4 ms·
Hm, a clever little theorem. I'm surprised it isn't recorded somewhere. Perhaps it's just under a different name because of it's relative... simplicity? Not to
by eximius 11y ago
Hm, a clever little theorem. I'm surprised it isn't recorded somewhere. Perhaps it's just under a different name because of it's relative... simplicity? Not to belittle her accomplishment. This is something you'd expect Euclid to write about or something.
- mchahn 11y ago> This is something you'd expect Euclid to write about Which is what makes the lack its history so fascinating.
- orbitingpluto 11y agoWell remember Euclid's geometry is fundamentally flawed. The parallel postulate, that fact that parallel lines never meet, can be be used to really mess with a grade 7-12 math teacher's head if you invest the time to teach a couple of tricks to your kid. The simplest example is to show that an obtuse angle and a right angle are the same through a construction. The placement of points is crucial to make that work.
- catpolice 11y agoIt follows from the statement that if two circles intersect in more than two points, they're identical. Which seems like a familiar theorem, that I can't seem to place. So it's more like a corollary. It would be remarkable if this hadn't come up before, but I want to believe because of how good a story it makes...
- javajosh 11y agoHa, that's a good one, actually. Quite elegant.
- mixedmath 11y agoYour fact probably doesn't have a name. But a good statement would be that a quadratic polynomial has at most two roots. To see this, suppose that the two circles are (x-a)^2 + (y-b)^2 = r and (x-c)^2 + (y-d)^2 = R. Subtracting gives an equation of the form (linear function in x and y) = r - R. This means that y is a linear function of x, and so we can use this to substitute in the equation for the first circle to get something of the form (quadratic function in x) = r. Then as there are at most 2 solutions for x, and each gives the corresponding solution for y, we see that two distinct circles intersect in at most 2 points.