## 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

## The cube root of 1

After a week of doing a little bit of practice for the AMC 8, my older son has returned to Art of Problem Solving’s Precalculus book. The chapter he’s on know is about trig identities.

Unrelated to his work in that book, the cube root of 1 came up tonight and he said “that’s just 1, right?” So, we chatted . . .

First we talked about the equation $x^3 - 1 = 0$:

For the second part of the talk, we discussed the numbers $\frac{1}{2} \pm \frac{\sqrt{3}}{2}$ and their relation to the equation $e^{i\theta} = \cos(\theta) + i\sin(\theta)$

Finally, I connected the discussion with the double angle (and then the triple angle) formulas that he was learning today. You can use the same idea in this video with $\cos(5\theta)$ to find that $\cos(75^{o}) = (\sqrt{5} - 1)/4$:

So, a lucky comment from my son led to a fun discussion about some ideas from trig that he happened to be studying today 🙂

## Exploring Newton’s method with kids

Yesterday we had about a 30 min drive and I had the boys open up to a random page in this book for a few short discussions in the car:

There were some fun topics that were accessible for kids, but then Newton’s method came up. Ha ha – not really drive time talk 🙂

It did seem like it could be a fun project, though, so I took a crack at it today. The goal was not computation, but mainly just the geometric ideas. Here’s how we got started:

Next I asked the boys if they could find situations in which Newton’s method wouldn’t work as nicely as it did in the first video. They were able to identify a few potential problems:

Now I had both kids draw their own picture to play out what would happen when you used Newton’s method to find roots. I think there’s a lot of ways to used the exercise here to help older kids understand ideas about tangent lines and function generally. I mostly let the kids play around here, though, and the results were actually pretty fun:

Finally, we went to Mathematica to see some situations in which Newton’s method produces some amazing pictures. Here we switch from real-valued functions to complex valued functions. Since I wasn’t going into the details of now Newton’s method works, rather than using some easier to understand code, I just borrowed some existing code from here:

The page from A. Peter Young at U.C. Santa Cruz that gave me the Newton’s method code for Mathematica

The boys were amazed by the pictures. For example, (and this is one we looked at with the camera off) here’s a picture showing which root Newton’s method converges to depending on where you start for the function $f(z) = z^3 - 2z + z - 1$:

Definitely a fun project. Even if the computational details are a bit out of reach, it is fun to share ideas like this with kids every now and then.

## Using 3d printing to help explore a few ideas from introductory algebra

Last spring I was playing around with some different 3d printing ideas and found a fun way to explore a common algebra mistake:

Does (x + y)^2 = x^2 + y^2

comparing x^2 + y^2 and (x + y)^2 with 3d printing

Today I decided to revisit that project. We started by looking at the same idea from algebra:

Does $x^2 + y^2 = (x + y)^2$ ?

At first we talked about the two equations using ideas from algebra and arithmetic.

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Now I asked the boys for their geometric intuition and then showed them the 3d printed graphs of the two functions.

This part ran a little long while my younger son was stuck on a small but important point about the graph $z = (x + y)^2$ – I didn’t want to tell him the answer and it took a couple of minutes for him to work through the idea in his mind.

/

Next I showed them 3d prints of $x^3 + y^3$ and $(x + y)^3$ and asked them to tell me which one was which. It is really neat to hear the reasoning that kids use to go from shapes to equations.

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For the last part of the project I asked the boys to come up with their own algebra “mistakes” for us to explore. My older son chose to compare the graphs of $\sqrt{x^2 + y^2}$ and $x + y$.

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My younger son chose the two equations $x^2 - y^2$ and $(x - y)^2$. Changing the + to a – in our first set of equations turns out to have some pretty interesting geometric consequences – “it looks sort of like a saddle” was a fun comment.

One especially interesting idea here was exploring where $x^2 - y^2 = 0$. We used Mathematica’s ContourPlot[] function to explore those two lines because those lines weren’t immediately obvious on the saddle.

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I’m happy to have had the opportunity to revisit this old project. I think exploring simple algebraic expressions is a fun and sort of unexpected application of 3d printing.

