"There is no one way"

Tuesday, December 10, 2013

Asilomar report, Part 1

Once again, I had a great time at the Asilomar conference of the California Math Council. Here are some notes from the first two sessions I attended.

Scott Farrand (of Cal State Sacramento) and UC Davis's Rick West's presentation "Diophantine Equations Can Hide Geometric Surprises" was a fun way to start the day.

Think of two whole numbers. Do they satisfy (m+n)/2=m? Do they satisfy 1+(m+n)/2=m? Do they satisfy m – n = √m + √n? Do they satisfy (m – n)/2 = √m + √n? Each of these questions had a nice geometric interpretation (which is in fact where the problem originated in the mind of its creator.)

For example, for m – n = √m + √n:
 
The figure shows that if m and n are consecutive perfect squares, the equation will hold. 

And for a great punchline, the last problem is a bit tougher: find two whole numbers that satisfy (m – n)^2 = m + n. This yields a nice graph if you use x and y for m and n. (In fact, if you don't want to do a lot of trial and error in searching for number pairs that work, you might make the graph first, look at where it passes through lattice points, and use that as a hint towards the geometric interpretation.)

For more fantastic Farrand material, see this short video.

Oakland USD's Scott Farrar (no relation) shared some good ideas on using GeoGebra in the classroom. I have in the past written against the model "do this, do that, what do you notice?" which is so often followed when using technology in math education. In general, I stand by that position (as articulated for example here) but as always in teaching, there are no absolutes, and Scott showed some good examples of excellent uses of that approach. It can be a good way to generate conjectures.

For example, make a random polygon in GeoGebra, measure its area and perimeter, and graph the point with coordinates (perimeter, area). Then distort your polygon in various ways, and trace the movements of the point. Here's the trace for a quadrilateral with two fixed equal perpendicular sides, and the fourth vertex moving about:


In other words, I started with a square, with three fixed vertices, and moved the fourth vertex. (Remember: the perimeter is on the x-axis, and the area on the y-axis.) What points are impossible? What points are on the boundary between possible and impossible? I don't really know the answer, but this This is wonderful stuff.

 --Henri

PS: I also reported on Asilomar last year. To go to the first post of that series, click here.
 For the continuation of this year's report, click here.

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