Recently my friend Tina, who writes the blog Diotima’s Ladder, asked me if I could help her with a diagram for her novel. (Apparently all the math posts I’ve written gave her ideas about my math and geometry skills!)
What she was looking for involved Plato’s Divided Line, an analogy from his runaway bestseller, the Republic (see her post Plato’s Divided Line and Cave Allegory for an explanation; I’m not going to go into it much here). The goal is a geometric diagram proving that the middle two segments (of four) must be equal in length.
This post explores and explains what I came up with.
[Update 5/24/2022] See my updated post, Back to Plato’s Line for a better proof!
Our starting point is what Plato, in the Republic, says to Glaucon about the line. He has introduced the idea of splitting the world into two classes of experience, the visible and the intelligible (also called opinion and knowledge). Plato then says:
Let us represent them as a divided line, partitioned into two unequal segments, one to denote the visual and the other the intelligible order. Then, using the same ratio as before, subdivide each of the segments. Let the relative length of these subdivisions serve as indicators of the relative clarity of perception all along the line.
A question arises from that last sentence. It can be easily demonstrated that the middle two segments must have the same length, but Plato’s further explanation of all four segments seems to imply an inequality, that each higher segment, representing greater clarity of thought, should be larger than the one below it.
A good starting point for something like this is throwing some numbers at the problem and seeing what happens. I started by trying some simple fractions:
In all cases, the left column is the initial division, as Plato said, “into two unequal segments.” I used one-third, one-fourth, and one-fifth, respectively, for the lower segment. The upper segment, then, is whatever remains (one minus the lower segment).
The middle column represents the subdivision “using the same ratio as before” applied to both the upper and lower segments. To get the actual values for the sub-segments we multiply, giving us the right-hand column. For example, one-third of one-third is one-ninth. Note that the fractions in the right column sum to one (as do the fractions in the first column).
What the examples show is that the middle segments are equal in all three cases.
Of course, these numbers might just happen to work out that way, so the next step is abstracting this to a general case. In the examples above, note that we start by picking some fraction, 1/d, for the lower segment (1/3, 1/4, 1/5). Then the upper segment is:
This might seem to exclude a fraction such as two-fifths (or three-eighths), but in fact:
So, our abstraction works for any fraction because, if necessary, we can normalize it to 1/d. Our abstraction is then:
And, again, the middle two segments are shown to be equal.
That seems conclusive, but if we’re skeptical, we can simplify by assigning length a to the lower segment. Then the upper segment has length 1-a, and therefore:
And, once again, the middle segments are equal.
We can simplify it even more by using 1-a=b. Then, using just a and b, we have:
Which is about as simple as it gets. Bottom line, if a line is divided according to some ratio (b/a), and the divisions are sub-divided by that same ratio, the middle two segments must always be equal in length.
Which is well and good, but totally algebraic. Algebra, as we learned it in school, doesn’t show up until about the 9th century, long after Plato. The Greeks did have some geometric algebra as well as facilities with numbers, so it’s not hard to believe the math above may have been perfectly well known to Plato.
But at that time, the Greeks were more about geometry, and, in any event, what’s needed for Tina’s book is a diagram — something geometric and visually appealing. Something sufficiently interesting for Plato scholars and not pure gibberish to dismay the mathematically (and especially geometrically) inclined.
The math above provides confidence that a geometry-based exploration must agree, if not hopefully somehow prove, that the middle segments must be equal. It turns out to be slightly tricker than it is with algebra. Similar to the example fractions above, we can show that any geometric example we make has equal middle segments but finding a proof they must be equal using only geometry is a bit harder.
A key limit in how we approach the geometry is that our only tools are the ability to draw a straight line (using a straight edge) and the ability to draw a circle of any radius (using a compass). There are two important tricks I want to cover before I proceed.
The first trick: Given any straight line, draw a perpendicular line (a line at right angles) intersecting the first line at some point, P.
Figure 1 illustrates this trick.
The black horizonal line is our initial straight line. We wish to create a perpendicular line (blue) that intersects the first line at point P.
Step 1: Use a compass to draw a circle of some radius, r1, with point P as the center point of the circle. This creates two new points, A and B, where the circle intersects the horizontal line.
Step 2: Use a compass to draw two new circles, one with point A as its center, the other with point B. Both circles must have the same radius, r2. The circles intersect at two points, C and D.
Step 3: Use a straight edge to draw a line from point C to point D. That line passes through point P and is perpendicular to the horizontal line.
