Wednesday, July 19, 2017

Open Ended Problems

I've been thinking more about open ended problems after reading a couple of different posts recently. Full disclosure: I actually engage in problem solving exercises every week when we meet and believe problems are the core of Mathematics teaching.  But I still don't know if I reached the final idea of how to use them: See http://mymathclub.blogspot.com/2017/02/making-explorations-successful.html for some older thinking.

What started this thought chain off was a post offering that the answer to differentiation for a mixed ability classroom is to use open ended problems. This seems to be a widespread consensus based on how often I've read the same idea. That feels problematic to me from several angles.

1. I'm not sure if there are enough open ended problems that provide different levels of depth and easy entry that span the entire curriculum.  I can see such problems as being useful some time but potentially needing to be interleaved a lot of the time with less flexible material more focused on concepts needing to be gone over.

2. These type problems are almost always prescribed in mixed random groups.  How far can one student pull such a group if they are diving deeper and what kind of experience does this provide for all the participants? For that matter, if a student is behind the others will they more or less rely on the group to peer teach whatever underlying concept is involved? How well does that work?  While the problems themselves are good the overall structure aims at collaboration and creating a common experience for everyone rather than differentiating per se.

In search of more ideas on what an open ended problem / curriculum might look like I was reading through this paper:  http://math.sfsu.edu/hsu/papers/HsuKyshResek-RichProblems.pdf


"Working in well-facilitated small groups on rich problems that are accessible puts students in the position of differentiating the content, processes, and product of their own work. When students are empowered to make natural choices as they work on rich problems together, there are almost always surprises for teachers and often for the students themselves. One of the most important surprises is who comes up with interesting ideas; it is not always who a teacher would have predicted. In this article we discuss what makes a problem rich enough to allow facilitation of this self-differentiated student work."
I did like some of the offered ideas and was thinking about this one:

"Which positive integers can be written as the difference of two squared integers? For example, 17 = 92 – 82?"

The authors caution that the teacher must be ready with questions to keep students going as they get frustrated. Based on my experiences I wonder a bit of how this would work in practice? I can easily picture kids trying out sample numbers and noticing patterns but I don't see them getting from there to concrete reasoning on why this is generally true without a sufficient bit of number theory/modular arithmetic.  i.e. all odd numbers 2n + 1 squared are 1 mod 4 and all even numbers 2n are 0 mod 4 and the only combinations you can get subtracting them are  0,1,or -1  which implies all odds and multiples of 4 but not the non multiple of 4 even numbers i.e. 2 mod 4.

I also don't quite see the pattern noticing section as being particularly fine grained in its accessibility. The most likely outcome would be some kids noticing it after writing a table of numbers and everyone else going "oh that looks true.."

If I were giving this problem I might start with some modular arithmetic background to provide some tools for the kids. Or perhaps treat this like a "3 act" exercise. Working on it long enough to deduce patterns and try to find reasons for them, providing the info on modular arithmetic as a group and then resuming in groups to see what could be discovered more formally.

So I found something that I might use but I'm still not certain that this is differentiating in any significant manner.  I also need more observation to see how it goes with real kids. My main feeling is that group work on interesting problems remains a good practice but one that aims at collaboratively learning as a group. It still suffers from weakness when the members of the group have different amounts of "ability" or experience especially as the gulf widens.


Friday, July 14, 2017

Double angles

I've been thinking about a generalization of the 15-75-90 construction over the last few days and have realized there are a lot more interesting consequences in it that are fairly pleasing.

First construct a rectangle ABED and divide it into 4 isosceles triangles.  My home made puzzle from this construction is: find an expression for the length of CD in terms of AC and AB.


Some angle chasing will show show that the lower and upper triangles are closely related.


If \(\angle{CAB} = \theta\) then \(\angle{DCH} = 2\theta\). What's more given the lengths of the triangle CDH sides (a,b and c) you can find the lengths of the triangle ACG in terms of them i.e.

Triangle ACG has side lengths of a, c - b and \(\sqrt{a^2 + (c-b)^2}\).

I've split the 30-60-90 before to get the 15-75-90 so I'll demonstrate splitting the 45-45-90 to get the 22.5 67.5 90 ratios.   

In this case,  let a = 1 then b = 1 and c =  \(\sqrt{2}\) and from that you get the 22.5, 67.5 90 triangle is in the ratio  \(1 : \sqrt{2} - 1 : \sqrt{4 - 2\sqrt{2}}\)


Essentially this is the  geometric equivalent of the trig double angle formulas:

  • \(\sin2\theta = 2\sin\theta\cos\theta\)
  • \(\cos2\theta = \cos^2\theta - \sin^2\theta\)
and allows you to calculate the ratios of the \(\theta\) triangle from the \(2\theta\) triangle (or vice versa).

In fact its easy to verify the equivalence:

\( \sin2\theta = 2\sin\theta\cos\theta \)
\( \frac{a}{c} = 2\frac{a}{\sqrt{a^2 + (c-b)^2}} \frac{c-b}{\sqrt{a^2 + (c-b)^2}} \)
\(                    = 2\frac{a(c - b)}{a^2 + (c-b)^2} \)
\(                    = 2\frac{a(c - b)}{a^2 + c^2 -2bc + b^2} \)
\(                    = 2\frac{a(c - b)}{2c^2 -2bc}  \)  (Pythagorean Theorem)
\(                    = \frac{a(c - b)}{c^2 -bc}  \)  
\(                    = \frac{a(c - b)}{c(c - b)}  \)
\(                    = \frac{a}{c}  \)


Some interesting relationships fall out when you play with integer ratios.  For example
the 1:2 triangle double angle cousin is the  3 : 4 : 5 which is consistent with their structure.

My next thought is there is probably another general construction for the triple angle formulas which would be nifty since it would allow you to generate ratios for a lot more common angles but I haven't gone farther down this track yet.

Tuesday, July 4, 2017

And yet more 15-75-90 fun

Continuing on the theme of 15-75-90 triangles (See: Last time and First Time) several interesting riffs on 15-75-90's in a box have come up recently.

Example dividing a square with four 15-75-90 triangles:


As is often the case, finding the relative area of the  triangles and square is straight forward using trigonometry:

Let s be the length of the  sides of the square:

The area of each triangle = \(\frac{1}{2} s^2 cos(15)sin(15) \) and using the double angle formulas
\(sin(30)=2sin(15)cos(15)\) so  after substitution and knowing sin(30) = \(\frac{1}{2}\) out pops the area = \(\frac{1}{8}s^2\)

But why is this happening? As usual a 30-60-90 triangle is usually lurking around that allows a Euclidean explanation.


What's particularly interesting about this is that it hints that dissections exist to transform a 1/4 or 1/8 of the larger square into the triangles and sure enough you slide the 1/4 triangle ABO until it becomes 2 15-75-90'!



But let's return to the original problem. There's another easy explanation of what's occurring that just uses the ratios of the triangle:


1. Note the  area of this triangle is \(\frac{1}{2}(2 - \sqrt{3})\)
2. Squaring the hypotenuse you get \(4(2 - \sqrt{3})\) which is 8 times the triangle area.
3. Or in other words each triangle is 1/8 of the square made on the hypotenuse.

And we've refound our original result.

Further questions: Are there other common triangles that divide the square into a unit or "simple" fraction.


I'll leave it to the reader to decide which problem based on this property is more fun (from @eylem and @sansu-seijin):

Given the square of length 6cm, how large is the shaded region?