# Some beautiful geometry in a challenge problem from Alexander Bogomolny

I did the project below with the boys on Sunday before they went off to camp for a week. The idea wasn’t to get into heavy math, but rather just a relaxed walk through some fun shapes. We got one detail wrong in the 4th video which I was sort of kicking myself for, but then I saw a tweet from Nassim Taleb showing some of the geometry in a different problem that Alexander Bogonolny had posted and it made me realize the connection between the algebra and geometry in our problem was still fun to show:

So, despite the error I thought I would publish the project anyway.

Here’s the original problem:

Below are the videos showing our walk through the geometry. First, though, here’s the quick introduction to the problem:

After that intro we looked at the region described by the constraint in the problem. We have to thicken up the region a little bit using the absolute value function in order to see it, so the Mathematica code looks a bit more complicated than in the problem, but that extra complexity is just to make the picture easier to see.

One cool thing about our discussion here is that my younger son thought there should be 3 fold symmetry in the shape because there was 3 fold symmetry in the equation 🙂

Now we looked at the situation in which the surface achieves the maximum value subject to the constraint in the problem. My younger son made the nice observation that the two surfaces appeared to be “blending together” at certain points. That “blending” is an important idea in Lagrange Multipliers – though, don’t worry, we aren’t going down that path today.

Next we looked at the minimum value of the surface subject to the constraint in the problem. The error I made here was accidentally reversing the two surfaces. The fixed surface – the one describing the constraint – is now on the outside rather than the inside.

Finally, I asked the kids to pick a value smaller than 45/4 for the curve so that we could see what happened. Unfortunately they picked 7 which is too small – there’s no surface! – so they chose 10 and that allowed us to see that the shrinking surface inside of the original shape. Also we can see fairly clearly (after some rotation) that the two shapes do not intersect.

Definitely a fun project showing the boys a beautiful side of a really challenging problem.

# Looking at Dave Richeson’s “Euler’s Gem” book with kids

I stumbled on this book at Barnes & Noble last week:

It is such a delightful read that I thought the kids might enjoy it, too, so I had them read the introduction (~10 pages).

Here’s what they learned:

Next we tried to calculate Euler’s formula for two simple shapes – a tetrahedron and a cube:

After that introduction we moved on to some slightly more complicated shapes – an icosahedron and a rhombic dodecahedron. The rhombic dodecahedron gave the kids a tiny bit of trouble since it doesn’t have quite the same set of symmetries as any of the Platonic solids:

Now we tried two very difficult shapes:

We studied these shapes last week in a couple of projects inspired by an Alexander Bogomolny tweet:

Working through an Alexander Bogomolny probability problem with kids

Connecting yesterday’s probability project with a few old 3d geometry projects

I suspected that this part would be difficult, so I had them count the faces, edges, and verticies of the two shapes off camera. Here’s what they found:

So, since the boys couldn’t agree on the number of verticies, edges, and faces of one of the shapes, I had them build it using our Zometool set to see what was going on. The Zometool set helped, thankfully. Here’s what they found after building the shape (and we got a little help from one of our cats):

Definitely a fun project. It was especially cool to hear the kids realize that the shape they were having difficulty with was (somehow) a torus. Or, as my older son said: “In the torus class of shapes.” I’m excited to try to turn a few other ideas from Richeson’s book into projects for kids.

# A short continued fraction project for kids

I woke up this morning to see another great discussion between Alexander Bogomolny and Nassim Taleb. The problem that started the discussion is here:

and the mathematical point that caught my eye was the question -> which positive integers are close to being integer multiples of $\pi$?

One possible approach to this question uses the idea of “continued fractions.” I learned about continued fractions from my high school math teacher, Mr. Waterman, who taught them using C. D. Olds’s book.

So, today I stared off by talking about irrational numbers and reviewing a simple proof that the square root of 2 is irrational:

Next we talked about why integer multiples of irrational numbers can never be integers. This I think is an obviously step for adults, but it took the kids a second to see the idea:

Now we moved on to talk about continued fractions. I’m not trying to go into any depth here, but rather just introduce the idea. I use my high school teacher’s procedure: split, flip, and rat 🙂

We work through a simple example with $\sqrt{2}$ and also see that the first couple of fractions we see are good approximations to $\sqrt{2}$.

With that background work we went up to use Mathematica to explore different aspects of continued fractions quickly. One thing we did, in particular, was use the fractions we found to find multiples of $\sqrt{2}$ that were nearly integers.

Finally, we wrapped up by using continued fractions to find good approximations to $\pi$, $e$ and a few other numbers.

