Wednesday, March 22, 2017

3/21 Graph Pebbling

This week I went back to a pure math circle format with my favorite activity from the recent Julia Robinson Festival: Graph Pebbling. Based on my experiences at the festival I thought it would occupy 30-40 minutes so I decided to do a warm up puzzle as well. Initially I had considering doing a battleship puzzle (see: but I found a tweet from Sarah Carter that looked interesting about slant puzzles:  These have a fairly simple set of rules: put a line through every cross and make sure to have the requested number of lines connecting to each square with a number. Unmarked square are free and can have any number of connections.

Simple is often good though. All the kids really liked them:

We then transitioned to graph pebbling: The full rules are here:   A series of graphs are included as well as 5 variations. For Math Club I used lima beans again as "knights"

My only issue was I have one table of boys that are harder to keep on task. I tried separating them a bit this time which didn't quite work but I may do it again next week but from the start. They're not disruptive per. se but they are distracting each other and only stay on task when I come over and work with them.


A fun factoring / number theory problem for this week:

Wednesday, March 15, 2017

3/14 Pi Day

Every 7 years or so accounting for leap years, Pi day actually occurs on a Tuesday. Yesterday was the first time that occurred while I've  been running the Math Club. Because most of the kids were here last year I did not go over my usual conceptual question "Why is the circumference of a circle in a constant ratio with its radius, and why such a funny value?"


I fall into the camp that its fun to celebrate as long as something mathematically meaningful occurs during the party. I also try to de-emphasize anything to do with memorizing digits. So due to all the apple pies being taken this year I picked up a strawberry rhubarb pie at the local super market which I served as everyone arrived in the cafeteria. This kept the mess to a containable minimum and as expected the kids were all very excited by the treat.

Like last year I decided to also do a pi day themed video after the following one showed up in one of my feeds:

After we were done I had another NASA packet to try out:
I tried this type material once before (space map session).  Since some of the kids liked it before, I thought 20-25 minutes would be about the right amount of time to try a similar activity again. I'm not completely keen on the formula plugging involved but in watching the kids, its actually useful every once in a while to use real, messy physical values and reason a bit how to apply basic geometry.

Overall everything went smoothly including setting up the video (cabling + wifi). The setup time did mean the kids fooled around for the 2 minutes before I could start but that just took a little extra talk to get the room's attention and settle in.

A not too hard but perhaps counter-intuitive circle property from

Wednesday, March 8, 2017

3/7 Olympiad #5

Today started with a small mix-up. A boy I recruited at the Julia Robinson Festival to join Math Club showed up. But the next quarter doesn't start for 3 weeks. I offered to let him join us anyway but I think he was too embarrassed. Hopefully, he'll still come on the real first day. The whole incident is a reminder that even though I assume I know most of the "mathy" kids in the grade, hidden depths are out there.

After that, the rest of the day went  more smoothly and had several small rewarding moments. We started by running down the  Problem of the week as a group.  I only had one student demonstrate how to divide the boards (its a stair like cut) and unfortunately this didn't generate as much problem solving discussion as I prefer.  From there, we completed the last MOEMs Olympiad for the year. Looking this one over, I thought it was among the trickiest of the series. We'll see how the scores go but several of the problems had fairly complex instructions to deduce the answers and I think the general trend will be a bit lower than the last one. I did have a good group problem solving session afterwards and had the kids show solutions for all the problems.  One small tweak I've implemented is to write the problems on all the whiteboards while the kids are working so we're set to go for the group discussion.  Kids were well focused through the entire time with the only extra chatting being about how to solve the problems differently. The one future topic I noticed among the problems was to work a bit on explaining how choosing unordered sets work i..e  \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)  This pairs well with a dive into Pascal's triangle. I'm going to take a look at Arthur Benjamin's book to see if he has an approach that is adaptable for a group.

For the light activity I had all the kids who finished early working on an Euler path exercise from "This is not a Maths Book"

The kids found this very interesting and it again could be a topic for a whole session.

Other Ideas from around the web I'm thinking about for future meetings:

Problem of the Week

Wednesday, March 1, 2017

2/28 Infinite Series

For this session of Math Club I wanted to revisit one of the ideas from the "free the clones" games: (See:

What is the sum of the infinite series  1 + 1/2 + 1/4 + 1/8 ...

