The armature of the rectangle uses a network of intersecting diagonals to identify fractional divisions of the rectangle. The intersections corresponding to fractional divisions along the horizontal x-axis are labeled in the diagram above.
The 1/2 diagram is constructed simply of two diagonals from corner to corner; it's intuitively obvious that their intersection locates the center point, and therefore the halfway mark on each side. But what about the 1/3? Why should it be that the intersection of one of the main diagonals with a line drawn from the halfway point of the top side to the bottom corner magically locates a point one third of way along the x axis? The 1/4 is more familiar, as it bisects the segment already cut by the half, but the 1/5? Is that even for real?
Let's start with the simple 1/2 diagram. Even though it's obvious that it's dividing the x-axis by 2, solving it mathematically will lay the foundation for doing the more complicated ones.
It turns out that the easiest way to think about this is in terms of graphing functions, where you have a square grid with a horizontal x-axis and a vertical y-axis. Points on the grid can be located with pairs of coordinates, similar to a map, and lines can be described by mathematical expressions in the form x=y or y=x, with some operation on either both of the terms. Typically a graph will be measured off in numbers, but because we are concerned with proportions, we'll use fractions of a variable maximum length for either axis: xmax and ymax. The 0 point for each axis is the lower left corner, with the maximum length for x at the lower right corner and for y at the top left corner.
The diagonal from the lower left to upper right is defined y=x. The line simply goes up the same distance for every step it goes horizontally. The line from top left to bottom right is defined y=-x + ymax. Here, the angle or slope of the line is the opposite, going down the same distance for every step it takes, so the x in the equation takes a minus sign. "+ ymax" means that it starts in the upper left corner; each value of y gets subtracted from that.
To find the location where they intersect, take the two expressions that appear after "y=" and set them to equal each other:
|-x + ymax = x|
Solving this equation will tell us the point where the function returns the same value for each line; that is, the intersection.
We wish to solve for x, the position along the horizontal side, so we do some algebra like this
|-x + ymax = x||the combined function we just created|
|-2x + ymax = 0||subtract x from both sides, so that x only appears on the left|
|2x + -ymax = 0||multiply both sides by -1 to make the x term positive|
|2x = ymax||add ymax to both sides to reduce the left side to x terms only|
|x = ymax/2||divide both sides by 2 to isolate x completely|
x = ymax/2 * (xmax/ymax)
We can go through this procedure, with appropriate tweaks, for the others:
For the 1/3 diagram, the initial slope equations are:
As we did above:
-x+ymax=2x the combined function we just created
-3x+ymax = 0 subtract 2x from both sides, so that x only appears on the left
3x+ -ymax = 0 multiply both sides by -1 to make the x term positive
3x=ymax add ymax to both sides to reduce the left side to x terms only
x = ymax/3 divide both sides by 3 to isolate x completely
x = ymax/3 * (xmax/ymax) give it the ratio treament:
For the 1/4 diagram, the initial slope equations are:
y=-x + ymax
As we did above:
x+ymax/2=-x+ymax the combined function we just created
2x+ymax/2=ymax add x to both sides, so that x only appears on the left
2x=ymax/2 subtract ymax/2 from both sides to reduce the left side to x terms only
x = ymax/4 divide both sides by 2 to isolate x completely
x = ymax/4 * (xmax/ymax) give it the ratio treament
For the 1/5 diagram, the initial slope equations are:
y=-x/2 + ymax/2
combined: -x/2 + ymax/2=2x
Should be pretty much S.O.P. by now:
-x/2 + ymax/2=2x the combined function we just created
-5/2x+ymax/2 = 0 subtract 2x from both sides, so that x only appears on the left
5/2x+ -ymax/2 = 0 multiply both sides by -1 to make the x term positive _
5/2x=ymax/2 add ymax/2 to both sides to reduce the left side to x terms only
5x = ymax multiply both sides by 2 to simplify the x term
x = ymax/5 divide both sides by 5 to isolate x completely
x = ymax/5 * (xmax/ymax) give it the ratio treament
So now I can be confident that the armature does indeed divide by five. It turns out it can be extended to divide by any number, and each instance can be proven by the same math, but this is enough for now.
If anyone detects errors or has ideas for how to explain this stuff more clearly, please share them.