Definitely a fun project, and one that makes me especially happy because of the connection to Mr. Waterman. Hopefully the boys will want to play around with this idea a bit more tomorrow.

# A terrific example for calculus students from Nassim Taleb and Alexander Bogomolny

I saw a wonderful exchange on twitter yesterday on a problem posted by Alexander Bogomolny:

At first this problem didn’t really jump off the page as a good first year calculus example, but then I as the solution that Nassim Taleb posted:

I’m a tiny bit time constrained this morning and can’t get the Taleb tweet to embed right, so here’s the solution a second time just in case the embedding remains broken:

So, Taleb reduces the difficult-looking limit and sums to two integrals. The ideas underlying this reduction are both beautiful on their own and fundamental in calculus.

A few questions that I think would be worth discussing with calculus students are:

(1) [this one was discussed in the twitter thread] Why did the integrals start at 0, and does that matter?

(2) Why is ratio of the integrals equal to the ratio of the sums? This answer to this question is related to the answer to (1). It is also an excellent way to reinforce some of the main ideas behind Riemann sums.

(3) Probably less mathematically interesting, but a good challenge exercise for students is evaluating Taleb’s integral formulation of the problem using l’Hospital’s rule. I say “less mathematically interesting” because you have to evaluate the same integrals in both approaches, but the approach via l’Hospital’s rule allows you to discuss the Fundamental Theorem of Calculus and also review the chain rule. The arithmetic here requires you to be extra careful, but I think the other ideas outweigh the annoying arithmetic.

Too bad that the school year is over – but this is a great example to keep in your back pocket for next year’s calculus classes!

# Connecting yesterday’s probability project with a few old 3d geometry projects

In yesterday’s project we were studying a fun probability question posed by Alexander Bogomolny:

That project is here:

Working through an Alexander Bogomolny probability problem with kids

While writing up the project, I noticed that I had misunderstood one of the
geometry ideas that my older son had mentioned. That was a shame because his idea was actually much better than the one I heard, and it connected to several projects that we’ve done in the past:

Learning 3d geometry with Paula Beardell Krieg’s Pyrmaids

Revisiting an old James Tanton / James Key Pyramid project

Overnight I printed the pieces we needed to explore my son’s approach to solving the problem and we revisited the problem again this morning. You’ll need to go to yesterday’s project to see what leads up to the point where we start, but the short story is that we are trying to find the volume of one piece of a shape that looks like a cube with a hole in it (I briefly show the two relevant shapes at the end of the video below):

Next we used my son’s division of the shape to find the volume. The calculation is easier (and more natural geometrically, I think) than what we did yesterday.

It is always really fun to have prior projects connect with a current one. It is also pretty amazing to find yet another project where these little pyramids show up!

# Working through an Alexander Bogomolny probability problem with kids

Earlier in the week I saw Alexander Bogomolny post a neat probability problem:

There are many ways to solve this problem, but when I saw the 3d shapes associated with it I thought it would make for a fun geometry problem with the boys. So, I printed the shapes overnight and we used them to work through the problem this morning.

Here’s the introduction to the problem. This step was important to make sure that the kids understood what the problem was asking. Although the problem is accessible to kids (I think) once they see the shapes, the language of the problem is harder for them to understand. But, with a bit of guidance that difficulty can be overcome:

With the introduction out of the way we dove into thinking about the shape. Before showing the two 3d prints, I asked them what they thought the shape would look like. It was challenging for them to describe (not surprisingly).

They had some interesting comments when they saw the shape, including that the shape reminded them of a version of a 4d cube!

Next we took a little time off camera to build the two shapes out of our Zometool set. Building the shapes was an interesting challenge for the kids since it wasn’t obvious to them what the diagonal line segments should be. With a little trial and error they found that the diagonal line segments were yellow struts.

Here’s their description of the build and what they learned. After building the shapes they decided that calculating the volume of the compliment would likely be easier.

Sorry that this video is a little fuzzy.

Having decided to look at the compliment to find the volume, we took a look at one of the pieces of the compliment on Mathematica to be sure that we understood the shape. They were able to see pretty quickly that the shape had some interesting structure. We used that structure in the next video to finish off the problem:

Finally, we worked through the calculation to find that the volume of the compliment was 7/27 units. Thus, the volume of the original shape is 20 / 27.

As I watched the videos again this morning I realized that my older son mentioned a second way to find the volume of the compliment and I misunderstood what he was saying. We’ll revisit this project tomorrow to find the volume the way he suggested.

