For two days, I've had a tab open with a neat Futility Closet post. (So many clever bits of mathematics and reasoning there.) It has this image:
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| A pleasing fact from David Wells’ Archimedes Mathematics Education Newsletter |
Matt and John jumped in. And then Henrí...
I love the cycle of generalization in math! Get rid of this restriction, and that restriction.
Get rid of the 2nd line.
Get rid of the shared vertex.
And then the what ifs. What if we restricted a vertex in the preimage?
Surprise!
Lines and circles to lines and circles... must be complex. But John had already gotten there!
Then Simon found circles another way!
Now I'll go find someone to talk with locally about the complex transformations here. I could do it alone, but I prefer math in dialogue! I think I want to see someone else get excited the math, too. I also enjoyed this coming up so soon after the post on Willingham's 4C's of story. Great illustration of the causality and complications inherent in an interesting mathematics situation.
It's available in GeoGebra if you want to play, too.
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You can get this without complex numbers.
ReplyDeleteLet's call the vertex that is being moved F, and the one that is being traced G. G is the image of F in a rotation (if the triangle is equilateral), or more generally, a rotation followed by a dilation. Both the rotation and the dilation are centered on the third vertex.
The locus of G is simply the image of the locus of F.
Since both rotations and dilations preserve collinearity, if F moves on a line, then G moves on a line, (the image of F's line in those transformations.)
Since the image of a circle in a rotation and in a dilation is a circle, if F moves on a circle, then G moves on that circle's image, another circle.
If you want some related amazing connections, let points A and B lie on concentric circles, and be vertices of equilateral triangle ABC. Rotate A and B around the center in opposite directions so that they complete orbits at the same time. Trace C. See what you get.
ReplyDeleteNext: let A and B rotate in the same direction, with the point on the smaller circle having an orbital time half as long. See what C gives you.