# Dan Meyer’s Geometry Problem

NOTE – 7:50 am on Tuesday February 25.  I’ll be editing this quite a bit over the next day or so, but just wanted to get the videos up before running off to work.  Sorry for the likely high amount of typos in this first draft . . . .

NOTE – 7:50 pm on Tuesday – a little more explanation behind each of the videos.

I woke up this morning to see a pretty interesting geometry problem for kids posted on Dan Meyer’s blog.  The article is here:

http://blog.mrmeyer.com/2014/confab-circle-square/

and the problem is this:

“Given an arbitrary point P on a line segment AB, let AP form the perimeter of a square and PB form the circumference of a circle. Find P such that the area of the square and circle are equal.”

I start with my kids every day at 6:15 am so that I can try to get two hours in with them before heading off to work.  I liked this problem when I saw it and asked my older son if he’d be interested in working through it.  He was, so off we went.

(1) The first part was just talking through the problem to see what his initial reactions would be.  I was happy that he was able to understand the setup for the problem right away.  In the post about the problem Dan Meyer was particularly concerned about students being able to  “mak[e] sense of precise mathematical language.”   I wasn’t sure about the wording either and am interested to see if there are better suggestions posted on his blog, but we were lucky this time and the problem’s wording wasn’t a stumbling block.

What did prove to be an interesting stumbling block was dealing with a $\pi$ as the coefficient of a polynomial.  After the fact I realized that I should have expected this, but while we were talking about the problem I was caught a little off guard.  A short diversion switching out $\pi$ from the equation and replacing it with 25 seemed to help my son understand how to get past this problem.

(2) In the second 5 minutes we focus in on his moving from the geometric equations he’s written down to finding the point on the line segment.  We build on the formulas that we found in the last video and ended up finding a formula for the length of the line in terms of the radius of the circle.  I was really happy that he was able to get to this stage of the problem without much help from me, since the rest of the solution is just calculating.

(3) Because all of the $\pi's$ were giving him problems, I went and got a calculator so that we could see the actual numbers.  I wish I hadn’t done that without asking him to approximate the values, but these are the decisions you make at 6:00 am . . . .  Looking back on this about 12 hours later, I have no idea why I decided to go get a calculator.  Ugh.  Oh well, we recovered a little during this segment anyway, and I thought finding an approximation for $\sqrt{\pi}$ was actually a fun little diversion even if I’d already sort of given the answer away.   At least get to do a little estimating . . . .

The main point of this section was to try to remove some of the confusion that was coming from all of the $\pi's$.  As I mentioned above, during the talk I was taken a little off guard by this problem and it seemed like the best way around this problem was to switch over to decimal numbers.  Unfortunately, though, making this switch only helped a little, and it wasn’t until we did an example with integers that we got around the problem.   One thing I learned from working through this problem was that I need to incorporate a few more problems where some of the numbers aren’t integers.

(4) Having finished up with his solution, I wanted to show him a different way to solve the problem (I especially wanted to do this because we just finished a section on quadratic equations).   This was actually the approach I was expecting him to take, so I’m actually pretty happy that he went in a different direction at the start.  It is always nice to be able to see multiple solutions to the same problem, even if I’m mostly leading the charge on the second solution.  The two things I was trying to accomplish in this part were (i)  working through a little arithmetic with $\pi$ because of the confusion in the first part, and (ii) introducing the idea that there is a second solution that comes in from solving the quadratic equation:

(6) Now for some extra fun – let’s do the same problem in 2 dimensions rather than in just one!!  This is the part I was really looking forward to when I read Dan’s blog this morning.  I actually considered doing this part as a separate excercise, but we were on a roll so we kept going.  It would have been fun to try to spend a little time thinking through what shape we’d expect to see before diving into the calculation, but now we only had about 5 minutes left at this point, so we just jumped right into the calculation.

(7) Finally, since the 2D equation looks like it is a little rough, let’s just end the morning on Wolfram Alpha to see the nice surprise:

All in all, I really liked this problem.  Love finding nice math surprises on the internet!