We used one of the problems for our project this morning:

We stared the project today by having the boys read the problem and talking about some potential strategies for how to approach it:

They started by looking at some simple examples. These examples helped them get a feel for how the problem worked, but didn’t quite point them toward the solution.

From the last discussion they began to believe that they game would always end the same way no matter how you played it. So they decided to see what would happen if they ran through the version of the game presented in the problem.

Their way of running through the game was fascinating to me

Now we talked about why there was only one way to end. They found their way to the idea that the even and odd numbers of white beans was the key to the problem.

Finally I showed them one way of thinking about how the odds and even number of white and black beans changes each turn.

I really like this problem. It is a super fun problem for kids to explore.

It looked like a fun activity to try, so we spent 20 minutes this morning going through it. It was nice to year what my son son thinking about fractions throughout the activity. The first 4 videos below show his work and the last is some quick thoughts from him on the morning:

One of my older son’s homework problems asked him to find 3 digit multiples of 7 whose digit sums were also multiples of 7. I was puzzled by this problem had it on my mind most of the day today.

I hoped that talking through it would help me understand what the math idea was behind the problem. Sadly no, but we still had a good talk.

Here’s the problem and the work my son did:

So – still quite puzzled about the problem – I decided to see if there was anything quirky that came up looking at a divisibility rule for 7 with 3 digit numbers. This gave us a nice opportunity to talk about modular arithmetic:

Finally, since I wasn’t making any progress seeing the point of the original problem, I had him talk about other divisibility rules that he knew:

So, a nice conversation, but I’m actually baffled. I’ll have to ask the author of the problem what he was trying to get at – I feel like I’m missing the point.

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

(i) , and

(ii)

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.

My older son had a terrific homework problem in his enrichment math class. I wanted to walk through the problem again today so that my younger son could see it and also to highlight some of the lessons in the problem.

To start the project we revisted a fun geometry problem that will make a surprise appearance at the end of the homework problem:

Next I introduced the homework problem. My older son is familiar with this problem, but my younger son is seeing it for the first time. In this video my older son highlights the main ideas that we need to solve the problem (well . . . see the next video for the one we forgot!):

Here’s the one extra piece that we missed from the last video:

Next with the triangles labelled properly, we worked to see how we can use the Pythagorean theorem to help solve for the values of the two unknowns. I used this section of today’s project to give my younger son a little algebra practice:

Now comes the task of simplifying the two complicated equations. Hopefully that will help us make some progress towards solving them.

After the simplifying in the last video we are now ready to take a crack at solving for the radius of the smaller circles. Solving the equation involves solving a quadratic and that gave us a chance to talk about factoring.

Finally, we went back to the picture from the homework problem. We hadn’t solved for x in the project, but now we can use the pictures to help us find x’s value. We see an 8-15-17 triangle and also a 3-4-5 triangle. We also see the 5-5-8 triangle from the beginning of the project!

So, a fun project connecting a neat geometry problem that we’ve studied before with a new homework problem.

Today I had the kids read the post. Here are some of the ideas that they thought were interesting:

With the camera off I started talking about some of the ideas from math that I knew that weren’t “useful” originally but became useful later. So, I turned the camera on and talked about it live. I would not be surprised if several of the things I said in this part are not historically accurate – this wasn’t prepared and was off the top of my head.

Next I asked each of the boys to design their own Engel machine. My older son went first and we counted to 10 using his machine.

My younger son went next. He had a different design and we used it to count to 10, too:

I’ve really enjoyed sharing Propp’s essay with the boys. It is a great way for kids to explore different bases and also a great introduction to some fun advanced ideas from math. I also love that kids can play with the “machines” using blocks – that seems to really keep them engaged.

Originally I wanted to have the kids read the essay and give some of their thoughts for part 2, but I changed my mind on the approach this morning. Instead I asked each of them to answer the question in the title of Propp’s essay -> How do you write 100 in base 3/2?

Propp points out in his essay that his approach to base 3/2 via chip firing / Engel machines / exploding dots is not what mathematicians would normally consider to be base 3/2. The boys are not aware of that statement, though, since they have not read the essay yet.

Here’s how my younger son approached writing 100 in base 3/2. The first video is an introduction to the problem and, from knowing how to write numbers like 100 (in base 10) in other integer bases.

I think the first 3 minutes of this video are interesting because you get to hear his ideas about why this approach seems like a good idea. The remainder of this video plus the next two videos are a long march down the road to discovering why this approach doesn’t work in the version of base 3/2 we are studying:

So, after finding that the path we were walking down led to a dead end, we started over. This time my son decided to try to write 100 as 10×10. This approach does work!

Next I introduced the problem to my older son. He also started by trying to solve the problem the same way that you would for integer bases, though his technique was slightly different. He realized fairly quickly (by the end of the video, I mean) that this approach didn’t work:

My older son needed to find a new approach, and he ended up finding an idea different from my younger son’s idea to find 100 in base 3/2. His idea was to use chip firing:

I thought that today’s project would be a quick reminder of how base 3/2 works (at least the version we are studying). That thought was way off base and was completely influenced by me knowing the answer! Instead we found – by accident – a great example of how to explore a challenging problem in math. Sometimes the first few things you try don’t work, and you have to keep trying new things.

I’m hoping to have time to spend at least 3 days playing around with Propp’s latest blog post. Today we had 20 min free unexpectedly in the morning and I used that time to introduce two of the ideas. They haven’t read the post, yet, but instead I started by having them watch Propp’s short video about the binary Engel machine:

After watching that video I had the boys recreate the idea with snap cubes on our white board. Here’s that work plus a few of their thoughts on the connection with binary:

Next I challenged the boys to draw the base 3/2 version of the machine. After they did that we counted to 10 in base 3/2 and talked about what we saw:

I was happy that the boys were able to understand the idea behind the base 3/2 Engel machine. With the work from today giving them a nice introduction to some of the ideas in Propp’s essay, I think they are ready to try reading the essay tomorrow. It’ll be interesting to see what ideas catch their eye. Hopefully we can do another short project on whatever those ideas are tomorrow morning.

The video is amazing (as usual) and I wanted to be able to share it with the boys. This one is a bit hard than usual – the topic is pretty advanced to begin with and is also pretty far outside of my own knowledge – but we gave it a shot.

Here’s what the boys thought after seeing the video:

Next we reviewed how Houston-Edwards divided the numbers from 0 to 1 into buckets. The boys didn’t quite have the details right, but that actually made talking through the idea pretty easy – I learned from their explanation what points needed to be re-emphasized.

Now we talked through the really challenging part of the video -> creating the set with no size. Given the challenge of explaining this idea to kids, I’m pretty happy with how the conversation went here. Also, I only finally understood the argument myself while I was explaining it to them!

Now we backed away from the complexity of the Axiom of Choice and reviewed two other slightly easier ideas that came up in our discussion. Here we discuss why is irrational:

Finally, we wrapped up by discussing why the rational numbers are countable:

Although kids will have a hard time understanding all of the ideas that Kelsey Houston-Edwards brings up in her Axiom of Choice video, I think it is fun to see which ideas grab their attention. The idea that you can have a set that doesn’t have a size is pretty amazing. I was pretty happy with how things went today – exploring the ultra complex idea first and then backing off to discuss slightly easier ideas involving infinity. Definitely a fun set of ideas to plant in the minds of younger kids 🙂