# An instructive counting / probability question from Alexander Bogomolny

I saw this question from Alexander Bogomolny today through a tweet from Nassim Taleb:

It reminded me a bit of a fun question I saw from Christopher Long a few years ago:

A great introductory probability and stats problem I saw from Christopher Long

The first solution that came to mind for the dice question involved generating functions. Here’s the code I wrote in Mathematica:

The general idea was to look at powers of the polynomial $x + x^2 + x^3 + x^4 + x^5 + x^6$ and keep track of the coefficients for powers greater than $x^{12}$. The one tiny bit of difficulty is that you also have to strip off the powers of $x$ greater than $x^{12}$ after each stage (since you only want to count the first roll giving you a sum greater than 12). Here’s that polynomial cubed, for example:

The coefficients for $x^{13}$ through $x^{18}$ tell us the number of different ways to get a 13 through an 18 with three rolls.

The results of counting the polynomial coefficients are given below (columns give the number of ways to roll 13 through 18, and the rows are the dice rolls 3 through 13):

These counts don’t have the right weights, though, since 21 ways of getting a 13 on the 3rd roll have a much higher chance of happening than the 1 way of getting an 18 on the 13th roll. In fact each row has a weight 6x greater than the row immediately below it. Weighting the rows properly we get the following counts:

Now we just have to add up the columns and divide by $6^{13}$ to get the probabilities of ending on 13 through 18 (and do a weighted sum to get the expected final number) -> the probability of ending on a 13 is about 27.9% and the expected value of the final number is roughly 14.69:

This is a nice problem to show how generating functions can help you find exact answers to problems that seem to require simulations. It was fun to think through this one.

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# Playing with 3d printed versions of shapes theorized by Hermann Schwarz

Saw a neat tweet earlier today about 3d printing, math, and engineering:

I recognized some of the shapes in the article as ones that we’d played with before:

The grey shape displayed in the article is a “made thicker for 3d printing” version of the surface $\cos(x) + \cos(y) + \cos(z) = 0.$ I thought it would be fun to print that shape today and use it for a little project with the kids tonight. Here’s the Mathematica code and what the print looks like in the Preform software:

8 hours later the print finished and I asked the boys to describe that shape plus the gyroid. It is always fascinating to hear what kids see in unusual shapes. My younger son went first:

Here’s what my older son had to say (and he’s starting to study trig, so we could go a tiny bit deeper into the math behind the shape I printed today):

Next we watched the video about the shapes made by Rice University:

After watching the video I asked the boys to talk about some of the things they learned:

Of course, mostly they didn’t want to talk about the shapes – they wanted to stand on them! So much for an 8 hour print and 45 min of trying to clean out the supports . . .

Here’s how the standing went:

Definitely a fun project and a fun way to show kids a current application of both theoretical math and 3d printing!

# A fun math surprise with a 72 degree angle.

We’ve been talking a lot about 72 degree angles recently. Yesterday’s project was about a question our friend Paula Beardell Krieg asked:

Paula Beardell Krieg’s 72 degree question

In that project we learned that a right triangle with angles 72 and 18 (pictured below)

Is nearly the same as a right triangle with sides of 1, 3, and $\sqrt{10}$

Today I wanted to show the boys a neat surprise that I stumbled on almost by accident. The continued fraction expansion for the cosine of the two large (~72 degree angles) are remarkable similar and lead to the “discovery” of a 3rd nearly identical triangle.

We got started by reviewing a bit about 72 degree angles:

Now we did a quick review of continued fractions and the “split, flip, and rat” method that my high school teacher, Mr. Waterman, taught me. Then we looked at the continued fraction for $1 / \sqrt{10}$:

Now we looked at the reverse process -> given a continued fraction, how do we figure out what number it represents? Solving this problem for the infinite continued fraction we have here is a challenging problem for kids. One nice thing here was that my kids knew that they could do it if the continued fraction had finite length – that made it easier to show them how to deal with the infinitely long part.

Finally, we went to the computer to see the fun surprise:

Here’s that 3rd triangle:

I love the surprise that the continued fractions for the cosine of the (roughly) 72 degree angles that we were looking at are so similar. It is always really fun to be able to share neat math connections like this with kids.

# Paula Beardell Krieg’s 72 degree question

A few weeks ago I got this question from Paula Beardell Krieg on Twitter:

Today I went through this problem with the boys – the difficulty of this exercise surprised me a bit. They really struggled to see how you could tell if an angle was 72 degrees.