Note: The radius of the circles isn’t important, but larger circles make for more accuracy. It is necessary that r1<r2, usually by a fair amount (in Figure 1 r2 is a bit more than twice r1). The crucial thing is that the two circles drawn from points A and B must have the same radius.
The second trick: Given two lines, prove they are (or are not) parallel. Alternately, given some straight line, construct a second line parallel to it at some given distance.
Figure 2 illustrates.
The two black lines slanting down to the right may, or may not, be parallel. We want to find out which.
Step 1: Using a straight edge, draw a line (blue) that intersects both black lines. It intersects them at point P and point A.
Step 2: Using a compass and point P as the center, draw an arc that intersects both black lines. Set the radius, r, of the compass such that it is the distance from point P to point A along the blue line. The arc intersects the lower black line at point B.
Step 3: Using the compass set at the same radius, r, draw two new arcs using point A for one and point B for the other. The arcs will intersect at some point C. If point C lies on the upper black line, then the black lines are parallel. If it lies off the line, they are not.
Note: What we’re doing is creating a parallelogram. If we succeed, the original lines are parallel (and so are both blue lines). We can use the same technique to create the upper parallel line (an exercise left to the reader).
This is the part of geometry that concerns itself with what can be done using only those two tools, the straight edge and the compass. Note that a straight edge is not a ruler — this type of geometry doesn’t allow measuring lines or angles other than proportionally.
It’s a fun kind of geometry, often taught early in school, and it’s surprising how much is possible using only those two simple tools.
Knowing we’re on firm ground mathematically, let’s see what we can do geometrically.
Step 1: We again start as simple as possible with a straight vertical line “partitioned into two unequal segments” as shown in Figure 3.
We label the bottom point a, the top point b. We pick an arbitrary point c to divide the line unequally. (In fact, everything we do here works fine in the degenerate case of equal segments.)
We label the lengths of the bottom segment X and the top segment Y. This gives us a ratio Y:X for the two segments. The total length of the line is X+Y.
Note that a ratio, Y:X, is the same thing as the fraction, Y/X. Here we always put the larger upper length above (or first, in a ratio) and the smaller lower length below (or second) to reflect the idea behind Plato’s vertical line.
Step 2: Using the trick described above for creating perpendicular lines, draw three straight horizontal lines perpendicular to points a, b, and c, as shown in Figure 4.
We also pick an arbitrary point d on the lower line and draw a straight vertical line perpendicular to it. The new vertical line intersects the middle horizontal line at point e.
(We care only about the part of those lines to the right (so that’s all I’ve shown). The perpendicular trick obviously results in some extension to the left and, in the case of the new vertical line, below. For clarity, I’m not showing the arcs of the trick or the extra bits beyond the parts we need.)
Note that the distance ad (that is, from point a to point d) can be anything, but for precision should be as long as convenient and comfortable.
Step 3: Draw a straight (slanted) line connecting point b to point d, as shown in red in Figure 5. This line intersects the horizontal line ce at point g.
Draw a new vertical straight line perpendicular to line ce such that it intersects point g. The new vertical line intersects the lower horizontal line ad at point f.
This subdivides the horizontal by same ratio (but not necessarily the same lengths) as the vertical. This gives us the following equal ratios:
The horizontal ratio matches the vertical one because point g projects down to point f. Think of the slanted red line bd as a mirror reflecting the vertical line ab proportionally onto the horizontal line ad. (Because the “mirror” is straight, the reflection can shrink or expand but must remain proportional.)
(If we happened to use the same length ad as we did for ab (by using a compass to make them equal), then ac=df and bc=af. The diagram would be square, and the “mirror” would have a 45° angle, making the reflection necessarily the same size.)
Step 4: Now we subdivide the vertical line segments and complete Plato’s Divided Line.
Draw a straight slanted line from point b to point e and another from point c to point d (shown in red in Figure 6). The upper one intersects the middle vertical line at point k, the lower one at point h.
Draw two new straight horizontal lines perpendicular to the vertical lines, one to intersect point k and the other to intersect point h. These intersect the original vertical line (ab) at points m and n.
The new diagonals, line be and line cd, create two new “mirrors” that each reflect the entire horizonal line ad onto, respectively, the upper and lower segments of line ab. Since we know that line ad has the same ratio af:df as the original vertical line (bc:ac), we also know the reflections must be proportional.
Therefore, we have subdivided the segments according to the original ratio as specified by Plato. We label the segments, bottom to top, as A, B, C, and D.
Now all we have to do is prove that B=C.
My first attempt to find a proof didn’t give me the strong result I wanted. It’s not difficult to show B=C in drawings with different sizes and ratios but proving it requires an abstraction as it did for the math above. I’ll get to that proof, but first a small detour…
Below is the second candidate diagram I came up for Tina’s novel (see below for the first candidate):
The goal is showing that the triangles P, Q, and R, are equal to each other. In particular, to show that P=R, because that means t=u (in this early version I labeled the four segments s, t, u, and v, rather than A, B, C, and D).
It’s easy to use a compass to show that the longer lines (the horizonal baselines or the diagonal hypotenuses) are the same length, but we could have done that with the B and C (or t and u) segments. There is some value in having longer lines to minimize error, but nothing here requires the equality.
And I know from drawing this diagram many, many times that precision is a problem. If I wasn’t very careful — especially when not using graph paper — I got a diagram in which they are not equal.
There is some logic we can apply, though. Figure 7 continues the diagram we began above (and may make the construction of the triangles more apparent than in the prototype above).
Step 5: Extend the horizontal line mh to the right such that hp=mh. A compass set to the distance mh allows determining the distance hp.
Step 6: Draw a (new) straight line from point n through point g to the extended line mhp. We expect it to intersect at point p. (In any carefully drawn diagram, it will.)
Note that the line ngp forms another “mirror” that reflects the horizontal line mhp onto the vertical line mcn. Such a mirror always reflects the same proportion (although it may shrink or expand the whole), so mh:hp=mc:cn.
Since we explicitly set mh=hp, therefore mc=cn (or B=C).
On the other hand, if our attempt to construct the triangles fails, if the line from point p to point m does not intersect point g, or if a line from point m through point g does not intersect at point p, then B≠C.
We also note that only in the case the segments are equal are the diagonal lines parallel to each other. This gives us one more way to demonstrate B=C. If we create the three triangles regardless of the lengths (that is, we create the lines mhp and ngp regardless of anything else — possibly ending up with the left or right example in Figure 8), if we can show the diagonals to be parallel, then P=Q=R and B=C.
Every carefully drawn diagram will demonstrate this just as every fraction we chose at our starting point does. We have a proof by example, but this is akin to always finding white swans. We can’t be sure that the next diagram (or fraction) won’t be a black swan.
We require a geometric abstraction just as we required an algebraic one.
Back from our detour, Figure 9 picks up where Figure 6 (from Step 4) above left off.
Note that Figure 9 shows the Y and X lengths and assigns lengths U and V to the subdivided horizonal line ad. We know from Step 3 above that Y:X=V:U (or, as fractions, that Y/X=V/U).
We start by focusing on the larger upper triangle formed by points c, b, and e, and the smaller lower triangle formed by points a, c, and d.
Since they share the same base length (U+V) and have heights according to our ratio Y:X (literally, the height of the upper triangle is Y and of the lower one X), the area of the triangles has the same ratio:
We can factor out the (U+V)/2, which leaves Y and X, our ratio.
Secondly, more importantly, note that the (red) diagonal lines (be and cd) have a slope, which is their vertical distance divided by their horizontal distance (or height divided by their width). We call these Sy and Sx, respectively:
We can factor out the 1/(U+V), again leaving Y and X, so these slopes also have the same ratio, Sy:Sx, as our line segments (Y:X).
Now we’re ready for…
New Step 5: Reflect the diagonal line cd by drawing a straight line from point c to a new point r that we create using a compass set to the distance ed (so that ed=er). Because this line has the same slope as cd, but in the opposite direction, we label the slope of the new line -Sx. (As shown in Figure 9.)
Note that this line intersects point k, and we have another demonstration that B=C, because, if ed=er, then gh=gk, because the triangle crd is an isosceles triangle.
What’s more important is the new triangle formed by cre, which by construction is identical to the original triangle acd. We know the triangles cbe and acd have the Y:X ratio to each other, so the triangles cbe and cre must also have that ratio.
Our advantage now is that their bases overlap and their heights (which have the ratio Y:X) start at the same baseline and extend upwards. The proof we seek is now in our hands.
Figure 10 focuses on these two triangles, cbe and cre. Crucially, we have the respective slopes Sy and Sx and these intersect each other.
New Step 6: The slopes intersect at point k, which obviously means they have equal heights above the baseline at that point. The “mirrors” formed by Sy and Sx project point k down onto point g, which we previously gave the lengths V and U.
We can think of line slope as akin to (constant) velocity. If the “slope” is 60 MPH, it means we’d go 60 miles in one hour, but it also means we’d go 30 miles in one-half hour (or 15 miles in one-quarter hour). We just multiply the “slope” by the time interval.
