## Playing With Exeter Math 2

July 29, 2017

I have been at TMC17 in Atlanta since Tuesday, and my morning session has been “Playing With Exeter Math”. In my group, we have been working on the Math 2 problems, which although I have worked through these before, I have enjoyed thinking about them more indepth.

I have been sitting next to James Cleveland, and he has a knack for asking interesting questions, which really make me think. For example, on problem 33, we have to show that a triangle is isosceles by setting the two distance formulas equal to each other. James realized that once you square both sides, you have two circle equations, and connected that to how we construct perpendicular bisectors by drawing two circles and marking their intersections. It became even more clear when I graphed the equality on Desmos, and Desmos produced it as a straight vertical line. Cool!

We also looked at problem 21, which asked us to divide the shape into two congruent shapes. We conjectured that any shape with a 180° rotational symmetry would have an infinite number of ways that it could be divided. In order to satisfy the conjecture to my satisfaction, I set it up in Sketchpad:

Yesterday, we looked at problem 45, in which the corner of a paper is folded down to the bottom of the paper. The original problem asks us to find the area of the triangle created. It then asks us to find the value of x that would maximize the area of the triangle. Working it out algebraically, we found that for a side length of 10, x would be $\frac{10}{3}$. This led James to conjecture that it would always be $\frac{2}{3}$ of the length of the side. After a lot of hairpulling, I got the following set up in Desmos that shows this to my satisfaction (I used a slider for $t$, and the equations were $y=\frac{1}{2}\left(t-x\right)\left(\sqrt{2tx-t^2}\right)$ and $x=\frac{2}{3}t$ along with a parabola to fit the maxima as $t$ varied.):

Today, which was our last day, I had thought to finish up problem 45 (I added the vertical line at $x=\frac{2}{3}t$.) Once again, James was working on a problem, but this time, he got almost the entire group intrigued by it. Problem 133 sounds simple: Dissect a 1-by-3 rectangle into three pieces that can be reassembled into a square.

We quickly realized that since the area of the rectangle was 3, that the length of each side of the square would need to be $\sqrt{3}$. We immediately went to 30-60-90 right triangles, but we couldn’t figure out a way to make the cuts. Finally, about 20 minutes before we were supposed to leave, Annie Perkins figured it out using Sketchpad. This is my re-interpretation of her solution:

I’ll probably write up a full reflection on my experience at TMC, but I wanted to get my morning session experiences written down before I forgot about them.