On reflection, I decided this would make a nice connection with converting repeating decimals back to fractions. I had actually tried this 2 years ago and it went okay. Most kids can convert fractions to decimals but can only handle non repeating decimals in the other direction. But in the intervening time I had lost the worksheet I used back then.  This time, I wanted to risk it and just work on the whiteboard, have the kids go off and experiment and come back and discuss what they found.

Planned Questions

1. What is .999999... equal to and why?

2. How can we represent .99999.... as a series of fractions.

3. Warm up with some easier ones.

S = 1 + 1/2  +1/4 ....
S = 2/1

S = 1 + 1/3 + 1/9 ....
S = 3/2

S = 1 + 1/4 + 1/16
S = 4/3

4. Find the pattern and then come up with the general case:

S = 1 + 1/n + 1/n^2 .....
S = n/n-1

5. Ok let's go back to decimals

S = 1/10 + 1/100 + 1/1000 just like above. Can you use the same technique?

How about if the digits differ

S = 12/100 + 12/10000 + 12/100000

Final Conundrum

1 = 2 / 3 - 1 vs  2 = 2 / 3 -2 as continued fractions.


I actually started by having everyone talk about the Julia Robinson festival. A couple kids mentioned the final flatland talk and this was of sufficient interest that I ended up spontaneously repeating a huge section of it for those who weren't there.  My retelling was accurate except I didn't have any klein bottle pictures on hand other than one on my phone. This ended up taking at least 10 minutes and I would repeat and make a day of it based on how it well it was received.

Basically you have a town in a 2 dimensional world  and the inhabitants assume they live in an infinite plane but have never explored it. Then finally one tries it out and discovers if he goes north and leaves a trail he arrives back in the town from the south side etc. Given the behavior when the inhabitants go N, E and then NE you conjecture the existence of a sphere, torus and then klein bottle. 

As I result I ended up skipping my planned kenken warm up. We made it through about question 5 from above but by this time I had exhausted the focus of the group, it was getting harder to keep everyone on task. So I made the executive decision to pull out the kenken puzzles after all and "cool off". Fortunately, that pulled everything together again.   My take away from this is:

  • Kids were aware that .9999 =  1 but the explanation was a bit fuzzy (no numbers between 9 and one) but I didn't have enough time to circle back at the end and show why this must be the case.
  • This was still too much material, I need to break it up with something "lighter" if I try again. I think I want either a visual interlude (color in one of these infinite series?) or to gameify the middle somehow.


I went with this puzzle from The Guardian:

Julia Robinson Festival

(The flatland talk at the end of the afternoon.)

For the second year in a row, I volunteered at the Julia Robinson Math Festival over the weekend. This is among my favorite mathematical activities to do for the whole year. This time around  I went to the training session before hand. That was useful, since I had a chance to look at the problems I would be facilitating prior to actually jumping in.

My first one was a bit daunting from the perspective of maintaining interest. The first part was to figure out the brain teaser: What comes next in this sequence?

      1 1
      2 1
   1 1 12
   3 1 1 2
2 1 1 2 1 3
3 1 1 2 1 3

This took me almost 25 minutes to see by myself and I worked through a bunch of different ideas. My goal was document all my wrong approaches so I could anticipate what students my do. I also knew it involved some lateral thinking. As I remember my main thought was "Gosh I hope this isn't something silly like number of curves and lines in the numbers."

At any rate, I was pleasantly surprised during the actual Festival.  Based on the prep work I managed to keep multiple students occupied for 30+ minutes in the productively stuck state. The main thing I did was to have folks work together, keep close tabs on everyone and ask about what they were trying. I also tried to emphasize regrouping the pyramid as a triangle and looking for patterns.

My second table was a really cool graph theory game.  I'm going to use this in Math Club and I will talk about it more then.

Tuesday, February 21, 2017

My own Geometry Puzzle

(This is based on my previous explorations of the @solvemymaths problems. As far as I know its a new so I'm very happy with it. Usually I just collate problems.)

Monday, February 20, 2017

Mid-Winter break Geometry

By tradition, I'm going off on some problem solving walk-throughs:

- Courtesy of @solvemymaths

This problem is a good example of the power of working backwards.

