Friday, September 25, 2015


I just received the roster list. So some quick demographics for the beginning of the year.
It looks like as expected the kids skew towards 4th grade this time. Unfortunately I did not maintain gender parity. I may come back to this next quarter and see if I can recruit some more girls with the teachers' help.

13 4th graders
 2  5th graders

11 Boys
4 girls

 And unlike last year I actually know 7 of the kids before starting.  That includes my own son who will be joining me this year now that he's in 4th grade.

Update: There are now 2 girls on my waiting list. I've put out a call for another volunteer to help.

Thursday, September 24, 2015

Procedural Updates

Starting Up

The roster filled up after a week and I have a small waiting list. Unfortunately I don't have the actual names yet so I can't start emailing the parents or looking for volunteers. If a few more students do show up I am going to try to find another volunteer so we can serve more kids.

In the meantime I have registered for MOEMS again, am about to start the AMC8 process and have had an independent email from the Washington Student Math Assoc. about having someone come out and speak to the math club later this fall.

Here's my intro email draft which I'm trying to decide if I can make any more fun:

Thanks for signing up for the Math Club.  I'm excited to start up a new year. In this mail I'm going to go over the basic procedures.

1. Please fill out my own membership form which I've enclosed in this mail and return it to me.

2. We meet after school finishes in the cafeteria.  New this year: there is no food allowed in the classroom. So please bring your own snacks and be prepared to eat them in the 10 minutes before we walk up to the room.  

3. Besides any snack, also bring a notebook and pencils each week. Especially in the beginning I'd appreciate it if you can remind your kids about supplies.  Also, I've found electronics to be very distracting for everyone. So I recommend not bringing any ipads/phones etc. on Math Club days or talking about putting them away for the afternoon beforehand with your kids. 

4. Parent pickup will be at the classroom: N204 at 4:50.  There will be a notebook that you should sign out from. If you haven't arrived by 5:00 I will bring your children down to the KidCo room by the Cafeteria. Be aware: Vanessa has informed me that there will be less leniency on charges this year if that happens.

5. The first day is Tue. 10/6. I've also enclosed the year calendar.

* I really like having other adults helping out in the room and more adults let me do some different activities. Please email me back if you can come  one of the afternoons. No special math experience is necessary. 

* This year I'm thinking about commissioning a T-Shirt for the club. If you have any graphic design talent and want to help out please also send me a mail.

Fun Stuff: I've re-enclosed my original flyer. For those who are chomping at the bit to get started please take a look at the problems. Then if you have any questions or have a solution send me an email back.

See you soon

I've also been updating the resource page: Resources a bit recently. At some I'm going to reindex by category as well.

Recent Interesting Problems:

I love medians and this problem has a lot of them. This would work great scaffolded with work noticing properties of medians i.e. they divide the area of the triangle in half, and the cool rotations you can do with them by splitting to form another related triangle.   Its a bit of a cheat but if you notice the quadrilateral is not strongly specified in the problem. As that suggests this is a general property. So you can start by considering the case where the quadrilateral is a square and find an easier solution for this subcase which will hold for more complicated versions. (In fact the general proof which involves triangles and medians is a bit less messy to draw in the square version as well.)

This problem showed up again.  Mikesmathpage has a fun treatment of this:  
Basically if you  model with some sort of dowel you can find a right triangle and then easily apply the pythagorean theorem.  I really like the idea of the physical modelling and it would fit really well after proving the theorem if we do that again this year.

Tuesday, September 15, 2015

Arthur Benjamin's talk: Exploring the Hidden Magic of Math

Last night I attended Arthur Benjamin's lecture circuit promoting his new book: The Magic of Math: Solving for x and figuring out why.

Walking up to town hall:

Oh wait - math is stuck in the basement again: (The noble prize for literature is totally overrated)

Once in the space it was great to see a full room out on a Monday night for a math talk. This is one of the reasons I love living in Seattle.

It turns out the talk was sponsored by my arch nemesis: Xeno  They still are totally whomping me on google showing up at least 2 pages earlier if you search for "math club blog" despite not even being a blog.  All kidding aside they're a great organization which I haven't had much of chance to interact with in person yet.

The talk itself was a lot of fun. I brought my son and the material was very accessible. There was a bit of warm up mental math tricks for doing squares, two fairly length sections going into Triangle numbers (see my take from last year: and  Fibonacci sequences. None of the subjects were particularly new to me but Arthur Benjamin was very entertaining and some of the development paths look like fertile ground..

For instance, I really liked his presentation of how to develop squaring tricks.  One of my favorite quotes: "And who doesn't love squaring a sequence of numbers?" So I plan to  read the book and do a full review some time in the future at which time I expect I will find some ideas to leverage for the math club on a close reading.


The teacher strike looks like it is coming to close and kids have started signing up for the club. Hopefully I'll have a roster in a week or so.

Friday, September 11, 2015

Using Symmetry: More complex proof walk through

This is a continuation of the last 2 posts exploring using symmetry in proofs. This example is the most complicated yet. Symmetry via reflection produced the base of the proof but it then required recognizing an in-center to complete the process.

