Giving a 1-hour talk to highschool math club: any topic suggestion?

Question

I've been invited (by my kid) to give a one hour talk to her highschool math club. Last year (right before the pandemic hit) I did two such talks on probability, and they loved it. I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$. These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard. The talk will be on March $1$st, $2021$, and will be on zoom. I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas. E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle. Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc. However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend. :)

Some ideas:

• modulo arithmetic esp. finite fields: Perfect shuffle is fun. Would cryptography (e.g. RSA) be too difficult? What about codes e.g. Hamming? Any other fun examples of modulo arithmetic?

• non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points". Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

• finite geometry: Personally I find Finite projective planes very beautiful. Do they (or finite affine planes) have any application? (Besides the kiddie game Spot It?)

• algorithm: This is actually my work area, but I don't know what I can cover in an hour. Maybe the $O(N \log N)$ lower bound for sorting? Some of these kids don't even have coding experience... :(

• graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs. Eulerian tour. Euler characteristic (of a planar graph) is related. What else? Shortest path would require discussion of algorithm. Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

• combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure. What are your favorite (elementary) examples? Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)

Answer 1

Have you considered game theory (e.g. Nim or Nim-like games)? Will students have the chance to interact and play a few games against each other?

Since you say it's a "math club", are these kids AMC 10/12 level?

Among your suggestions, I find combinatorics the most promising. There are lots of angles of attack: 1. story-problem/story-proof, 2. algebraic proof, 3. generating functions.

Aw man you're making me miss my high-school times. Can you record this and put it on the internet?

Answer 2

Again, thanks all of you for answering and commenting. I thought some of you might be interested in what actually happened.

I decided to go with a theme of "finite mathematics" and talk mostly about modulo arithmetic, using it to illustrate $3$ different topics:

(A) I started by performing a crowd-pleasing "magic trick" related to the perfect shuffle. I actually cannot do a perfect shuffle at all, so instead I did a sort of reverse: I dealt a deck of cards into 4 piles, re-assembled them into a deck, and repeated the process $3$ more times. As expected, the kids were very surprised to see that the original deck order was restored! Which I then explained via $4^4 = 256 = 1 \pmod{51}$ as in the wikipedia article. (BTW if some of you are not aware of this trick, it's actually quite neat and also easy to do in practice.) This took about $15$ minutes.

(B) Then I talked about error correction codes, starting with the background: why you need it, message, encoding, codeword, error, decoding, etc. For motivation, I found out that the ASCII for lower-case "l" and lower-case "d" differ by just $1$ bit, so I made up a story how, with a $1$-bit error, "Trump lied" would become "Trump died". :D

Anyway, I discussed the Hamming $(7,4)$ code in detail, with actual examples of encoding and decoding, doing parity checks (which I called mod $2$ addition) by hand, and introducing the concept of Hamming distance. I did not use linear algebra, as most kids haven't learned that.

Then I discussed the crucial fact that, when $N$ is prime, mod $N$ arithmetic supports division (I did not use the term finite field), which in turn means you can solve systems of equations. This segued into polynomials and Reed Solomon codes, which I covered briefly using the approach of this excellent blogpost by @JyrkiLahtonen - thanks!

The whole error correction code segment took about $30$ minutes.

(C) Lastly, I talked about the kiddie game Spot-It - I have asked my kid to poll her friends before hand and many are familiar with it. While the game itself is uninteresting, the math behind the card design is fascinating: it's based on the finite projective plane of order $7$. I spent many slides using modulo $5$ arithmetic to construct the finite affine plane of order $5$, then showing how it satisfies some of Euclid's axioms and therefore deserves the fancy name of "finite plane". Then, I showed how it can almost be turned into the cards of Spot-It, discussing point-line duality briefly. Finally I added the points and line at infinity to turn it into a finite projective plane for the full solution. This segment took about $25$ minutes.

The whole thing took about $70-75$ minutes and was quite well received, both during the event, and when my kid told me the others' reactions afterwards. She said the older kids mostly got it, while the younger kids had some trouble but still got the flavor.

I spent way, waaaaay too much time over the weekend making the slides, as I was having such a great time! I just couldn't bear to leave out any of the $3$ topics since they're all fun (and even "educational" in the case of error correction codes) and also show a great variety between them. I ended up with about $60$ slides! Had to talk really fast to go through them in $70$ minutes. :)

Answer 3

Conway's rational tangles, continued fractions, and the "greedy" Euclidean algorithm!

Here are some Math Circles notes by Tom Davis on the topic. You're going to want to have two pieces of thick rope, at least 10 feet in length, and hopefully have at least 4 volunteers to hold the ends for a very fun interactive activity.

Pictured: construction of the tangle $$ -4 + \frac{1}{2+\frac{1}{3}} = -\frac{25}{7} $$

Answer 4

The $2$-dimensional case of Sperner's lemma is a good introduction to graph theory. The proof is self-contained aside from using the Handshaking lemma, which is also easy to show. Take a triangle and subdivide it into several smaller triangles. Call a triangle minimal if it contains no other triangles. If whenever two minimal triangles touch, they share a side, then we call this subdivision a triangulation. (This is to avoid the possibility that the side of one minimal triangle is part of a side of another triangle.)

