## Tuesday, April 28, 2015

### 4/28 Magic Squares

After a very challenging week last time, my goal for this session of the math club was to do something fun/low key and then go over some of the solutions to the purple comet meet as  a group. I started by debriefing the kids about the contest. The biggest issues they had were around working as a team which is fairly typical especially with coed groupings like the ones we used last time.  I'm leaning towards doing some of the problems in a non-competitive setting next year rather than the full contest but I'll make a judgment based on next years kids. I had one new student who just started last week which was unfortunate timing for forming a first impression. On the other hand, I had one of the stronger kids who missed the contest complete most of the problems at home.

While finishing up snack I handed out today's warm up a magic square progression. I found a web site that generates some worksheets: http://www.worksheetworks.com/puzzles/magic-squares/integer.html going from a sample to a 3x3, 4x4 and then 5x5. On the back I added a magic hexagon from: http://fivetriangles.blogspot.com/2015/04/235-hexagon-puzzle.html. First we talked about the basic rules and did a few squares from the sample. Then I let kids start working on their own with a request to see if they could see any relations between the first 2 3x3 squares. Victory, first I had one girl notice that all the squares in each one were multiples of the same factor. Then a second realized you could transform each square from the first into the second by multiplying by the same ratio. 3x3 squares have only 1 family of solutions unlike all the other larger versions.   This generally worked really well. Everyone was quiet and focused and the kids had started to attempt the hexagon right about the time we had to move on.

We then walked through the first 10 problems using the overhead projector. This was a bit more limited than the normal discussions with only 3-4 kids volunteering solutions and it was about the upper limit for how much I can keep the kids focused on each other. Improving my ability to facilitate peer discussion is one of my long term goals.  I don't really plan to do the back half of the questions in class. Instead I will email the pdf of the solutions since I think these were  the ones that were mostly too difficult and least likely to be productive as a discussion.

Looking forward I've found another interesting activity. https://news.cs.washington.edu/2013/05/11/the-washington-state-algebra-challenge/ I have to do a bit more investigation but hopefully this will pan out.

## Thursday, April 23, 2015

### 4/21 Purple Comet Online Meet

This week of math club we tried out the online Purple Comet Math Contest. http://purplecomet.org. Running the contest required a few modifications to our normal format. We worked in the library so we could have access to the computers and skipped any snacks or warmups since the lasted the entire 60 minutes. I've also been delaying posting of this entry until after the contest ends so I can discuss a few of the problems.

Overall the contest worked out fairly well. Of 20 questions most of the first 15 were potentially doable with their knowledge base .  A few in between I think were most likely to need algebra and then their were the remainder that made me raise my eyebrows. Each team of 4-5 kids solved 4-6 problems out of 20 which was about what I expected although I was hoping for a little closer to 8. The hardest part for the kids was keeping focus after about 30-40 minutes. Its really hard to keep going and not just give up and start chatting after you've tilted at problems that need a clever solution.  I'm going to debrief everyone next time and see how they liked the experience. In the future we could do some of the problems as individual items on a week to week basis like I sometime use the five triangle problems instead of actually competing.   If done that way they could be potentially scaffolded a bit.  On the other hand sometimes the element of competition is motivating.  The other logistical problem I have is that more kids were out than normal and I plan to go over the problems as a group next session. I'm going to email out a pdf of the contest so those who were absent can look at them. The dilemma is that those who were here deserve a group discussion / understanding of solutions  to the problems they tried but talking about a large group of  problems that  were unseen tends to be counter productive for the rest. Also I don't plan to discuss most of the back problems unless prompted by the kids since I think they were mostly out of scope.

#### Selected Problem Discussion

6Find the least positive integer whose digits add to a multiple of 27 yet the number itself is not a multiple  of 27. For example, 87999921 is one such number.

My first thought was if the digits add to 27 then its definitely a multiple of 9 and that the example is more a red herring.  This is actually quite easy to find via a guess and check strategy. First there are no 1 or 2 digit numbers whose digits sum to 27 at all and there is only one 3 digit number which is 999 but its also divisible by 27 and not the answer. So then onto the 4 digit cases. Since we want to start with the smallest candidate set the 1000th place to 1 and maximize the lower places i.e. the next number is 1899.  1899 / 9 is 211 which is not divisible by 3 using the divisibility test and thus the answer.  As I remember both groups arrived at the solution.

12.
Right triangle ABC with a right angle at A has AB = 20 and AC = 15. Point D is on AB with BD  = 2
Points E and F are placed on ray CA and ray CB respectively such that CD is a median of triangle CEF,  Find the area of triangle CEF.

The hardest part of this problem is that the drawing was incomplete and I completely misinterpreted what they were asking on my first read. I assumed E was in the middle of CA and F in the middle of CB and that CD went along the median as so:

This figure as I found out is unconstrained and has multiple solutions. What I completely missed was they really meant ray not segment where it was written and CD is the exact median.