To start off with like all of these type problems, I draw the center of the circles in and connect all the tangent points to find the inner structure and look for triangles.

One immediate simplification is to only find the ratio of BI to BK since its the same as the larger rectangle (1:2 scaling).  Secondly the inner right triangle EGJ is ripe for the Pythagorean theorem.

Before going any farther I noted some expressions:

  • BI = 2R + T 
  • BK = 2S + T
The required ratio to prove is \(BI= \sqrt{5}BK\) so squaring each side to get rid of the radical you get  \(BI^2= 5BK^2\) or \(4R^2 + 4RT +  T^2 = 5(4S^2 + 4ST + T^2) \)  This simplifies to \(R^2 + RT = 5S^2 + 5ST + T^2\)

For the rest of the exercise I kept this in mind as the target (although as you'll see I adjusted as I noticed more).

The second thing to immediately try was what fell out of the Pythagorean relationship in the triangle EGJ. Using \((R + S)^2 = (S+T)^2 + (R+T)^2\)   That simplifies to: \(RS = ST + RT + 2T^2\).  Which unfortunately doesn't look much like the target.  For one there is no R^2 or S^2 term and there is an extra RS and none of the coefficients are near yet.

I then munged around a bit and tried algebraically manipulating this expression to get it closer with no luck. So I looked back the drawing and noticed something I had missed initially  BK = 2S + T but it also is the radius of the large circle in other words 2S + T = R.  This immediately simplifies the target of   \(BI^2= 5BK^2\) to  \((2R + T)^2  = 5R^2\) or \(R^2 = 4RT + T^2\) which already looks closer to the Pythagorean expansion. But what's nice is you can also rewrite that as well with the segment  GJ = R - S rather than S + T.  

So I redid the Pythagorean relationship and found \((R + S)^2 = (R-S)^2 + (R+T)^2\) which simplifies to \(4RS = (R+T)^2\) Again this looks more regular than our starting point but still not exactly the same. Then since our target is only in terms of R and T we need to substitute out the S which we can do given 2S + T = R so 2S = R - T and applying that you now have \(2R(R-T) = (R+T)^2 \) or \(2R^2 -2RT = R^2 + 2RT + T^2\).    Combining like terms you get \(R^2 = 4RT  + T^2\) which is what we needed to show!

However what i actually did for the last step was the exact opposite of that explanation. Instead I took the target and put it into a form closer to what we had to see what was missing i.e. 
$$R^2 = 4RT + T^2$$
$$R^2 = (R + T)^2 + 2RT$$  (Completing the square)
$$R^2 - 2RT = (R+T)^2$$
It was this final form that reminded me to substitute back in for S since it was so close. And note how it was much easier to match the two expression after simplifying both of them rather than just going with the Pythagorean relation and trying to end at the initial goal.

5 Squares

Also from @solvemymaths.  Prove the area of the square is equal to the triangle.

This one was is closely related to and both rely on the  fact that the triangles formed between touching squares have equal areas.   See the previous link for the proof. 
The 4 key observations here are the

1) bottom two triangles around the square are congruent. This is the start of a Pythagorean Theorem proof in fact.  (See below if KH = a and JL =  b then each of the triangles is an a x b and FI = c where \(a^2 + b^2 = c^2\).

2) Each of the lower and middle triangles pairs have the same area because they are formed between squares. (i.e. CDF and FHI)

3) So all the lower and middle triangles have the same area (1/2 ab)!

4) You can create a new triangle with the same area as ABC that's easier to work out.

That's pretty nifty but I noticed something interesting when modelling a bit in Geogebra. If you let the 3 generator squares be a Pythagorean triple i.e. a = 3, b = 4, c = 5 all of the points in the model and all the areas are also integral.  That didn't look like a coincidence.   In fact I could roughly see the upper 2 squares had areas \(2(a^2 + c^2) - b^2\) and \(2(b^2 + c^2) - a^2\). But why was this happening?

The key idea I first came up with was squaring or boxing off the figure and finding the new triangles.

1. First I found the base of the new triangle and then the height.

[More variations on  square boxing problems]

These all revolve around boxing or squaring off a square with 4 congruent right triangles.


Most elegant solution comes through boxing the large square.


All the triangles are isosceles and all the quadrilaterals are rhombi. Find the  area of the square at the top.