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A proof from the mouth of a babe

One of the high points of this last academic year has been volunteering in Katya's 2nd/3rd grade English and Math classes. I was suitably rewarded for this just a few weeks ago with a delicious treat of an elegant mathematical argument from a 2nd grader (Katya's in 3rd). Here's the puzzle and its solution (names of individuals have been changed).

It all started with the teacher, Judith, posing a seemingly simple problem to the kids. She was teaching fractions that Thursday. In the latter half of Math class, the kids were each given rectangular cards, and asked to imagine they were cookies (or pizzas). Since classes at Ohlone are always mixed between two grades, the problems are always posed in a range of difficulties. With this one, the kids had to divide up their cookies into 2, 4, 8 and 3 equal parts. The more different ways you could slice it up, the more points you get.

The puzzle I want to talk about concerns the division of the rectangle into 4 equal parts. Most kids did it the conventional way, by which I mean folding the rectangle in half, first vertically, then horizontally, and cutting along the creases. Edgar (name changed), however, adopted a different tack. He folded the rectangle along its two diagonals.

The cool thing about this is that it's not immediately obvious that this way of cutting the rectangle produces 4 equal cuts, especially not to 2nd or 3rd graders. See the figure below, for instance:

rectangle-1.jpg

Can we tell immediately that Slice A is the same size as Slice B? Edgar claimed they were, but wasn't able to explain it - he just felt it in his gut. Judith, awesome teacher that she is, immediately recognized the opportunity to enrich and picked up on this issue, calling the attention of the entire class to the problem. She wrote it up on the white-board and named it "Edgar's challenge". Anyone who was done with the rest of the problems was welcome to give it a shot. Can they show that the four slices were the same size?

It turns out that a great many of the kids were done with the "rest of the problems" really quickly, and were intrigued by Edgar's challenge. They all tried to prove it, and mostly with what I might call the straightforward and more "boring" proof. It uses the formula for the area of a triangle, which is equal to half its base times its height. The base of triangle A is equal to the width of the rectangle (if you rotate A by 90 degrees). Its height, clearly is half the length of the rectangle. Thus it's area must be one quarter the product of the rectangle's height and width. By a similar argument, you can show that the area of triangle B is also exactly that quantity.

I was checking to see if there were any other interesting arguments and questions, when Mary ran up to me and asked if I could check her answer to Edgar's challenge. What she had was different from everyone else, and so she couldn't tell if it was right or not. So I went, and was suitably impressed. Mary took the 4 triangles and assembled two rhombuses (or is it rhombi?) from them.

rhombus-1.gif
rhombus-2.gif

As you can see from the above pictures, the first rhombus was made with triangles A and C, and the second with triangles B and D. She laid one over the other and said they're the same size. Since each of these is made of two "equal" triangles, she said, the 4 triangles also have to be of the same size each. No formulas, no equations, just a plain and simple, albeit elegant geometric proof, stated with child-like innocence. Pretty awesome, considering that that's how one comes up with the formula for a triangle's area in the first place! Somewhere along the continuum, one loses sight of the beginnings until teachers like Judith and schools like Ohlone teach us the value of the past!

That weekend, I remember recounting this marvelous experience to Chris Nguyen over a breakfast catchup. And he followed up with an equally interesting tidbit his daughter had just discovered a week ago at a play structure - What single formula gives the area of a rectangle, trapezoid and a triangle? The answer, not so surprisingly (or surprisingly to some) is quite simple - It's the product of the height and the average length of the two parallel sides.

There you go - two nice little rewards in return for volunteering in a child's class, which is rewarding in itself. Not too shabby.

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This page contains a single entry from the blog posted on May 8, 2011 8:49 AM.

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