That means point F has to be out beyond the original triangle like the above picture and this is quite a different problem. Luckily this version is much simpler than my first interpretation and very solvable using similar triangles. Some initial observations: triangle CEF is made up of 3 triangles  one of which we already can calculate: CBD. Also because CD is the median of the triangle CBD + BDF is one half of the median and CDE is the other and they both have an equal area. So our goal is to to find either one of the remaining triangles and the area will fall out.

Now comes the  use of similar triangles. Since CD is a median that means ED is the same length as DF and that triangle EDA is similar to EFA'  scaled up 2x. So since DA is 18 FA' is 36 in length. Then we also know triangle CBA is similar to CFA' and we can scale again CA' / CA = FA'/BA or CA'/15 = 36/20 and CA' = 27.    That gives us all the dimensions needed to find requested triangle. CBD is  1/2 * 15 * 2 = 15 based on the original dimension. BDF 1/2 * 2 * (AA')  AA' = (27-15) = 12. so BDF is 12. That means CFD the sum of the two previous triangles is 15 + 12 = 27 and since that's also one half of the median of CEF, the final area must be 54.  Overall doable but a full diagram would improve the question making it less about what I found a subtle visualization.

19.  Find the real solution for the following:
$a^2 + 3b^2 + \frac{c^2 + 3d^2}2 = a + b + c + d - 1$

At first glance this looks ill-defined given the four variables. A little bit of experimentation shows that the left hand side is almost always larger than right and that there are no real solutions if you zero out any of the terms (although that requires alg2 analysis and some understanding of the quadratic formula)

The key here is realizing that this is actually the sum of four parabolas:

$a^2 -a, 3b^2 - b, \frac{c^2}2 - c$ and  $\frac{3d^2}2 - d$ each of which must have a minimum value.  The problem as stated strongly suggests that the sum of the minimums is likely to be the solution or otherwise there would be multiple answers but in any case its interesting to determine the floor for the total value.

Now either you know the formula for the vertex of a parabola $ax^2 + bx + c$ is  $-b/2a$ or set the derivative of each parabola function to zero to solve (Advanced Geometry or 1st year Calculus) for each term and you derive

$a^2 -a$ has a minimum of -1/4 at a =1/2
$3b^2 - b$ has a minimum of  -1/12 at b = 1/6
$\frac{c^2}2 - c$ has a minimum of -1/2 at c  = 1
$\frac{3d^2}2 - d$ has a minimum of -1/6  at d = 1/3

Lo and behold -1/4 + -1/12 + -1/2 + -1/6  =  -1!  and its done.  I'm waiting to see the official contest solutions but I don't see any obvious shortcuts for the geometry/quadratic analysis besides maybe graphing the parabolas and guessing the vertex from the graph. And even that would still require insight to split the function apart and that look at the sub functions. All of which I think makes this a great maybe 10-11th grade problem in my book rather than even a MS level question.

## Monday, April 20, 2015

### Book Review: Love and Math -The Heart of Hidden Reality

Spring break brought a lot of time to catch up on my reading this year.  Best of all, one my reserves
Love and Math - Edward Frenkel  came into the library before we left for the break.  Overall, it was good read and led me to think more about abstract algebra than I usually do in my day job.  The book interleaves Frenkel's autobiography with some of the math he was encountering along the way described in broad and fairly accessible strokes.  I found this fascinating from several angles. There are the lengthy descriptions of growing up in 1980's russia pre-glasnost.  One of the hardships he had to overcome was antisemitism that prevented Frenkel from attending the premier university for theoretical math and instead he ended up in  an oil and gas applied math program, sneaking into the lectures with other fellow Jews.

On the flip side the description of his math education was quite positive. By 16 he was done with what we consider the normal elementary math sequence and had found a math mentor who introduced him to abstract algebra.  Prior to that he was more interested in quantum physics (although I think the divide between the two is overstated)  He then moved to college and was taken in wing by a very strong group of math professors and fellow students. Its hard to imagine a similar environment here in the U.S.

Also interspersed are longer discussions about topics such as loop theory and symmetries.  These are done at the descriptive level and not suitable for really learning the subject. But its so rare to find any such material and it was great as a tour of what topics concern modern mathematical research. I found the descriptions of the Langland Program to be especially interesting. http://en.wikipedia.org/wiki/Langlands_program

In sum, I'd definitely recommend this to others. One of Frenkel's main goals is to give a clearer picture of what mathematicians really do and here he succeeds brilliantly.  For future years I'm going to think about how I could connect material such as this to the fifth graders. So few kids at this point have any idea of the general terrain of mathematics and what a further career might concern.  More immediately, they know they will study  trigonometry and calculus eventually but at this point they couldn't really tell you what they concern in even very general terms. This is quite different from how we approach almost every other subject in the primary grades.