Likewise, we can determine the height of any point along the line slope by multiplying the slope by the length (along the baseline) at that point. Specifically, we can determine the height of point k on the slope Sx by:
And the height of point k on slope Sy (starting from point e and heading left because we must start from zero) by:
As we did with the slopes above, we can factor out the 1/(U+V), leaving us with:
We know from above that V/U=Y/X and that leads to the following identity:
Because we multiply both Y/X and V/U by X×U, which cancels out the lower part of the fraction leaving us with the crucial identity Y×U=X×V.
Crucial because we just established that the height of point k is, in one case X×V and in the other Y×V. Since those equal each other, we know that point k is a necessary height equality between the two slopes where they intersect.
And since triangle cre is a mirror version of triangle acd, that means the line segments gh and gk must be equal. Or in other very important words: B=C.
This makes perfect sense. Given the proportional ratio between the slopes, it seems almost obvious that the rise of slope X over distance V must be the same as the rise of slope Y over distance U.
Put in terms of velocity, if we go half as fast for twice the time, we cover the same distance as if we went twice as fast for half the time. Concretely, if we go 30 MPH for one hour, we go 30 miles, and if we go 60 MPH for one-half hour, we also go 30 miles.
We no longer depend on drawing precise diagrams, numeric examples, or even abstract algebra, but on simple logic. In Plato’s Divided Line, the middle segments, B and C, have to be equal.
I set a limit of 4000 words for this post, and since I’m under 3500 at this point, I’ll include a few extras. I mentioned above there was an earlier candidate diagram I came up with for Tina’s novel. Here is one version of it:
It illustrates a given ratio (in this case Y:X=2:1) and includes some bits of math illustrating the necessary algebraic equality of the middle segments. It doesn’t prove anything, and I couldn’t come up with a good way to show a proof.
The bold vertical lines are intended as segments (B:A or D:C) and combining appropriate pairs (indicated by the blue rectangles joining pairs) would give the entire Divided Line. The red shading demonstrates that the B segment (upper part of a left member of the pair) equals the C segment (lower part of the right member). Since the slopes of the connecting lines are all parallel, we see that B=C in all cases.
There might be a way to demonstrate geometric necessity, but if so, it eluded me. In any event, I sense that any proof would require a fair bit of algebra, and what I came up with above seems much simpler and clearer.
(I think Fred Brooks, in The Mythical Man-Month (1975), was mostly right when he wrote that designers should be prepared to throw away their first design because it’s likely to be wrong anyway. I’ve found over and over that one needs to chew on a problem for a while to really see the shape of it.)
For fun, here’s a version of the proof done using only a compass and straight edge (no graph paper). This is how Plato or Glaucon might have done it.
As you see, the compass work makes it a little messy. (What you don’t see is all the attempts that weren’t precise enough that lines didn’t intersect correctly. I generated a lot of scrap paper on this project!)
Lastly, we look at the diagram as a sort of mechanical framework we can move around for different subdividing ratios (see Figure 11).
Imagine that the lines can telescope to fit appropriately as we move the mechanism around. Also imagine that the green circles are joints that bind the lines but allow them to slide through them. The two green squares (upper left and lower right) are fixed points that only allow pivoting.
Note the shaded red “kite” shape in the lower half. It appears to be symmetrical about its horizontal axis (and, as we’ve shown above, it truly is).
We can consider what happens when we resize the mechanism or move the middle horizontal line up or down to change the ratio between the top and bottom.
The green circle on the blue diagonal line fixes the intersection with the middle vertical and horizontal lines, so moving the horizontal line down moves the vertical one right and vice versa. Because all our lines are straight, the movement is linear.
As those two lines move, the red diagonal lines also move in sync. Here’s what different configurations of the mechanism look like:
The symmetry remains constant, and this gives us a mechanical sense of how and why B=C.
We can take it a step further. The top row of Figure 12 shows the degenerate case where the top and bottom segments are (contrary to Plato) equal in length. In this case it’s obvious that B=C because A=B=C=D by construction.
Not shown (because it would be boring) is the other degenerate case where the lower segment collapses to zero length, giving us A=B=C=0 (and D=the whole line). Again, by construction, B=C.
Since the mechanism moves in linear fashion (no curves), and the degenerate end cases have B=C, we are further assured that B=C for all intermediate cases.
And on that note, we’re done. Quod Erat Demonstrandum!
Stay geometric, my friends! Go forth and spread beauty and light.