Talking through the 4×4 case of Larry Guth’s “No Rectangles” problem with my younger son

A twitter conversation from last week got me thinking about advanced ideas in math that are accessible to kids AND do not require proficiency in arithmetic. Yesterday I went through one of my favorite projects that fits this idea -> Larry Guth’s “No Rectangles” problem. That project is here:

Revisiting Larry Guth’s “No Rectangles” problem

In the project we played with a 3×3 grid and tried to fill in as many of the squares as we could without forming a rectangle with 4 corners filled in. For the 3×3 grid the maximum number of squares you can fill in without forming a rectangle is 6. My son was able to find a good proof of that fact.

Today we looked at the 4×4 grid. The question here is a bit harder, but still accessible to kids (and turns out to come with some additional questions to explore that my son found interesting).

I started the project today by reviewing the problem and introducing the 4×4 case:

After guessing that the maximum number of boxes we could fill on the 4×4 grid without forming a rectangle was 8, we set out to see what arrangements we could find. My son’s first approach was to fill in all of the squares and then start creating holes.

This approach was initially a little difficult, but it did actually lead us to find an arrangement that had 9 squares filled in.

One of the things my son noticed about the arrangement of 9 filled in boxes was that one diagonal of the square was filled in. My son wondered if you needed the diagonal to be filled in, and I thought that would be a fun idea to explore. Working through this problem actually turned out to be more difficult for him than I was expecting, but he ended up having a really great idea at the end of this video that answered the question.

I think this video is a great example of a kid using mathematical reasoning (with no arithmetic!) to approach a pretty advanced problem about symmetry.

Finally, we wrapped up by trying to figure out if an arrangement with 10 boxes filled in could have no rectangles. The argument is similar to the 3×3 case but maybe is one step up in complexity – but importantly still accessible to a kid.