Solutions to recent puzzles

Chalkdust

10 January 2019

• Click to share on X (Opens in new window)
• Click to share on Facebook (Opens in new window)
• Click to share on Reddit (Opens in new window)
• Click to share on LinkedIn (Opens in new window)
• Click to share on Threads (Opens in new window)
• Click to share on Mastodon (Opens in new window)

Many a Chalkdust article started life as a puzzle. What’s the intersection of these two lines? How many of these cups would it take to make a sphere? How can I cut my birthday cake? …How else can I cut my birthday cake? Writing and doing puzzles is great but, well, sometimes you don’t feel satisfied unless you’ve got the solution. Never fear, we’re here to give you the answers to three puzzles that were set in recent articles!

Who got the grant? (Yuliya Nesterova)

We start with the results for the third and final Chalkdust Christmas competiton, Yuliya’s Eggnog mystery. This holiday puzzle took place at a festive social, where the attendants were all jostling to work out who won a big academic grant. Readers were invited to have a go at being sleuths themselves, piecing together the following logical clues to deduce the grant winner. We’ll let Meg, the protagonist, explain how it went down.

Here are the clues in full. Remember, exactly one clue is false. Among five offices in a row, where the office at each end is a corner office…

• Dr. Eric Jun’s office is next to the grant winner.
• Moe owns a cat.
• Ester owns a parrot.
• Ness took an office that is not adjacent to the cat owner’s to protect her fish.
• The dog owner’s office is next to Moe’s.
• The leftmost office is mine.
• Exactly two neighbouring doors’ name labels have the same first initial.
• Moe is midway between two corner offices.
• The fish owner is 3 doors to the right of my office.
• The grant winner has a corner office.

We’ll let Megan, the protagonist, describe the solution:

All ten of these couldn’t hold true at once, that’s for sure: there was a liar out there, and not the out-lier kind we see in statistics. A fiendish fibber. A precautious prevaricator. A mathematician you wouldn’t want to seat at your truth table. Luckily, the culprit did not lie well, or rather she did not lie in a logically undetectable fashion. And lucky for me, Herc picked upon it.

Statements 2, 4, 6, 8, and 9 are mutually exclusive, so the single lie must be contained therein. If statement 2 were false, among other matters, statement 7 couldn’t hold: no two neighbouring names begin with the same letter.

If statement 4 was false, … well, we’ll get to that in a minute. If statement 6 were false, the fact that I didn’t win wouldn’t help us pick a winner.

If statement 8 were false, there’d be three options based on Moe’s potential location:

a contradiction with 1 and 10: Eric can’t be near the winner.

is a contradiction with 7: two pairs of neighbours share a first initial. Also, I don’t own a dog, last I checked. And something sure smells a whole lot of fishy with all these pet fish.

is a contradiction with statement 9: the fish cannot be three doors from me.

If statement 9 were false, why, there’d be no two neighbours with the same first initial, and statement 7 couldn’t hold. Ness almost got away with it, too.

Ness Newton lied about ever owning a fish and about her office’s location. Statment 4 is false. To be frank, she was at variance with the whole idea of us looking for the grant winner: Ester is her friend and wanted to be out of the spotlight for one lovely evening. Well, news sure travels fast in closed connected networks!

Congratulations to competition winners Jonathan Winfield, Rachel Long and Alana Huszar. A Chalkdust t shirt will be on its way to you shortly!

A complex integral (Jamie Handitye)

Jamie’s article from August 2018 looked at the history of complex numbers, and some of the ways that they helped mathematics develop. It ended with this challenging question: Use complex numbers to evaluate the following integral,

$$\int_{0}^{1}\frac{\sin(\log x)}{\log x}dx.$$

The procedure for doing this is a little involved, and uses several clever tricks that are handy to have in your back pocket! First, use Euler’s formula to re-write $\sin(\log x)$ in terms of complex exponentials:

$$\sin(\log x) = \frac{x^i – x^{-i}}{2i}$$

Then comes a clever bit. Introduce a new variable $b$, and define

$$I(b) = \int_{0}^{1} \frac{x^i – x^{-b i}}{2i \log x}dx$$

so that our problem is to find $I(1)$. This technique is known as Feynman’s method. Although it doesn’t seem to have simplified the problem we can, through Liebniz’s rule, compute the derivative
$$I'(b) = -\frac{1}{2} \frac{1}{b i – 1},$$
so that
$$I(b) = \frac{i}{2} \log(bi – 1) + c,$$
where $c$ is a constant of integration. From the integral definition, $I(-1) = 0$ and so $c = -i \log{(-i – 1)}/2$. Plugging this back into our expression for $I$ gives the answer as
$$I(1) = \frac{i}{2} ( \log(i – 1) – \log(-i – 1) ),$$
which, wouldn’t you know it, turns out to be $\pi/4$. Lovely stuff.

Tic-tac-toe (Alex Bolton)

Alex’s article in issue 08 looked at playing tic-tac-toe on various topologically interesting boards. The article also featured various puzzles of the form; given some positions on the board, how can X win? Assume that O always plays optimally.

The cylinder:

The Mobius strip:

The torus:

The Klein bottle:

The projective plane:

The adjacency map for the plane looks like this: