Monday Math: A First Pass
Sep. 3rd, 2012 08:08 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowLet's put what we've built to use, under the cut.
We'll continue looking at quadrilaterals. Each quadrilateral has, associated with it, three numbers z1, z2, z3 and their conjugates w1, w2, w3. We're going to look at products of these, with the same number of z's as w's. The simplest set of such products contains one of each, and there are nine such products.
We have three things we can do to the quadrilateral which have interesting effects on these functions: shifting the initial vertex by one, reversing the order of the vertices, and reflecting the quadrilateral. We'll call these operations r, f, and c respectively. r multiplies the function by a power of i; we get the following division.
Multiplication by 1: z1w1, z2w2, z3w3
Multiplication by i: z2w1, z3w2
Multiplication by -1: z3w1, z1w3
Multiplication by -i: z1w2, z2w3
Look first at the products that are multiplied by 1 or -1. What do f and c do to them? I'll list each of them, followed by the result of applying f and the result of applying c.
z1w1: z3w3, z3w3
z2w2: z2w2, z2w2
z3w3: z1w1, z1w1
z1w3: z3w1, z1w3
z3w1: z1w3, z3w1
What we're going to construct now are functions which are multiplied by 1 or -1 when f and c are applied. This gives us the following five functions, listed according to the effects of r, f, c, in that order.
1,1,1: z1w1+z3w3, z2w2
1,-1,-1: z1w1-z3w3
-1,1,1: z1w3+z3w1
-1,-1,1: z1w3-z3w1
Now comes the hard part. We define our classes by making one of these functions be 0; but how do we interpret them geometrically? That takes some work, but here's what comes out.
z1w1+z3w3 is proportional to the sum of the squares of the diagonals. So, it's only 0 if the diagonals both have length 0; in other words, the quadrilateral has the form ABAB. (This is not an interesting class....)
z2w2 is proportional to the square of the distance between the midpoints of the diagonals, so it's 0 if the diagonals bisect each other. That means that the quadrilateral is a parallelogram.
Since these two are affected in the same way by r, f, c, we can actually combine them, as a(z1w1+z3w3)+bz2w2, where a and b are any two numbers, but the results are harder to describe, and I won't go into that.
z1w1-z3w3 is proportional to the area of the quadrilateral. There are some subtleties, though. A quadrilateral (or any other polygon) divides the plane into pieces, one of which is infinite; but there may be more than one finite piece. (Think of an hourglass-shaped quadrilateral, like 0, 1, i, 1+i, or of a pentagram.) If the polygon goes around a piece counterclockwise, we count the area positively; if clockwise, negatively. So the signed area of the quadrilateral 1, i, -1, -i is 2; the signed area of 1,-i,-1,i is -2; and the signed area of 0, 1, i, 1+i is 0. The quadrilaterals with signed area 0 are the ones we get from this function; they're either flat, or they cross themselves, creating two triangles with the same area. These I call null quadrilaterals.
z1w3+z3w1 turns out to be proportional to the difference between the squares of the diagonals; so this gives rise to what I call equidiagonal quadrilaterals, whose diagonals have the same length. Rectangles are equidiagonal, but there are many others.
z1w1-z3w3 is proportional to the dot product of the diagonals, if we think of them as vectors. This gives us the (already known) class of orthodiagonal quadrilaterals, whose diagonals are perpendicular.
What about the functions which are multiplied by i or -i when you apply r? That's... a bit more complicated, and I'll save it for next time.
We'll continue looking at quadrilaterals. Each quadrilateral has, associated with it, three numbers z1, z2, z3 and their conjugates w1, w2, w3. We're going to look at products of these, with the same number of z's as w's. The simplest set of such products contains one of each, and there are nine such products.
We have three things we can do to the quadrilateral which have interesting effects on these functions: shifting the initial vertex by one, reversing the order of the vertices, and reflecting the quadrilateral. We'll call these operations r, f, and c respectively. r multiplies the function by a power of i; we get the following division.
Multiplication by 1: z1w1, z2w2, z3w3
Multiplication by i: z2w1, z3w2
Multiplication by -1: z3w1, z1w3
Multiplication by -i: z1w2, z2w3
Look first at the products that are multiplied by 1 or -1. What do f and c do to them? I'll list each of them, followed by the result of applying f and the result of applying c.
z1w1: z3w3, z3w3
z2w2: z2w2, z2w2
z3w3: z1w1, z1w1
z1w3: z3w1, z1w3
z3w1: z1w3, z3w1
What we're going to construct now are functions which are multiplied by 1 or -1 when f and c are applied. This gives us the following five functions, listed according to the effects of r, f, c, in that order.
1,1,1: z1w1+z3w3, z2w2
1,-1,-1: z1w1-z3w3
-1,1,1: z1w3+z3w1
-1,-1,1: z1w3-z3w1
Now comes the hard part. We define our classes by making one of these functions be 0; but how do we interpret them geometrically? That takes some work, but here's what comes out.
z1w1+z3w3 is proportional to the sum of the squares of the diagonals. So, it's only 0 if the diagonals both have length 0; in other words, the quadrilateral has the form ABAB. (This is not an interesting class....)
z2w2 is proportional to the square of the distance between the midpoints of the diagonals, so it's 0 if the diagonals bisect each other. That means that the quadrilateral is a parallelogram.
Since these two are affected in the same way by r, f, c, we can actually combine them, as a(z1w1+z3w3)+bz2w2, where a and b are any two numbers, but the results are harder to describe, and I won't go into that.
z1w1-z3w3 is proportional to the area of the quadrilateral. There are some subtleties, though. A quadrilateral (or any other polygon) divides the plane into pieces, one of which is infinite; but there may be more than one finite piece. (Think of an hourglass-shaped quadrilateral, like 0, 1, i, 1+i, or of a pentagram.) If the polygon goes around a piece counterclockwise, we count the area positively; if clockwise, negatively. So the signed area of the quadrilateral 1, i, -1, -i is 2; the signed area of 1,-i,-1,i is -2; and the signed area of 0, 1, i, 1+i is 0. The quadrilaterals with signed area 0 are the ones we get from this function; they're either flat, or they cross themselves, creating two triangles with the same area. These I call null quadrilaterals.
z1w3+z3w1 turns out to be proportional to the difference between the squares of the diagonals; so this gives rise to what I call equidiagonal quadrilaterals, whose diagonals have the same length. Rectangles are equidiagonal, but there are many others.
z1w1-z3w3 is proportional to the dot product of the diagonals, if we think of them as vectors. This gives us the (already known) class of orthodiagonal quadrilaterals, whose diagonals are perpendicular.
What about the functions which are multiplied by i or -i when you apply r? That's... a bit more complicated, and I'll save it for next time.
