b) and hypotenuse *c*. Below these triangles are what the students called a “waffle cone” shape, formed between the extensions of one of the top triangle’s hypotenuse and a line that’s perpendicular to the other hypotenuse.

In their proof, they compute distances *p* and *q*, where *p* extends from the leftmost vertex of the two triangles to the intersection of the lines, and *q* extends from the topmost vertex of the two triangles to the same intersection.

Your challenge is to determine expressions for *p* and *q* in terms of *a*, *b* and *c*. However, in doing so, you *absolutely cannot* use the Pythagorean theorem in any of its forms (e.g., the so-called “distance formula,” etc.). After all, solving for *p* and *q* is a key step toward *proving* the Pythagorean theorem.

*Extra credit:* Once you’ve determined *p* and *q*, try completing a proof of the Pythagorean theorem that makes use of them. Remember, the students used the law of sines at one point.

## Solution to the last Riddler Express

Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â Max Chai ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Foster City, California, winner of last week’s Riddler Express.

Last week, you and your family decided to decorate 10 beautiful Easter eggs. You pulled a fresh carton of eggs out of your fridge and removed 10 eggs. There were two eggs remaining in the carton, which you returned to the fridge.

The next day, you opened the carton again to find that the positions of the eggs had somehow changed — or so you thought. Perhaps the Easter Bunny was snooping around your fridge?

The 12 slots in the carton were arranged in a six-by-two array that was symmetric upon a 180-degree rotation, and the eggs were indistinguishable from each other. How many distinct ways were there to place two eggs in this carton? (Note: Putting two eggs in the two leftmost slots was considered the same as putting them in the two rightmost slots, since you could switch between these arrangements with a 180-degree rotation of the carton.)

First off, how many ways were there to place two eggs in a carton with 12 slots? That was 12 choose 2, or 66. Since some of these 66 ways were equivalent to each other after a 180-degree rotation, that meant the answer had to be less than 66.

The majority of these 66 arrangements could be paired up so that they turned into each other upon a 180-degree rotation of the carton. However, arrangements that were already symmetric were not paired up, as they merely turned back into themselves upon being rotated. There were six such symmetric arrangements, as shown below:

The remaining 60 arrangements formed 30 pairs, which meant the number of distinct ways to place two eggs in the carton was 30 + 6, or **36**.

For extra credit, you had to determine the number of distinct arrangements for other numbers of indistinguishable eggs between zero and 12. The “Packsize Riddle Solving Team” from Salt Lake City, Utah, extended the approach for two eggs to *x* eggs. There were 12 choose *x* ways to place the eggs in the carton, many of which paired up.

Now if *x* was odd, there was no way for the arrangement to turn back into itself after a 180-degree rotation. One way to convince yourself of this was that one half of the carton (e.g., the left half) had to have an even number of eggs while the other half had an odd number. After the rotation, that first half now had an odd number while the second half had an even number. There was no way for these arrangements to be the same. So when *x* was odd, the number of distinct arrangements was simply **(12 choose ***x***)/2**.

But when *x* was even, you had to subtract the symmetric arrangements before dividing by two. There were 6 choose (*x*/2) different ways to place half the eggs on the left half of the carton, and each of those had a single symmetric way to place the remaining eggs on the right half. So when *x* was even, the number of distinct arrangements was ((12 choose *x*) − (6 choose (*x*/2)))/2 + (6 choose (*x*/2)), which simplified to **((12 choose ***x***) + (6 choose (***x***/2)))/2**.

The maximum number of distinct arrangements (a whopping 472) occurred when *x* was 6. And across all possible values of *x* from 0 to 12, there were a grand total of 2,080 arrangements.

## Solution to the last Riddler Classic

Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±âIzumihara Ryoma ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Toyooka, Japan, winner of last week’s Riddler Classic.

Last week, you were the captain of a three-member crew (not including yourself): Geordi, Sidney and Alandra. Your ship had been captured by a previously unknown foe, who decided to return your ship if you could win a simple game.

Each of the three crew members was to be issued a number between zero and one, randomly and uniformly picked within that range. As the captain, your objective was to guess who had the highest number.

The catch was that you could only ask one yes-or-no question to each crew member. Based on the answer to the question you asked the first crew member, you could update the question you asked the second. Similarly, based on the answers to the first two questions, you could update the third question you asked. But in the end, you still had to guess which crew member had the highest number.

What was your optimal strategy, and what were your chances of regaining your ship?

Several readers interpreted the puzzle to mean that each crew member knew the numbers assigned to the *other* crew members. In this case, you could simply ask each crew member, “Do you have the largest number?,” thereby guaranteeing you’d know who had the largest number.

Yawn. The puzzles in this column are all about “math, logic and probability,” like it says at the top. So if you have some trivial interpretation, try reading the puzzle a different way or reach out to seek further clarification.

Now, this puzzle became *very* interesting when you assumed that each crew member knew *their own* number, but *not* those of their fellow crew mates. To see why, suppose for now that there were only two crew mates — say, Geordi and Sidney — instead of three.

In this case, you’d approach Geordi and ask him the only reasonable question you could: Is your number greater than *x*? I’m not saying (yet) what that number *x* is, but presumably, there was some value that optimized your overall chances of figuring out whether Geordi’s number or Sidney’s number was greater. If Geordi said yes, then you’d turn to Sidney and ask her if her number was greater than *y*. If she said yes, then you’d pick Sidney; otherwise, you’d pick Geordi. If Geordi said no, then you’d ask Sidney if her number was greater than *z*. Again, if she said yes, then you’d pick Sidney; otherwise, you’d pick Geordi.

But what were these values of *x*, *y* and *z*? If Geordi said yes, then his number was equally likely to be anywhere between *x* and 1, so the optimal value of *y* was halfway between these extremes, or (*x*+1)/2. And if Geordi said no, then his number was equally likely to be anywhere between 0 and *x*. Once again, the optimal value of *z* was halfway between these extremes, or *x*/2. With Geordi’s and Sidney’s numbers equally likely to be between zero and one, the diagram below highlights which coordinate pairs would lead you to guess *incorrectly*:

The area of this highlighted region was (*x*/2)^{2} + ((1−*x*)/2)^{2}, or (2*x*^{2}−2*x*+1)/4. This was minimized when its derivative was zero, i.e., when *x* = 1/2 and the area was 1/8. So when there were only two crew members, the cutoff value for Georgi’s question was 1/2 and the cutoffs for Sidney were 3/4 and 1/4. After all of this, your chances of correctly identifying who had the greater number was 7/8.

Whew! All that would have made for quite a challenging Riddler Classic. However, last week’s puzzle included a *third* crew member, Alandra, about whose number you could additionally inquire. Instead of a square, you now had a cube to consider.

In the end, Izumihara, this week’s winner (and the only person to solve the puzzle by the submission deadline, I might add) was able to identify a series of cutoff values that maximized your chances of regaining your ship. Below, I list the approximate values and the questions they’d correspond to:

- Geordi, is your value greater than 0.624334?
- If no: Sidney, is your value greater than 0.460442?
- If no: Alandra, is your value greater than 0.347818?
- If no: Guess Geordi
- If yes: Guess Alandra

- If yes: Alandra, is your value greater than 0.730221?
- If no: Guess Sidney
- If yes: Guess Alandra

- If no: Alandra, is your value greater than 0.347818?
- If yes: Sidney, is your value greater than 0.824920?
- If no: Alandra, is your value greater than 0.813443?
- If no: Guess Geordi
- If yes: Guess Alandra

- If yes: Alandra, is your value greater than 0.918159?
- If no: Guess Sidney
- If yes: Guess Alandra

- If no: Alandra, is your value greater than 0.813443?

- If no: Sidney, is your value greater than 0.460442?

With this set of questions, your chances of correctly identifying the crew member with the greatest number were approximately **82.395 percent**.

If you ever find yourself in this situation where your ship is captured, you have to play this game to secure your ship’s release and you’re short on time in deciding your strategy, you’d better hope you don’t have any more than two or three crew members. Could you imagine solving this puzzle with *four* crew members?

## Want more puzzles?

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.