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One of JD2718’s students has a question

May 10, 2006 am31 3:34 am

Prize question is at the bottom of this post.

When the kiddies are having fun with a topic in math, why not keep it front and center as a motivator? So today's "multiplication of radicals and rationalizing denominators" was relegated to about 5 brief minutes.

Instead, they explored a question posed by John in the comments for "Delayed Gratification" about some funny Pythagorean Triples. (I need to install that LaTex).

So they got x, 4, 5. And the work showed that 4+5 = 9, and the square root of 9 = 3
they got x, 12, 13. And 12 + 13 = 25, sqr of 25 = 5 (5, 12, 13 really is a pythagorean triple)
and they got two more examples. And the awful direction: Explain what is wrong with this work. So, bright kiddies, they created some counter-examples. And then three independently guessed that it works when one leg is 1 less than the hypotenuse. They stayed with me as we explored this algebraically, and learned to generate our own triples: pick an odd number, square it. Half (the square – 1) is the other leg, Half (the square + 1) is the hypotenuse.

OK, so a girl calls me over. 15, 112, 113 is a triple she tells me. Good I say. No, you don't see she says. If you take a "1" away from each, it's still a triple.

Wow. I set the girl and some other kids on exploring this further. I don't know what is going on (which of course I tell them. They like, I think, that I need to go ask people for help). We do find that 29, 420, 421 is a triple, and that if you strike out those 2's, 9, 40, 41 is also a triple. What is going on here?

3 Comments leave one →
  1. May 10, 2006 pm31 8:45 pm 8:45 pm

    is it somehow related to the fact that 10*10 = 100 and 20*20 = 400 and those are the digits you’re removing?

  2. May 10, 2006 pm31 11:22 pm 11:22 pm

    That’s my working theory, but I need to extend it, generalize it, and justify it algebraically. And I have so far been unable.

    Last night I asked for help (and recieved it) from some graduate students in doing this problem. I hope it is not an awful spoiler, but the better of the two solutions we found used Heron’s formula. (this is a neat formula for finding the area of a triangle using all three sides. Click Heron’s Formula to learn more.)

    Hey, Heron’s formula is more square roots. And I’ve been telling the kidlets that it’s ok to attack hard problems, even if you don’t solve all of them, and here’s one I couldn’t solve without help, so boom, huge motivation for right topic wrong lesson. My kids now know Heron’s Formula, and a few think it is sort of cool. And everyone has extra practice factoring and simplifying square roots.

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