Color the nodes of the original triangle three colors, say red, blue, and green. Then color all of the other nodes with those colors, under the condition that if a node lies on a side of the original triangle, then it has to be given one of the colors used for the original two nodes of that side. Nodes inside the triangle can be given any of the three colors. See the picture below. (Apologies for the poor quality- this was made in mspaint and then hastily resized.) Observe that there's a red-blue side, a red-green side, and a blue-green side.

There are five minimal triangles that have a vertex of each color. Sperner's lemma states that there must always be a minimal triangle with a node of each color, and in fact, there must be an odd number of such triangles. To see this, we will construct a graph from this triangulation. We do so by associating a vertex to each face, including the outside face, and draw an edge between two vertices if the side shared by their corresponding triangles has two different colors. See the picture below.

The internal vertices can have degree either $0$, $2$, or $3$. The degree is $0$ if the nodes of the corresponding triangle all have the same color, $2$ if the nodes use two different colors, and $3$ if the nodes all use different colors. We want to show that there are an odd number of internal vertices of odd degree. The Handshaking Lemma tells us that a graph must have an even number of vertices with odd degree, so if we can show that the outside vertex has odd degree, then the lemma follows. For example, the outside vertex in our picture has degree $7$.

To see this, look at any side of the original graph, say the red-green side. Moving from the red node to the green node, that side contributes $1$ whenever the colors switch from red to green or green to red. Since we start with a red node and end with a green node, there must be an odd number of switches. Hence, each side of the original triangle makes an odd contribution to the degree of the outside vertex so the outside vertex has odd degree. By the Handshaking Lemma, Sperner's Lemma follows.

Answer 5

When the numbers are not too bad, we can just do a one-step calculation over a few times: Here I am using decimals; for worse numbers like $\sqrt{61}$

Lubin posted an algorithm (using whole numbers only) that appears to be what Fermat used; I then programmed that. There is a method due (largely) to Gauss and Lagrange that produces a chain or cycle of indefinite quadratic form, I programmed that long ago. Again, no decimal, all whole number, allows for any $\frac{A+\sqrt B}{C}.$ Finally, Conway's Topograph method. For that, I can recommend Weissman for self study or parent-child team study of some sort

Full treatment of $\sqrt{19}$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 19} = 4 + \frac{ \sqrt {19} - 4 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {19} - 4 } = \frac{ \sqrt {19} + 4 }{3 } = 2 + \frac{ \sqrt {19} - 2 }{3 } $$ $$ \frac{ 3 }{ \sqrt {19} - 2 } = \frac{ \sqrt {19} + 2 }{5 } = 1 + \frac{ \sqrt {19} - 3 }{5 } $$ $$ \frac{ 5 }{ \sqrt {19} - 3 } = \frac{ \sqrt {19} + 3 }{2 } = 3 + \frac{ \sqrt {19} - 3 }{2 } $$ $$ \frac{ 2 }{ \sqrt {19} - 3 } = \frac{ \sqrt {19} + 3 }{5 } = 1 + \frac{ \sqrt {19} - 2 }{5 } $$ $$ \frac{ 5 }{ \sqrt {19} - 2 } = \frac{ \sqrt {19} + 2 }{3 } = 2 + \frac{ \sqrt {19} - 4 }{3 } $$ $$ \frac{ 3 }{ \sqrt {19} - 4 } = \frac{ \sqrt {19} + 4 }{1 } = 8 + \frac{ \sqrt {19} - 4 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccc} & & 4 & & 2 & & 1 & & 3 & & 1 & & 2 & & 8 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 9 }{ 2 } & & \frac{ 13 }{ 3 } & & \frac{ 48 }{ 11 } & & \frac{ 61 }{ 14 } & & \frac{ 170 }{ 39 } \\ \\ & 1 & & -3 & & 5 & & -2 & & 5 & & -3 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 19 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 19 \cdot 1^2 = -3 & \mbox{digit} & 2 \\ \frac{ 9 }{ 2 } & 9^2 - 19 \cdot 2^2 = 5 & \mbox{digit} & 1 \\ \frac{ 13 }{ 3 } & 13^2 - 19 \cdot 3^2 = -2 & \mbox{digit} & 3 \\ \frac{ 48 }{ 11 } & 48^2 - 19 \cdot 11^2 = 5 & \mbox{digit} & 1 \\ \frac{ 61 }{ 14 } & 61^2 - 19 \cdot 14^2 = -3 & \mbox{digit} & 2 \\ \frac{ 170 }{ 39 } & 170^2 - 19 \cdot 39^2 = 1 & \mbox{digit} & 8 \\ \end{array} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$\sqrt{21}$ Simple continued fraction tableau:

4.58, 1.7, 1.39, 2.52, 1.89, 1.11,8.58

$$ \begin{array}{cccccccccccccccccc} & & 4 & & 1 & & 1 & & 2 & & 1 & & 1 & & 8 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 5 }{ 1 } & & \frac{ 9 }{ 2 } & & \frac{ 23 }{ 5 } & & \frac{ 32 }{ 7 } & & \frac{ 55 }{ 12 } \\ \\ & 1 & & -5 & & 4 & & -3 & & 4 & & -5 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 21 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 21 \cdot 1^2 = -5 & \mbox{digit} & 1 \\ \frac{ 5 }{ 1 } & 5^2 - 21 \cdot 1^2 = 4 & \mbox{digit} & 1 \\ \frac{ 9 }{ 2 } & 9^2 - 21 \cdot 2^2 = -3 & \mbox{digit} & 2 \\ \frac{ 23 }{ 5 } & 23^2 - 21 \cdot 5^2 = 4 & \mbox{digit} & 1 \\ \frac{ 32 }{ 7 } & 32^2 - 21 \cdot 7^2 = -5 & \mbox{digit} & 1 \\ \frac{ 55 }{ 12 } & 55^2 - 21 \cdot 12^2 = 1 & \mbox{digit} & 8 \\ \end{array} $$

=======================================

$ \sqrt{41}$

6.403, 2.48, 2.08, 12.40,2.48, 2.08, 12.39

$$ \begin{array}{cccccccccccccccc} & & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 6 }{ 1 } & & \frac{ 13 }{ 2 } & & \frac{ 32 }{ 5 } & & \frac{ 397 }{ 62 } & & \frac{ 826 }{ 129 } & & \frac{ 2049 }{ 320 } \\ \\ & 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 41 \cdot 0^2 = 1 & \mbox{digit} & 6 \\ \frac{ 6 }{ 1 } & 6^2 - 41 \cdot 1^2 = -5 & \mbox{digit} & 2 \\ \frac{ 13 }{ 2 } & 13^2 - 41 \cdot 2^2 = 5 & \mbox{digit} & 2 \\ \frac{ 32 }{ 5 } & 32^2 - 41 \cdot 5^2 = -1 & \mbox{digit} & 12 \\ \frac{ 397 }{ 62 } & 397^2 - 41 \cdot 62^2 = 5 & \mbox{digit} & 2 \\ \frac{ 826 }{ 129 } & 826^2 - 41 \cdot 129^2 = -5 & \mbox{digit} & 2 \\ \frac{ 2049 }{ 320 } & 2049^2 - 41 \cdot 320^2 = 1 & \mbox{digit} & 12 \\ \end{array} $$

============================

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 21 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 21 \cdot 1^2 = -5 & \mbox{digit} & 1 \\ \frac{ 5 }{ 1 } & 5^2 - 21 \cdot 1^2 = 4 & \mbox{digit} & 1 \\ \frac{ 9 }{ 2 } & 9^2 - 21 \cdot 2^2 = -3 & \mbox{digit} & 2 \\ \frac{ 23 }{ 5 } & 23^2 - 21 \cdot 5^2 = 4 & \mbox{digit} & 1 \\ \frac{ 32 }{ 7 } & 32^2 - 21 \cdot 7^2 = -5 & \mbox{digit} & 1 \\ \frac{ 55 }{ 12 } & 55^2 - 21 \cdot 12^2 = 1 & \mbox{digit} & 8 \\ \end{array} $$

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf

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Answer 6

My own math skill tends toward discrete: comp-sci, combinatorics, etc. However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend. :)

Some ideas:

• modulo arithmetic esp. finite fields: Perfect shuffle is fun. Would cryptography (e.g. RSA) be too difficult? What about codes e.g. Hamming? Any other fun examples of modulo arithmetic?

...

• algorithm: This is actually my work area, but I don't know what I can cover in an hour. Maybe the $O(N \log N)$ lower bound for sorting? Some of these kids don't even have coding experience... :(

You did probability before. Even without cryptography, I think it's fascinating that one can pretend to have probability using a pseudorandom number generating algorithm. The simplest algorithms are very simple, but still depend on basic results in modulo arithmetic, finite fields, etc. You could teach some basic statistical tests on PRNG quality if you want.

You can get what you need from Knuth's TAOCP vol. 2, ch. 3, or maybe Kneusel's Random Numbers and Computers (too shallow?) or Niederreiter & Winterhof (too deep?).

Another idea: Fourier analysis. Nice graphics. Sum cool curves, infinite series, integrals--or stick with discrete cases. You can even transform Boolean functions.

Answer 7

I've always been a fan of the coin problem. It's basically a generalization of the question, "what is the largest score in American football that cannot be achieved (say, ignoring safeties)?" It's got a built-in hook with the story of the "McNugget numbers". It very naturally motivates $\gcd(a_1, \ldots, a_n)$ and you can discuss existence of the Frobenius number. As I recall there's a nice graphical proof of the formula for two coins. The topic typically uses some basic number theory like Bezout's lemma, but if anything that's a feature and not a bug.