More on Duality
Feb. 13th, 2005 02:31 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowAs promised, under the cut.
In an earlier post, I described how to use a circle to construct a projective duality. It turns out that the size of the circle is not particularly important; changing the radius has the effect of scaling things by a certain factor, but nothing else happens. The center of the circle, which I'll call Z, is more important, and I'll refer to the duality as "duality with respect to Z".
The effect that dualizing has on distances is rather complicated, and I'll say no more than this. If the point P is dual to the line l, then the distance from Z to l is inversely proportional to the distance from Z to P. (Specifically, the product of the two distances is the square of the radius of the circle.)
Something interesting happens to angles, but to explain it I have to define directed angles. If l and m are two lines, let P be their point of intersection. If you rotate l counterclockwise around P, eventually it will coincide with m; the angle of rotation required is the directed angle from l to m. So, for example, if l runs east-west and m runs northeast-southwest, the directed angle from l to m is 45 degrees, and the directed angle from m to l is 135 degrees. Now, if l and m are two lines and their duals are Q and R, then the directed angle from l to m is equal to the directed angle from the line ZQ to the line ZR. In particular, l and m are parallel if and only if the points Q, R, and Z are collinear.
Now suppose that C is a conic section - an ellipse, a parabola, or a hyperbola. (Circles count as ellipses.) If you take the duals of all of the tangents to C, you get a bunch of points, which form another conic section C', the dual of C with respect to Z. If C is a circle centered at Z, C' is also a circle centered at Z. If C is a circle centered somewhere other than Z, then C' can be any sort of conic section, but one of its foci is guaranteed to be Z. (Conversely, if C has Z as a focus, then C' is a circle.)
Here's an interesting example. Suppose ABCD is a quadrangle (remember, I mean by that a four-sided figure considered in terms of its vertices) and Z is the intersection of its diagonals. Let lmno be the quadrilateral (four-sided figure considered in terms of its sidelines) dual to ABCD with respect to Z. Since Z lies on the line AC, the duals of A and C - that is, the lines l and n - must be parallel. Similarly, since Z lies on BD, the lines m and o are parallel. That is: the dual of a quadrangle with respect to the intersection of its diagonals is a parallelogram. Turning it around, every quadrilateral can be obtained as the dual of a parallelogram with respect to some point Z. (That's not quite true. There are quadrilaterals whose diagonals are parallel, and they can't be obtained in this way, but they're rare. An example: put the vertices A,B,C,D at (0,0), (1,0), (0,1), and (1,1), in that order. ABCD is shaped like an hourglass, and AC and BD are both vertical lines.) That suggests that the work I've done with class-one quadrilaterals - i.e., parallelograms - may have implications for quadrilaterals in general.
There's more to be said about quadrilaterals in this context, but this is getting long, and I'd rather talk about triangles. That'll be the topic of my next mathematical post.
In an earlier post, I described how to use a circle to construct a projective duality. It turns out that the size of the circle is not particularly important; changing the radius has the effect of scaling things by a certain factor, but nothing else happens. The center of the circle, which I'll call Z, is more important, and I'll refer to the duality as "duality with respect to Z".
The effect that dualizing has on distances is rather complicated, and I'll say no more than this. If the point P is dual to the line l, then the distance from Z to l is inversely proportional to the distance from Z to P. (Specifically, the product of the two distances is the square of the radius of the circle.)
Something interesting happens to angles, but to explain it I have to define directed angles. If l and m are two lines, let P be their point of intersection. If you rotate l counterclockwise around P, eventually it will coincide with m; the angle of rotation required is the directed angle from l to m. So, for example, if l runs east-west and m runs northeast-southwest, the directed angle from l to m is 45 degrees, and the directed angle from m to l is 135 degrees. Now, if l and m are two lines and their duals are Q and R, then the directed angle from l to m is equal to the directed angle from the line ZQ to the line ZR. In particular, l and m are parallel if and only if the points Q, R, and Z are collinear.
Now suppose that C is a conic section - an ellipse, a parabola, or a hyperbola. (Circles count as ellipses.) If you take the duals of all of the tangents to C, you get a bunch of points, which form another conic section C', the dual of C with respect to Z. If C is a circle centered at Z, C' is also a circle centered at Z. If C is a circle centered somewhere other than Z, then C' can be any sort of conic section, but one of its foci is guaranteed to be Z. (Conversely, if C has Z as a focus, then C' is a circle.)
Here's an interesting example. Suppose ABCD is a quadrangle (remember, I mean by that a four-sided figure considered in terms of its vertices) and Z is the intersection of its diagonals. Let lmno be the quadrilateral (four-sided figure considered in terms of its sidelines) dual to ABCD with respect to Z. Since Z lies on the line AC, the duals of A and C - that is, the lines l and n - must be parallel. Similarly, since Z lies on BD, the lines m and o are parallel. That is: the dual of a quadrangle with respect to the intersection of its diagonals is a parallelogram. Turning it around, every quadrilateral can be obtained as the dual of a parallelogram with respect to some point Z. (That's not quite true. There are quadrilaterals whose diagonals are parallel, and they can't be obtained in this way, but they're rare. An example: put the vertices A,B,C,D at (0,0), (1,0), (0,1), and (1,1), in that order. ABCD is shaped like an hourglass, and AC and BD are both vertical lines.) That suggests that the work I've done with class-one quadrilaterals - i.e., parallelograms - may have implications for quadrilaterals in general.
There's more to be said about quadrilaterals in this context, but this is getting long, and I'd rather talk about triangles. That'll be the topic of my next mathematical post.