Here’s the introduction. The boys noticed a few things about the picture and got some ideas with how to proceed:

Next we drew the two squares on a piece of paper and I let the boys explore the question. Here they struggled to make much progress beyond the things that they noticed in the first part of the project:

The thing giving them trouble was that they didn’t know any relationships between angles in a right triangle with a 72 degree angle. That left them completely stuck. Eventually they decided to measure the squares and found that they had something that looked like a triangle with side $1, 3,$ and $\sqrt{10}$.

Next we explored some of the ideas around $1, 3, \sqrt{10}$ triangles. After a little nudging from me they decided to measure the angles with a protractor.

Now I showed them my solution and let them see where the $1, 3, \sqrt{10}$ triangle comes up:

Finally, I let them play with two sets of triangles that I printed overnight. Two of these triangles are right triangles with 72 and 18 degree angles, and two of them are $1, 3, \sqrt{10}$ triangles. The question is -> are all 4 triangles the same?

Here are pictures (to scale) of the two triangles. You can see how similar they are.

First, the right triangle with a 72 degree angle:

Second, the $1, 3, \sqrt{10}$ triangle:

Tomorrow we’ll explore a second similarity between these two triangles. I found it playing around while I was making the triangles yesterday 🙂

# Playing with Cos(72) -> part 2

Last week we used an old AMC problem to explore Cos(72):

That project is here:

Finding cos(72)

Today we built a decagon with our Zometool set to see if we could approach the problem a different way:

I started by having the kids explore the decagon and having my older son explain where cos(72) and cos(36) were (roughly) on the shape:

Next we used a T-square to try to get good approximations for both Cos(72) and Cos(36). The T-square + Zometool combination was a little harder for the kids than I was expecting, but we got there.

Finally, I wrapped up with a challenge question for my older son. If we know that Cos(36) – Cos(72) = 1/2, find the value of Cos(36). He did a nice job working through this problem:

I’ve enjoyed playing around with properties of angles that arent usually part of the trig curriculum. We might have one more project on 72 degrees this weekend – I’m thinking of playing with the idea that Tan(72) is close to 3, but haven’t quite figured out that project yet.

# The “impossible” New Zealand geometry test

Saw this tweet yesterday:

Here’s a direct link to the Guardian article:

The Guardian’s article “Impossible New Zealand Maths Exam Even Flummoxes Teachers”

Glancing through the problems, it looked like a challenging but fair test other than the mysterious appearance of a trig problem. Maybe trig is introduced as part of the geometry curriculum in New Zealand, though.

This morning I had my older son (in 8th grade) go through the test. The trig problem and one of the angle problems gave him a bit of trouble. The rest he was able to solve and explain.

Here is his work (in the order that the problems appear on the test – my numbering doesn’t correspond to the numbering on the exam). Also, sorry about the camera not being in perfect focus. I set the focus before zooming in – whoops!

(1) Parallel lines and angles

(2) A problem about right triangles

(3) A problem about arcs in circles

(4) The mysterious trig problem + a problem about similar triangles and scaling

(5) A problem about similar triangles

(6) A problem about parallel lines

(7) A curious problem that seems to be an exact repeat of problem (3)

(8) A problem about arcs and angles in a circle

(9) A problem about arcs and angles that gave him some trouble

(10) An angle problem that he inadvertently solved struggling with the last problem

(11) A problem about angles in a kite

# Finding Cos(72)

My older son is learning trig out of Art of Problem Solving’s Precalculus book this year. Yesterday he was working on the “sum to product” section, which derives rules for expressions like Cos(x) + Cos(y). It reminded me of one of my all time favorite math contest problems:

Today I thought I would show him my solution to that problem. What we go through probably isn’t the best or easiest solution, but I think it is an instructive solution for someone learning trig.

We started by talking about the problem and how some of the ideas he was currently learning could help solve it:

At the end of the last video we’d found a nice equation that we derived from the original problem:

$\cos(36^o) - \cos(72^o) = 2 \cos(36^o) * \cos(72^o)$

Now we used the double angle formula to simplify even more and find a cubic equation satisfied by Cos(36):

Now we tried to find the solutions to the cubic equation we found in the last video. This part gave my son a bit of trouble, but he eventually got there.

Now we were almost home! We just had to compute the value of Cos(72) and we’d be able to solve the problem. That involved one last application of the double angle formula:

I think solving this problem from scratch would be far too difficult for just about any kid just learning trig. But, the fun thing about this problem is that the ideas needed to solve the problem are all within reach using elementary trig identities. So, I think that working through the solution to this problem is a nice exercise for kids.