## 3D Printing Paula Beardell Krieg’s dissected cube shapes

I’ve been thinking about exposing the boys to math through 3d printing lately. Today I decided to explore making Paula Beardell Krieg’s cube shapes with them. Here’s the exploration the boys did back in March when we first got them:

Even though we’ve played a bit with these shapes before I still thought that thinking through these yellow and pink shapes would be a fun challenge. The project turned out to be a tiny bit harder than I thought it would be, but it still was a nice conversation.

We started by first looking at the three pyramids that can come together to make a cube and continued by looking at what happens when you slice those shapes in half.

In the last video the boys were thinking about trying to describe these shapes by describing the lines that formed the edges. At the beginning of this video I told them that this particular approach was going to be tough since they didn’t know how to write equations of lines in 3 dimensions.

So, I had them continue to search for properties of the shapes that they could describe.

The boys were still struggling to find some ideas about the shape that went beyond the lines on the boundary, but we kept looking.

My older son hit on the idea that the shape was made from “stacking squares on top of each other.” We spent the rest of the video exploring that idea.

Now that we had the idea about stacking squares we went to Mathematica to try to create the shape. It took a few steps to move from the ideas about the squares to generating the code for the shape. We didn’t get all the way there during this video, but we did figure out how to make a cube.

Unfortunately I had to end the video since the camera was about to run out of memory.

While I was getting the videos off the camera the boys worked on how to change the cube shape to the pyramid shape. It was a good challenge for them and they got it. We talked about that shape for a bit and then moved on to the challenge of creating the “pink” and “yellow” shapes that Paula Beardell Krieg created from paper.

We had a little bit of extra time today and it was fun to walk through this challenging problem. I think creating shapes to 3d print is a really fun way to motivate math with kids. Can’t wait to use the printed shapes in a project tomorrow!

## Talking through 3 AMC 8 problems with my son

My son was working on the 1993 AMC 8 yesterday and had trouble with a few problems. Today we sat down and talked through those problem.

The first was problem 17:

Here’s what my son had to say:

Next was problem 19:

Next was problem 24:

I like these old contest problems. They lead to really nice conversations!

## A mistake that led to a great conversation

My older son had a homework problem that asked him to find the area of the region bounded by the two equations:

(i) $| 2x + 3y | \leq 6$, and

(ii) $| x - 2y | \leq 4$

Mathematica’s picture of that shape is here:

He told me that he used Pick’s theorem and found that the area of the shape was 13 square units.

There’s just one small problem – you can’t use Pick’s theorem to find the area of this shape since the corners of the shape are not lattice points of the grid.

What to do . . . .

I wrote a quick little program that picked 100 million random points in the 10×10 square centered at (0,0) and tested whether or not they were part of the shape. That program found that 13.71% of the points were part of the shape – that was enough to convince him that the area might be larger than 13 square units.

Next I had him re-read Pick’s theorem to see what went wrong. He saw pretty quickly that the shape didn’t meet the condition of having the corners lie on lattice points.

I really wanted to try to find a way to make Pick’s theorem work with this shape.  I had him determine the y-coordinate for the far right corner.  The value was y = -2/7.

After finding that value, we had a good talk about scaling.  To make the new grid larger we had to *divide* the x and y coordinates in the equations by 7.  Here’s Mathematica’s picture of the new shape and grid (note that the x and y values run from -25 to 25 in this picture):

With this shape we are able to use Pick’s theorem to calculate the area.   We counted 40 grid points on the boundary without too much difficulty.   Counting the ones in the middle was a little bit more of a pain, so we wrote a short program to perform that calculation for us.  Note that we have to change the “less than or equal to” from the original equations to “strictly less than” since we want to be inside the shape:

So, we have 653 lattice points in the inside and 40 on the boundary.  Pick’s theorem tells us that the area is equal to the number of lattice points on the interior plus half the number of lattice points in the boundary minus 1.  That’s 672 units.    In the picture above, 1 unit is equal to 1/49 of a unit in the original picture, so the original area is 96 / 7 or 13 5/7.   Close to what he found originally, but not equal!

Along the way we also talked about alternate ways to find the area – the easiest being dividing the shape into two triangles with a vertical line through the middle.

I’m really excited about the discussion that we had tonight.  Funny how many important ideas in math can come up from a problem about absolute value and inequalities.