I started by looking at problem 45 on gogeometry.

My first instinct here was to increase the overall symmetry in the drawing by reflecting everything.
Once that's done through a variety of different angle chases you repeatedly arrive at 2b + 2a + x = 90.
The easiest place to do this is in corner BAC once the perpendicular bisector is drawn.

I could see the parallel lines in my reflected drawing which gave me a few more x angles but not another triangle equation I could use to solve the problem.

I next tried a few unsuccessful techniques. I back-cracked from the answer 30 degrees and filled in the other angles to see if any other common triangles were there that I had missed on visual  inspection.

I then removed all the x's and replaced them with  90 - 2b - 2a in the total picture to see if this helped.
I looked around to see if there was any way to use the 30-60-90 properties of side length to deduce the angle measure but that looked unlikely given the constraints.

Then I took a break for a day after hitting a wall.  And suddenly on restarting during lunch I noticed that I had the  in-center  of the triangle when redrawing the picture of one side. That meant the third side was also an angle bisector and everything fell into place. [The need for a break before finding a solution on a stumper is pretty normal for me] 

Thursday, September 10, 2015

Using Symmetry: Geometry Walk Though #2

This is another in my series exploring geometric problem solving through rigid transformations (and creating diagrams with geogebra). See:

This problem starts with a very simple picture but I want to show my first attempt down a non-productive line of thought.

Given B is the median of AC and angle EAC is 30 and angle ECA is 1 5 find angle BEC.

Two thoughts came to me immediately. Angle EAC is almost part of an equilateral triangle if we reflect the whole triangle. Secondly ECA is half of EAC and it could be divided into two halves so we end up with many 15 degree angles. I pursued the second thought first because it created some attractive symmetry.

With some angle tracing EGF falls out as 30 degrees as well. So it looked like I just had to prove triangle EFG was isosceles (or find angle EFG) and here I became stuck. I started to break the triangle down further to tease out 30-60-90 triangles and side relationships. This didn't quite arrive at a solution. I also tried some more reflections and found a few 45-45-90 sub triangles. But that again didn't quite get me where I wanted to be.

So I went back to my first thought. Reflect the whole triangle AEC and create a giant equilateral triangle out of it.

I connected the two points C and C' and drew the full angle bisector in to where it intersected at D'.
Then I started angle tracing at which point I realized I didn't need the full equilateral triangle the bottom 30-60-90 was interesting by itself.
First I started finding congruent sides. CD was half of AC because of the 30-60-90 triangle. I had forgotten that B was the median by this point and actually started to draw one on for a minute and thought I would have to prove B was a median until I rechecked and realized it was given. I then connected BD to form the interior equilateral triangle since that seemed useful both in terms of symmetry and subdividing angles. For instance ADB is 30 degrees by angle chasing meaning  triangle ABD is isoscleses. By further angle chasing it also turns out that DCE must be 45 degrees (60 - 15) and then the final angle of the right triangle CDE must also be 45 and it is also isosceles. This produced the fifth congruent line DE. At this point I could tell I was very close. I now had proved a new isosceles triangle EBD which gave up EBD was 75 degrees.  This was the last unknown angles besides x itself in the triangle BCE!  x + 15 + 60 + 75 = 180. So x = 30.

This one fell out after trying 2 ideas. With some other problems I go down many more avenues. My general guiding principle is to try out reflections and angle division additions first around the figure always with an aim at more symmetry or new congruences.

Wednesday, September 9, 2015

Using Symmetry: Ratios in square problem

I've been trying to put together a tutorial for working on geometry problems over the summer based on the various problems I've tried out. The outline is only partly finished:

but over a few posts I want to explore one of the most interesting techniques: focusing on symmetry through rigid transformations. I find its very useful on many more complex problems which require auxiliary lines.  Thinking about the problem in terms of transformations makes it easier to find structure/solutions.  With all these examples I try to document my process in an effort to show how you often can try different approaches and experiment.

First Example

This was a problem posed by five_triangles: I think its good introductory example of the power of rigid transformations (reflection).

Square ABCD, diagonals meet at E, ∠BAC bisected, meeting BD&BC at X&Y. Prove 2XE=YC. For what school year?

To answer the question I gauge this at the advanced high school level due to its need for auxiliary lines. Here's the quick proof I sketched out:

My thinking on this problem was to increase the symmetry through reflection. That had the immediate benefit of creating a new triangle with one side of the required length (2XE).  I then started to look at the implications of the 2 similar triangles AXF and AYG. It was also fairly obvious that if I could find a number for the ratio that the base could be translated into the required side since they were both on a 45-45-90 triangle. I made an early error and assumed that triangle had a side length of exactly 1/2 the square's side. That quickly led to an erroneous answer. After I rechecked my steps realized I had made a bad assumption. From there some quick angle chasing confirmed my visual impression that all 4 triangle are congruent above.  Because I was aiming to use the similar triangles, to find a constant ratio and then transform that to refer to YC it was straightforward to do the computations in step 6.

Second Day: