Duality, how things are connected in unexpected ways. The simplest case is that of the five regular Platonic solids, the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron. They all look rather different, BUT…..
take any one of them and find the mid point of each of the faces, join these points up, and you get one of the five regular Platonic solids. Do it to this new one and you get back to the original one. Calling the operation “Doit” we get
tetrahedron –Doit–> tetrahedron –Doit–> tetrahedron
cube –Doit–> octahedron –Doit–> cube
dodecahedron –Doit–> icosahedron –Doit–> dodecahedron
The sizes may change, but we are only interested in the shapes.
This is called a Duality relationship, in which the tetrahedron is the dual of itself, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other.
Now we will look at lines and points in the x-y plane.
3x – 2y = 4 and y = (3/2)x + 2 and 3x – 2y – 4 = 0 are different ways of describing the same line, but there are many more. We can multiply every coefficient, including the constant, by any number not 0 and the result describes the same line, for example 6x – 4y = 8, or 0.75x – 0.5y = 1, or -0.75x + 0.5y + 1 = 0
This means that a line can be described entirely by two numbers, the x and the y coefficients found when the line equation is written in the last of the forms given above. Generally this is ax + by + 1 = 0
Now any point in the plane needs two numbers to specify it, the x and the y coordinates, for example (2,3)
So if a line needs two numbers and a point needs two numbers then given two numbers p and q I can choose to use then to describe a point or a line. So the numbers p and q can be the point (p,q) or the line px + qy + 1 = 0
The word “dual” is used in this situation. The point (p,q) is the dual of the line px + qy + 1 = 0, and vice versa.
The line joining the points C and D is dual to the point K, in red. The line equation is 2x + y = 3, and we rewrite it in the “standard” form as -0.67x – 0.33y +1 = 0 so we get (-0.67, -0.33) for the coordinates of the dual point K.
A quick calculation (using the well known formula) shows that the distance of the line from the origin multiplied by the distance of the point from the origin is a constant (in this case 1).
The second picture shows the construction of the dual point.
What happens as we move the line about ? Parallel to itself, the dual point moves out and in.
More interesting is what happens when we rotate the line around a fixed point on the line:
The line passes through the fixed point C. The dual point traces out a straight line, shown in green.
This can be interpreted as “A point can be seen as a set of concurrent lines”, just as a line can be seen as a set of collinear points (we have fewer problems with the latter).
It gets more interesting when we consider a curve. There are two ways of looking at a curve, one as a (fairly nicely) organized set of points ( a locus), and the other as a set of (fairly nicely) arranged lines (an envelope).
A circle is a set of points equidistant from a central point, but it is also the envelope of a set of lines equidistant from a central point (the tangent lines).
So what happens when we look for the dual of a circle? We can either find the line dual to each point on the circle, or find the point dual to each tangent line to the circle. Here’s both:
In this case the circle being dualled is the one with center C, and the result is a hyperbola, shown in green. The result can be deduced analytically, but it is a pain to do so.
The hyperbola again. It doesn’t look quite perfect, probably due to rounding errors.
The question remains – If I do the dualling operation on the hyperbola, will I get back to the circle ?
Also, why a hyperbola and not an ellipse ? Looking at what is going on suggests that if the circle to be dualled has the origin inside then we will get an ellipse. This argument can be made more believable with a little care !
If you get this far and want more, try this very heavy article: