Plato’s Divided Line

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.

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:

\displaystyle{1}\!-\!\frac{1}{d}\;\;=\;\;\frac{d}{d}\!-\!\frac{1}{d}\;\;=\;\;\frac{d\!-\!\!1}{d}

This might seem to exclude a fraction such as two-fifths (or three-eighths), but in fact:

\displaystyle\frac{a}{b}=\frac{{a}\div{a}}{{b}\div{a}}=\frac{1}{{b}\div{a}}\;,\;\;\;\mathrm{e.g.}\;\;\frac{2}{5}=\frac{1}{{5}\div{2}}=\frac{1}{2.5}

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.

Figure 1

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.

§

Figure 2

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.

§ §

Fig. 3

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.

Figure 4

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.

Figure 5

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:

\displaystyle\frac{Y}{X}\;=\;\frac{bc}{ac}\;=\;\frac{af}{df}

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.)

Figure 6

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):

Plato’s Divided Line potential diagram candidate #2

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.

Figure 7

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.

Figure 8. Three examples of reflecting the horizontal line through a “mirror” to the vertical line (or vice versa). Note how the proportion is preserved even though the size changes. Also note that only in the middle example, with equal length segments, are the diagonals parallel.

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.

§ §

Figure 9

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:

\displaystyle\mathrm{Area}(cbe)={A}_{Y}=\frac{1}{2}\left[{Y}\,(U\!\!+\!\!V)\right]={Y}\left(\frac{U\!\!+\!\!V}{2}\right)\\[1.2em]\mathrm{Area}(acd)={A}_{X}=\frac{1}{2}\left[{X}\,(U\!\!+\!\!V)\right]={X}\left(\frac{U\!\!+\!\!V}{2}\right)

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:

\displaystyle\mathrm{Slope}(be)={S}_{Y}=\frac{Y}{U\!\!+\!\!V}={Y}\left(\frac{1}{U\!\!+\!\!V}\right)\\[1.2em]\mathrm{Slope}(cd)={S}_{X}=\frac{X}{U\!\!+\!\!V}={X}\left(\frac{1}{U\!\!+\!\!V}\right)

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.

Figure 10

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:

\displaystyle\mathrm{Height}_{X}(k)={S}_{X}\!\times\!{V}=\frac{X}{U\!\!+\!\!V}\!\times\!{V}=\frac{X\!\times\!V}{U\!\!+\!\!V}

And the height of point k on slope Sy (starting from point e and heading left because we must start from zero) by:

\displaystyle\mathrm{Height}_{Y}(k)={S}_{Y}\!\times\!{U}=\frac{Y}{U\!\!+\!\!V}\!\times\!{U}=\frac{Y\!\times\!U}{U\!\!+\!\!V}

As we did with the slopes above, we can factor out the 1/(U+V), leaving us with:

\displaystyle\mathrm{Height}_{X}(k)={X}\!\times\!{V}\,,\;\;\;\mathrm{Height}_{Y}(k)={Y}\!\times\!{U}

We know from above that V/U=Y/X and that leads to the following identity:

\displaystyle\frac{Y}{X}=\frac{V}{U}\;\;\therefore\;\;{Y}\!\times\!{U}={X}\!\times\!{V}

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.

Geometrically speaking.

§

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:

Plato’s Divided Line potential diagram candidate #1

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.

Doing it the old-fashioned way without graph paper to help!

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!)

§

Figure 11

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:

Figure 12. Moving the mechanism around to different sizes and ratios. Each row represents a given ratio from 1:1 (top) to 7:1 (bottom). The columns show different sizes for the mechanism.

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.

About Wyrd Smythe

The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe

17 responses to “Plato’s Divided Line

  • Wyrd Smythe

    Appendix I

    As a reference point for what Plato’s Divided Line is all about, here is a table I made that tried to summarize the content I’d cribbed from other examples:

    On the edges are a bit of example math (left) and the abstract algebra (right) that, respectively, demonstrate and prove the equality.

  • Wyrd Smythe

    Appendix II

    If you want to explore the mathematical side, here is how to derive the necessary numeric values for each of the four segments:

    Given any desired ratio:

    \displaystyle\frac{b}{a}\;=\;\frac{b\div(a\!+\!b)}{a\div(a\!+\!b)}

    Calculate:

    \displaystyle\frac{b}{(a\!+\!b)}\times\frac{b}{(a\!+\!b)}\;=\;\frac{b^2}{(a\!+\!b)^2}\\[1.0em]\frac{b}{(a\!+\!b)}\times\frac{a}{(a\!+\!b)}\;=\;\frac{ba}{(a\!+\!b)^2}\\[1.0em]\frac{a}{(a\!+\!b)}\times\frac{b}{(a\!+\!b)}\;=\;\frac{ab}{(a\!+\!b)^2}\\[1.0em]\frac{a}{(a\!+\!b)}\times\frac{a}{(a\!+\!b)}\;=\;\frac{a^2}{(a\!+\!b)^2}\\[1.0em]

    And that provides the values for any b/a. (The middle two, of course, are equal.)

    For example, given 5/3…:

    \displaystyle\frac{5}{(3\!+\!5)}\!\times\!\frac{5}{(3\!+\!5)}\;=\;\frac{5}{8}\!\times\!\frac{5}{8}\;=\frac{5^2}{8^2}\;=\;\frac{25}{64}\\[1.0em]\frac{5}{(3\!+\!5)}\!\times\!\frac{3}{(3\!+\!5)}\;=\;\frac{5}{8}\!\times\!\frac{3}{8}\;=\frac{5\!\!\times\!\!3}{8^2}=\frac{15}{64}\\[1.0em]\frac{3}{(3\!+\!5)}\!\times\!\frac{5}{(3\!+\!5)}\;=\;\frac{3}{8}\!\times\!\frac{5}{8}\;=\frac{3\!\!\times\!\!5}{8^2}=\frac{15}{64}\\[1.0em]\frac{3}{(3\!+\!5)}\!\times\!\frac{3}{(3\!+\!5)}\;=\;\frac{3}{8}\!\times\!\frac{3}{8}\;=\frac{3^2}{8^2}\;=\;\frac{9}{64}

    QED! 😉

  • Wyrd Smythe

    Appendix III

    The debate over Plato’s Divided Line is from the perceived tension between the easy-to-show equality of the middle segments and Plato’s discussion of what each segment means. (As the table in Appendix I above shows, as Plato explores them, these segments are seen from multiple perspectives and are rich in meaning.) The question arises: Did Plato know the middle segments were equal? If so, why does he imply each level is greater than the one below it? Was he in error? Was he kidding? Was he creating a deliberate paradox to foment thought?

    Note, however, that Plato never explicitly says each segment is larger than the one below. That (as I understand it) is only implied by his statements (although, again as I understand it, the implication is quite strong, hence the debate).

    My take, for whatever it’s worth, is that, of course Plato knew the segments were equal. I don’t see how he would have gotten that wrong, the Greeks were plenty facile with numbers and geometry. Therefore, I think he was either teasing or, more likely, creating a deliberate paradox for followers to chew on.

    My (very uninformed) guess is that crossing the major line, the first division of unequal parts, from Opinion to Knowledge, causes a bit of a “reset” that puts one in new and unfamiliar territory — the realm of nontangible thought and knowledge. And while abstractions, understanding, and critical thinking, may morally and intellectually be closer to “the Good”, larger in spirit, that they are a new beginning reduces them in some sense creating the equality between our experience of the tangible world and this new experience of abstraction.

    I tried to echo this in the post, first moving from tangible example fractions to an abstraction that collects their meaning, and then moving from example geometry to an abstraction that doesn’t require examples.

    • Wyrd Smythe

      Up until last night, all the examples of the Divided Line I’d seen made the upper segments larger than the lower with the implication that moving towards the Good was greater. Tina sent me a link to a video where the lecturer had the lower segments larger (the middle ones still being equal, of course). He claimed doing it the other way was an error. For him, segment D, at the top, was the smallest. (Perhaps because it’s the rarest?)

      (I’ll have to watch that bit again to see exactly why. It was something about the proportions. At the time I was more interested in his geometry, which presented a different approach to subdividing the major segments. I plan to show that in an Appendix below once I create and scan a diagram.)

  • Wyrd Smythe

    Appendix IV

    I mentioned above that Tina sent me a link to a presentation that divides the line using a different, and rather interesting, technique.

    Step 1: Begin, as before, with a vertical line intended to divide and sub-divide. In this case, however, use a compass to draw a large circle (with center point p) that covers most of the line.

    The two points where the circle intersects the line become points a and b, the bottom and top, respectively.

    Step 2: Pick an arbitrary point c to divide the line ab “into two unequal segments.” Draw a horizontal line perpendicular to point c that intersects the circle (on the left side) at point d.

    Step 3: Draw straight lines from points a and b to point d. This creates triangle adb inscribed inside the semi-circle.

    Note that, because we inscribed the triangle inside a half-circle, by Thales’s theorem, this is necessarily a right-triangle. The angle at point d is 90°.

    Step 4: Draw two lines (red in figure below) from point c to each side of the triangle such that they are perpendicular to the side they intersect.

    The intersection points are e (lower) and f (upper).

    Step 5: Draw two horizontal lines, one from point e and one from point f, such that they are perpendicular to the vertical line.

    These lines intersect the vertical line at points g and h, respectively.

    And that’s it. We’ve divided and sub-divided Plato’s Line per his instructions.

    Why does this work? We have a large triangle adb that we know to be a right-triangle. Inside the large triangle, we created two smaller triangles cfb and aec. By construction, these are also right-triangles, and each shares an angle with the larger triangle. They also both contain a 90° angle, which means they necessarily share the third angle. Thus, they are equivalent triangles — the same triangle but scaled.

    We also know the hypotenuse (long side) of the main triangle is ac+cb. Our ratio is ac/cb, so the upper (red) triangle has the same ratio to the large triangle as the upper segment cb has to the whole line. Likewise, the lower (red) triangle has the same ratio to the larger triangle as does the lower segment to the whole line.

