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 http://mymathclub.blogspot.com/2015/05/cool-geometry-1problem.html 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.




1.

Most elegant solution comes through boxing the large square.

2. http://www.sineofthetimes.org/a-geometry-challenge-from-japan/

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










Wednesday, February 15, 2017

2/14 Valentine's Day Math Olympiad #4

By the luck of the draw (well really modular arithmetic), this year Valentine's day fell on a Math Club Tuesday. I don't really go in for holiday themed activities much but I was in the drugstore and in a fit of whimsy bought a bag of heart shaped gummies. So I ended up handing them out to the kids as they arrived yesterday which always makes the start of the session more exciting.  As I was going around the table, the thought crossed my mind "Gosh I hope they didn't eat a ton of candy already from their various class parties. If so some of the kids are going to bounce off the walls." That was fortunately not the case.

Thematically, I was in a bind again this week. We lost last week to the snow, next week is Winter Break and  I had to give another MOEMS Olympiad to stay on schedule. This made for a little too few free form sessions since the last one.  Looking forward, I'm going to try to fit in some kind of circle geometry oriented activity to build up to Pi Day. I have also been excited by some reading on function machines and am thinking if there is a fun game or activity inplicit in them.  That said, I appreciate the structure the MOEMS contest enforces. Done properly, this results in a lot of intense focus on the part of the kids on 5 problems over a half hour.   Providing this exposure to more challenging material is part of my over arching goals.

To start off the day, we went over the P.O.T.W (see: http://mymathclub.blogspot.com/2017/01/131-chessboard-problems-or.html)  The kids came up with two different approaches. The first leveraged guess and check and the fact that the overall perimeter was supplied to narrow down on the boxes dimensions. I wouldn't have thought to go this way, in fact I had considered removing the given perimeter since its not needed, but with it in hand this strategy works fairly efficiently.  The second was a more traditional completely Pythagorean Theorem based approach.

Moving on,  I proctored the MOEMS contest. Exponents reared there head again which seems to be a recurring theme for this year.  From what  I can tell so far, there was less conceptual issues with what does the notation mean.  But my work is not done.  Most kids given something like:

$$\sqrt{4^6}$$ will compute  \(4^6\) first and then search for a root manually rather than notice that this is the same as \(\sqrt{(4^3)^2}\) and thus the same as \(4^3\). I'm hoping calling these problems out on the whiteboard afterwards will lead to growth over time.

On the positive side, I had one student who usually has not talked much this year raising his hand frequently and volunteering to demonstrate solutions during our followup whiteboard session. Noticing that trend was my favorite part of the day.

I went with 2 KenKen puzzles of differing degrees of difficulty for the kids to work on if they finished early.  These worked well, but I'll bring 3 next time since 1 student actually managed to finish them both before I was ready to move on.

P.O.T.W:

(This is a slightly modified version of a twitter problem I found from @five_triangles)


2/3 of the kids in one classroom exchanged cards with 3/5 of the kids in a second classroom. What fraction of the total kids didn’t participate?

Planning

I'm still brainstorming about next year.  I'm not sure if its going to be easier or harder to keep 6th-8th graders on task. One of my thought experiments, is whether I could present circle activities at different levels on different weeks and have the kids who found it either too hard or too easy due to the age gap work on practice MathCounts based activities.  Its also quite possible to use the pre-canned MathCounts curriculum which I'll definitely experiment with and see how I and the kids find it.




Tuesday, February 7, 2017

Making Explorations Successful

In a fit of perhaps excessive caution, the district cancelled all after school activities today despite the snow being almost completely melted.  So I'm tabling my plans for Math club for this week.  I really look forward to working with the kids so I'll have to work some of that energy out with my own children instead. I'm particularly fond of watching Numberphile videos as a family.

In the meantime, I saw a quote that I wanted to bounce around:

"If you group kids by "ability", those who are struggling may never see the pattern. Groups need to be mixed"

My first reaction to this idea was contrarian. For one,  if you don't really believe in ability i.e. quote the word to signal skepticism, then why does it make a difference if you mix the kids or not?  If ability doesn't matter groups are basically random already.  Kids should succeed anyway based on their own potential regardless of which peers they are with.  If it does matter, then what kind of learning is happening exactly in these situations?  My fear would be that basically you end up with a set of kids forging ahead and a second set copying what the others have learned. For me this is a poor man's version of direct instruction.  Rather than having an adult who has specialized in instruction showing the way, you devolve to peer to peer tutoring.  And having a reasonable amount of experience, I can safely say even kids who really get a concept are usually not nearly as good at communicating it.

So how does this relate to my Math Club?  First, I do have semi-random groups since I let the kids self select who they work with. The clusters tend to be gendered as a result and split along lines of friendship not necessarily skill. The kids obviously have volunteered to join the club which correlates mostly to some passion for math but in practice there are differences among them that are still probably comparable to a classroom.  When we do non-trivial explorations or tasks which is most of the time, kids discover concepts at vastly different rates.   This is one of the great weaknesses of this structure. In a one on one setup, I could slow down and scaffold just the right amount to let each individual "get it".  In a group, I'm always balancing the needs of the many against each other.

I try to compensate for this by having group discussions where everyone shares and by working individually with clusters during any activity.  I also work really hard to focus on having everyone participate. Those mitigate to some extent, but I still don't achieve a truly even amount of learning. Some kids still regularly have more breakthroughs than others. In a way, I think this shows the need for individual tasks. There needs to be a space, where everyone can struggle with a problem without having  it short-circuited by a peer finding the answer.  I'm sensitive when giving advice to not do all the work. Friends on the other hand jump right to the answer.

But in the end of the day, group inquiry based learning works best for me the more level the playing field to start off with and I'm not sure I've found an entirely satisfying way to resolve the issues that arise when it really isn't. And in thinking about this more, to me this is the crux of why teaching is non-trivial in general.