I really enjoyed this project. It is fun to take challenging problems and find ways to make them accessible to kids. Also, geometric probability is an incredibly fun topic all by itself!

# A night with Cut the Knot, Nassim Taleb, and some Supernova

Please note the correction at the bottom of the post

A further correction – there is still an error. Ugh. This approach may not work, unfortunately . . .

Saw a neat problem from Alexander Bogomolny earlier today:

I actually missed the problem when it was initial posted, but saw it via Nassim Taleb’s solution:

The problem sort of gnawed at me all day and I figured it was in the maybe 1 in 10 problems that Bogomolny posts that I might be able to solve.

So, tonight I poured a glass of Supernova and gave it a go

One thing on my mind all day with this problem was Jensen’s inequality. What I would *love* to be able to do is say that by Jensen’s inequality:

$(1/3) \sqrt{x^2 + 3} + (1/3) \sqrt{y^2 + 3} + (1/3) \sqrt{xy + 3}$

$\geq \sqrt{ (1/3)( x^2 + y^2 + xy) + 3}$

Which is easily seen to be $\geq 2$ because of the constraint $x + y = 2.$ That work would show the original inequality was $\geq$ 6.

The approach has a tiny bit of merit since $\sqrt{x^2 + 3}$ is concave up for $x$ between 0 and 2 -> here’s a little Mathematica plot showing that the second derivative is indeed positive on 0 to 2:

But . . . the problem is that folding in the 3rd term in the sum is stretching the rules of Jensen’s inequality a bit, I think, since it is not of the form $\sqrt{a^2 + 3}$.

With the first two terms, though, applying Jensen’s inequality seems ok, but I now need (1/2)’s instead of (1/3)’s since there are only two terms. So, I’ll use Jensen’s on the first two terms only and try to show that

$(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3$

By Jensen’s inequality this new sum is

$\geq \sqrt{ \frac{x^2 + y^2}{2} + 3} + (1/2) \sqrt{xy + 3}$

A bit of algebra and the fact that $x + y = 2$ allows us to simplify this expression to:

$\sqrt{5 - xy} + (1/2) \sqrt{xy + 3}$

and then further to:

$\sqrt{ (x - 1)^2 + 4} + (1/2) \sqrt{4 - (x - 1)^2}$

Now we are just down to a fairly straightforward calculus problem, and I’ll let Mathematica do the heavy lifting since the algebra isn’t that interesting:

We can see visually that the minimum occurs at $x = 1$ from the plot, and the plot of the derivative further confirms that there is only one critical point. The value of the last expression at $x = 1$ is indeed 3 as we were hoping.

So, Jensen’s inequality, a bit of calculus, and a nice glass of scotch shows that the original inequality is indeed true.

Thanks to Alexander Bogomolny for the problem, and to Nassim Taleb for his solution that got me thinking about the problem.

Correction

I received a note from Alexander Bogomoly over night. He spotted an error in the calculation:

and I thought my kids having trouble sleeping and waking me up at 5:00 am today was a bad start to the day!

But it seems that I’ve gotten very lucky – both learning from my carelessness in applying Jensens inequality and that the path forward from Bogomolny’s correction is easier than the path I actually took.

Starting here – we wish to show that:

$(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3$

The correction shows that the expression on the left hand side is $\geq$ than

$\sqrt{ (\frac{x + y}{2})^2 + 3} + (1/2) \sqrt{xy + 3}$

but since $x + y = 2$, the first piece of this expression is equal to 2 and the 2nd expression simplifies as before. So we are left with

$2 + (1/2) \sqrt{4 - (x - 1)^2}$

and this expression has a maximum of 3 at $x = 1.$

That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6.

I’m grateful to Alexander Bogomolny for spotting the error in my original argument.

# A hand waving approach to a problem posted by Cut the Knot

Saw this tweet yesterday:

It was a fun problem to think about and the two solutions on the site use the Stolz-Cesaro Lemma, which is basically l’Hospital’s rule for sums.

Through the various Christmas preparations yesterday I was wondering if there was a simple way to see why the limit exists in the first place. What follows below isn’t a rigorous proof (or even close to one!) but instead how I convinced myself that the limit probably does exist.

