stoutfellow (![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
stoutfellow) wrote2004-12-11 11:50 am
stoutfellow) wrote2004-12-11 11:50 am
Entry tags:
On Parametrizing Polygons, 4
Here's the final installment.
I've been pretty vague about exactly what the parameters e and s are, in general. But in the specific case of triangles, I can give precise formulas. If the sides of a triangle are a, b, and c, then e=4(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)/(a^2+b^2+c^2)^2, and s=4(b^2+c^2-2a^2)(a^2+c^2-2b^2)(a^2+b^2-2c^2)/(a^2+b^2+c^2)^3.
The parameter space is bounded by the lines e=1 and s^2=e^3, as usual. That's a trumpet-shaped region, with corners at (0,0) (the equilateral triangle), (1,1) (a flat triangle two of whose vertices are identical), and (1,-1) (a flat triangle with one vertex halfway between the other two). The boundary curve s^2=e^3 corresponds to isosceles triangles; if the apex angle is less than sixty degrees, s is positive, and s is negative if the apex angle is more than sixty degrees.
Okay, so we know where isosceles triangles are; where are right triangles? Computation reveals that they lie along the straight line 3e-2s=1. Now that's intriguing: if a triangle satisfies a^2+b^2=c^2, it lies along that line, and if it satisfies a^2+b^2=2c^2, it lies along the line s=0. It turns out that if you pick a constant k which is at least equal to 1/2, then there are triangles satisfying a^2+b^2=k c^2, and those triangles also lie along a line. Not just any line, though; the lines that arise in this way are tangent to the boundary curve. (The reason for that is not too hard. If an isosceles triangle has b=c and satisfies a^2+b^2=k c^2, then it also satisfies a^2+c^2=k b^2. That is, it satisfies the equation twice over, and that's why the line touches the curve "twice over" at the corresponding point - i.e., it's tangent to the curve.)
Now, the interesting thing here is this. If you apply the median map, you just flip the parameter space over the line s=0, and if a line is tangent to the boundary curve, its image after the flip is still tangent. In other words, for any k there is an l so that, if a triangle satisfies a^2+b^2=k c^2, its median triangle satisfies a^2+b^2=l c^2. To be precise, l=(k+4)/(2k-1). In particular, the median triangle of a right triangle satisfies a^2+b^2=5 c^2. These triangles can also be identified by the fact that two of their medians are perpendicular. (I'm kind of tempted to call these "left triangles"...) (Note that if k=2, then l=2 - which is what you'd expect, since those are the symmetry-0 triangles, which are similar to their own median triangles.)
So. That's most of the content of my paper (leaving out boring stuff like calculations and proofs). It's nothing splashy, but I think it's kind of pretty, like a cameo. (There is that one rough spot with case-3 midgons, but every artwork has to have a blemish somewhere...)
I've been pretty vague about exactly what the parameters e and s are, in general. But in the specific case of triangles, I can give precise formulas. If the sides of a triangle are a, b, and c, then e=4(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)/(a^2+b^2+c^2)^2, and s=4(b^2+c^2-2a^2)(a^2+c^2-2b^2)(a^2+b^2-2c^2)/(a^2+b^2+c^2)^3.
The parameter space is bounded by the lines e=1 and s^2=e^3, as usual. That's a trumpet-shaped region, with corners at (0,0) (the equilateral triangle), (1,1) (a flat triangle two of whose vertices are identical), and (1,-1) (a flat triangle with one vertex halfway between the other two). The boundary curve s^2=e^3 corresponds to isosceles triangles; if the apex angle is less than sixty degrees, s is positive, and s is negative if the apex angle is more than sixty degrees.
Okay, so we know where isosceles triangles are; where are right triangles? Computation reveals that they lie along the straight line 3e-2s=1. Now that's intriguing: if a triangle satisfies a^2+b^2=c^2, it lies along that line, and if it satisfies a^2+b^2=2c^2, it lies along the line s=0. It turns out that if you pick a constant k which is at least equal to 1/2, then there are triangles satisfying a^2+b^2=k c^2, and those triangles also lie along a line. Not just any line, though; the lines that arise in this way are tangent to the boundary curve. (The reason for that is not too hard. If an isosceles triangle has b=c and satisfies a^2+b^2=k c^2, then it also satisfies a^2+c^2=k b^2. That is, it satisfies the equation twice over, and that's why the line touches the curve "twice over" at the corresponding point - i.e., it's tangent to the curve.)
Now, the interesting thing here is this. If you apply the median map, you just flip the parameter space over the line s=0, and if a line is tangent to the boundary curve, its image after the flip is still tangent. In other words, for any k there is an l so that, if a triangle satisfies a^2+b^2=k c^2, its median triangle satisfies a^2+b^2=l c^2. To be precise, l=(k+4)/(2k-1). In particular, the median triangle of a right triangle satisfies a^2+b^2=5 c^2. These triangles can also be identified by the fact that two of their medians are perpendicular. (I'm kind of tempted to call these "left triangles"...) (Note that if k=2, then l=2 - which is what you'd expect, since those are the symmetry-0 triangles, which are similar to their own median triangles.)
So. That's most of the content of my paper (leaving out boring stuff like calculations and proofs). It's nothing splashy, but I think it's kind of pretty, like a cameo. (There is that one rough spot with case-3 midgons, but every artwork has to have a blemish somewhere...)