Every Triangle corresponds to a Descartes configuration of Circles

Cake Baby

Very obvious fact to some, fascinating to me though.

Consider a triangle. Think of it as three points, where three lines intersect. Imagine that we are only dealing with Rational Points (points whose coordinates are rational numbers)

1

It appears to me, though I have no proof, that each triangle will have the potential to draw upon it an associated set of three circles, all touching each other at one point. (this is also called “mutually tangent” or even “kissing circles”).

2

I would imagine this is true for any triangle, including funky skinny thin ones

triangle with one tiny leg, two large legs, showing 3 circles drawn with centers at the triangles vertexesa triangle with one long leg, and two smaller legs almost the same length, so that the area of the triangle is quite small. two large circles are drawn at the far vertexes, and a tiny circle rests between them at the middle vertex

Even in the case where the triangle is just a line, where all points are collinear. There are three circles, it’s just that the middle circle has a radius of zero. Or, you might say, it has infinite curvature. (Curvature being the inverse of Radius, or 1 divided by Radius)

a line with three points, two circles drawn with centers at the far points, and a 'zero circle' indicated at the central point. all tangent.

Well…. what happens when you have…

View original post 572 more words

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