## Wednesday, April 8, 2015

### Simplifying a Proof

I found this problem online yesterday (briefly) via a post on the google+ k-12 education.  In the figure below

• $\bigtriangleup$ BCX has an area of 5
• $\bigtriangleup$ AXY has an area of 3
• $\bigtriangleup$ CDY has an area of 4
What is the area of $\bigtriangleup$ CXY in the middle?  So I have a reasonable solution for the problem but I'm still trying to see if I can find a way to eliminate the quadratic equation  from the process.

### My solution

First add in some perpendicular lines and divide the big rectangle into 4 quadrants.

Conveniently we know most of the quadrants areas from the  triangles.

• $\Box$ AXOY has an area of 6 since its two $\bigtriangleup$ AXY.
• $\Box$ BCXZ has an area of 10 since its two $\bigtriangleup$ BCX.
• $\Box$ CDYW has an area of 8 since its two $\bigtriangleup$ CDY.
Notice the last two boxes overlap in the upper right corner and I split that out into x. Now we have 2 sets of proportional rectangles going either horizontal or vertical. I'll just choose the vertical ones but you can go either way.

$\frac{\Box BWOX}{\Box AXOY} = \frac{\Box WCZO}{\Box OZDY}$

so

$\frac{10-x}{6} = \frac{x}{8-x}$ which once simplified produces the following quadratic:

$x^2 - 24x + 80 = 0$ or $(x - 4)(x - 20) = 0$  We choose the first solution where x = 4 since the second produces negative areas in the other boxes.   That means the entire rectangle has a size of

6 + 6 + 4 + 4 = 20 and

$\bigtriangleup$ CXY =  Area of the big rectangle minus the 3 triangles or  20 - ( 3 + 4 + 5) = 8.

#### Improvements?

The presence of the quadratic means I couldn't give this one to the kids but it seems like there should be  a refinement that lets me remove it. I thought about it a bit yesterday and couldn't come up with any simplifications. Jump in if you have any ideas.

## Tuesday, April 7, 2015

### 4/7 Spring Quarter Starts

Today was the first day of our last quarter for the math club. I lost 2 kids from the winter session that were replaced with 3 new ones. (I'm embarrassed to admit I didn't track/count my emails correctly so now I have 16 in the club)  So I started from square one again this session with introductions from everyone and a brief discussion of our goals and rules.

I've decided to summarize the rules as being about respect.
1. Respect the classroom which we are borrowing and leave it clean just as we found it.
2. Respect each other both in how we talk to each other and how we listen to each other's ideas.

Goals are pretty loose for this session,. I intend to mostly operate in math circle mode and explore interesting topics.

It was nice to hear from the kids when asked why they were here that several mentioned they were having fun or that the problems we covered were interesting.

While snacking on  cookies I also had everyone talk about the recent contests they went to or the Julia Robinson Math Festival. (Which I highly recommend: http://jrmf.org/) I was really please that almost everyone enjoyed the festival and had a favorite problem they tried there.

I then warmed up again with an easy/medium Sudoku combo. Since this was our second time I had everyone think about what strategies they were using and share them about 10 minutes into the process.  I had a few good suggestions including checking how a number in 2 columns and rows implied where it must be in the remaining one. I think I'd like to have the kids demo these further in front  of the group next time we try the puzzle. Maybe some partials on the projector would work well.

For our main section I had the kids try out a practice test from  the Purple Comet online contest.
http://purplecomet.org/home/home  I just found out about this a few days ago and we're able to squeeze it in. So I postponed what I had planned to do and observed how well the kids can handle the sample problems.   What's nice about them is that you'll have several minutes at least per problem and that they can be team solved. They also range nicely through most pre-algebra topics and had some good puzzles embedded in them. However, these are going to be hard for the kids based on the session. In practice they were able to solve 2-4 of them over 25 minutes working in groups of 2-3. And regrettably I didn't have time to go over any solutions as a whole. So for the real contest I'll make the groups larger 5-6 and talk about looking all the problems over and trying to do the easier ones first. I'm planning to frame this as a stretch for the kids where getting even a few correct is an accomplishment.

Parenthetically this session was a classic example of how when I plan I'm always worried that I won't fill the hour while in practice I find I usually could go one half to an hour longer. The sudoku by itself could fill an hour especially if I chose harder ones. And if I had enough time to review problems from the sample contest I could also have easily filled a large chunk of time.

## Thursday, April 2, 2015

### 4/2 Math Night

Tonight was the annual school Math Night. Since we didn't have a math club meeting until next week this was a fun filler activity to participate in.  I was also hoping to do a little recruiting for next year among the 3rd and 4th graders. About half of my kids volunteered to man the tables and I brought some easy kenken puzzles and a sample math olympiad. It turns out I vastly overestimated the number of handouts needed about 20 of each would have sufficed.  For future reference, we need a table that's closer to the library where everyone naturally drifted towards or in the Cafeteria where they were serving refreshments. Overall I think the kids had fun but we mostly saw 2nd graders and younger so it was not as targeted an event as I hoped.

On an unrelated but fun note I've found another triangle number problem to add to the set I've been collecting involving dominos and dot counting. So I'm leaning heavily towards that for the subject of next week.