Since seeing Tim Gowers “live blog” his solution to an old IMO problem, I’ve been interested in occasionally sharing the solution process rather than polished solutions to problems. Two examples of problems I’ve used for that idea are below:

A Challenge / Plea to math folks

A challenge relating to a few problems giving my son trouble

So, for the problem at hand, here’s my “hand waving” approach to convincing myself that the limit even existed:

We know that:

$\lim_{x\to\infty} E_n = \lim_{x\to\infty} (1 + \frac{1}{n})^n = e$

and that

$\lim_{x\to\infty} H_n = \lim_{x\to\infty} (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}) \approx \ln{n} + \gamma$

So I’ll approximate the difference we are considering like this for large $n$:

$(E_{n+1})^{H_{n+1}} - (E_n)^{H_n}$

$\approx e^{H_{n+1}} - e^{H_n}$

$\approx (e^{H_n})(e^{\frac{1}{n+1}} - 1)$

$\approx (e^{\ln{n} + \gamma}) (\frac{1}{n+1} + \frac{1}{2(x+1)^2} + \ldots )$

$\approx (e^{\gamma}) * n * (\frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots )$

$\approx e^{\gamma}$

The fact that this hand waving approach arrives at the “right” answer is just a coincidence as I’m playing pretty fast and loose with limit rules. But, at least I now have some indication that this strange (and lovely!) $\infty - \infty$ limit might actually exist.

Just for fun here’s what the expression looks like for n up to 1,000,000:

Definitely a fun little problem to noodle over 🙂

# Steven Strogatz’s circle area exercise

Saw this really neat tweet from Steven Strogatz yesterday:

I asked him if he had downloadable versions of the circle and he was nice enough to share the templates with me (yay!)

So, with a little enlarging and a little cutting we had the props ready to go through the exercise.

We began with a short conversation about circles. My older son knows lots of formulas about circles from his school’s math team practices, but my younger son doesn’t really know all of the formulas. The quick review here seemed like a good way to motivate Strogatz’s project:

Now we moved into Strogatz’s project – how do we show that the area of a circle is $\pi * r^2$? We cut the circle into the 16 sectors and rearranged them into a shape that was more familiar to us:

Next was the big challenge and the really neat idea in Strogatz’s first tweet – there is a different shape we can use to find the area. The boys were able to find this triangle fairly quickly, but then we had a really fun discussion about what the triangle would look like if we used more (smaller) sectors. So, the surprising triangle from Strogatz’s tweet led to a really fun and totally unexpected discussion! It is so fun to hear kids think through / wonder about math questions like the one they asked about the new triangles.

The last part of the project today was inspired by a tweet from our friend Alexander Bogomolny that was part of the thread Strogatz’s tweet started on Twitter yesterday:

I love it when Twitter writes our math projects for us 🙂

I had the kids look at the picture and describe what they saw. At the end I asked them why they thought the slanted lines in the triangle were lines and not curves – they had interesting thoughts about this little puzzle:

The amount of great math shared out twitter never ceases to amaze me. Thanks (as always!) to Steven Strogatz and to Alexander Bogomolny for inspiring this project about circles. Can’t wait to try out this project with other kids.

# 3 proofs that the square root of 2 is irrational

My younger son has been learning a little bit about square roots over the last couple of weeks and I thought it would be fun to show him some proofs that the square root of 2 is irrational. Because this conversation was going to explore some ideas in math that are both important and pretty neat, I asked my older son to join it.

I wasn’t super happy with how this little project went – it felt a bit rushed while we were going through it. Hopefully a few of the ideas stuck.

We started by talking about the square root of 2 and what basic properties the boys already knew about it:

After that short introduction we moved on to the first proof that the square root of 2 is irrational – I think this is probably the most well-known proof. The proof is by contradiction and starts by assuming that $\sqrt{2}$ = A / B where A and B are integers with no common factors.

The next proof is a geometric proof that I learned a few years ago from Alexander Bogomolny’s wonderful site Cut The Knot. It is proof 8”’ here:

Proof 8”’ that the square root of 2 is irrational on Cut the Knot’s site

If you like this proof, we have also explored some geometric infinite descent proofs in a slightly different setting previously inspired by a really neat post from Jim Propp:

An infinite descent problem with pentagons

Finally, we looked at a proof that uses continued fractions. It has been a while since I talked about continued fractions with the boys, and will probably actually revisit the topic soon. It is one of my favorite topics and always reminds me of how lucky I was to have Mr. Waterman for my math teacher in high school. He loved exploring fun and non-standard topics like continued fractions.

So, although I don’t go deeply into all of the continued fraction ideas here – hopefully there’s enough here to show you that the continued fraction for the $\sqrt{2}$ goes on forever.

So, although this one didn’t go quite as well as I was hoping, I still loved showing the boys these ideas. We’ll explore them more deeply as we study some basic ideas in proof over the next year.