    If point d divides the line by some ratio ac/cb, then points e and f necessarily also divide their respective segments by that ratio.

    In this case, rather than reflecting in a “mirror” we’ve scaled a triangle proportionally, but it’s the same basic idea. Start with the main division and then find a way to apply that to each segment formed by that division.

    It’s a clever trick. It does rely on Thales’s theorem, and that theorem does predate Plato as a theorem. Euclid proved it (in Elements) around 300 B.C. (after Plato), but the theorem, and perhaps even a proof, would have been known to him.

    What this technique doesn’t appear to do is prove the necessity of the equality B=C. It’s possible one exists, but nothing immediately springs out at me. The speaker in the video used algebra to prove the equality and used a compass to demonstrate the equality in the diagram but didn’t present a geometric proof.

    • Wyrd Smythe

      The technique above requires drawing a line from some point p such that it is perpendicular to a given line. The trick for doing that is similar to drawing a perpendicular line to some point on a given line (described in the post).

      Step 1: Using the point p as the center point, draw an arc that intersects the given line at points a and b.

      Step 2: Using points a and b as center points, draw two arcs that intersect on the opposite side of the line from point p.

      Step 3: Draw a line from point p to the intersection of the arcs. The new (blue) line will be perpendicular to the original (black) line.

  • Wyrd Smythe

    Appendix V

    Here is an algebraic proof of B=C.

    Step 1: Start with the four sub-segments (bottom to top), A, B, C & D. We also have the two original segments, X=A+B and Y=C+D, and that is our ratio Y:X (also known as the fraction Y/X).

    Thus, we have as axioms:

    \displaystyle\frac{Y}{X}=\frac{C\!+\!D}{A\!+\!B}=\frac{B}{A}=\frac{D}{C}

    Step 2: From the above, we can create two important equalities. The first is:

    \displaystyle\frac{D}{C}=\frac{B}{A}\;\;\therefore\;{D}\!\times\!{A}={C}\!\times\!{B}

    Because we multiply both sides by C and then by A, which cancels out the lower part of the fractions leaving the equality.

    The second equality is similar:

    \displaystyle\frac{D}{C}=\frac{C\!+\!D}{A\!+\!B}\;\;\therefore\;{D}\!\times\!(A\!+\!B)={C}\!\times\!(C\!+\!D)

    Because we multiply both sides by A+B and then by C, which again cancels out the lower part of the fractions leaving the equality.

    Step 3: The proof begins with the second equality above:

    \displaystyle{D}\!\times\!(A\!+\!B)={C}\!\times\!(C\!+\!D)

    Step 4: A bit of math gives us:

    \displaystyle({D}\!\times\!{A})+({D}\!\times\!{B})={C}\!\times\!(C\!+\!D)

    Because, on the left, we multiply D by (A+B).

    Step 5: Using the first of the two equalities in Step 2, we substitute D×A for C×B:

    \displaystyle({C}\!\times\!{B})+({D}\!\times\!{B})={C}\!\times\!(C\!+\!D)

    Step 6: A bit of math gives us:

    \displaystyle({C}\!+\!{D})\times\!{B}={C}\!\times\!(C\!+\!D)

    Because, on the left, we factor out the common B.

    Step 7: Finally, we see that:

    \displaystyle{B}={C}

    Because we can divide both sides by 1/(C+D), which leaves just B and C.

    QED

  • Anonymole

    Ugh. Looks like you had fun, in a masochistic kinda way.

    I kept reading those =! as Equal-NOT.

    • Wyrd Smythe

      Ha, yeah, or worse, the “factorial of equals” (whatever TF that is)!

      For me, exploring new territory like that (math, geometry, coding, a new language) is just fun. Effortful fun, to be sure, but it’s like runners who love running or those who love working out or those who love the effort of building something. There’s a joyful aspect to the progress and growth that more than compensates for the effort.

  • diotimasladder

    I’m lucky to have found someone who finds this sort of thing rewarding, and you’ve done a fantastic job!

  • diotimasladder

    Reblogged this on Diotima's Ladder and commented:
    Fantastic work on the mathematical features of Plato’s Divided line over at Logos Con Carne——using ancient methods!

  • diotimasladder

    I hope you don’t mind that I’ve reblogged this chez moi. I want as many people as possible to see it! 🙂

  • Wyrd Smythe

    I keep saying synchronicity rules my life. Here it is again. In this post, I casually mention The Mythical Man Month by Fred Brooks, and the next video by Computerphile is:

    I rest my case.

  • Back to Plato’s Line | Logos con carne

    […] Last February I posted about how my friend Tina, who writes the Diotima’s Ladder blog, asked for some help with a set of diagrams for her novel. The intent was to illustrate an aspect of Plato’s Divided Line — an analogy about knowledge from his worldwide hit, the Republic. Specifically, to demonstrate that the middle two (of four) segments always have equal lengths. […]

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