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Once the magical forthcoming Everything2 image embedding capability is working, I'll revise this, but for now, I've uploaded a diagram of how this all works to my website; here's the direct link.

To trisect line segment AB:

  1. Construct two circles, each with radius AB, one centered around A and one around B.
  2. Label the two intersections of the circles "C" and "D."
  3. Label the intersections of these new segments "E" and "F."
  4. Construct line segments connecting E and F to B and A. (They're already connected to A and B, so you just need to connect them up the other way.)
  5. Label the intersection of AD and BE "G". Label the intersection of BD and AF "H."
  6. Construct line segments connecting C to G and H.
  7. Label the intersections of CG and CH with AB "I" and "J." These points trisect AB.

An edit: I, in my infinite wisdom, assumed I had determined a general method for trisecting an angle based on looks alone. I've withdrawn this, and turned in my mathematician's badge and gun. I'm sorry.

I enjoyed Major General Panic's exposition on trisecting a line. Looks cool and is interesting!

His claim that trisecting a line is equivalent to trisecting an angle, though, is... optimistic, to put it kindly. Two non-expert refutations come to mind:

  • If trisecting an angle really is that easy, then surely he's not the first to consider using a trisected line to trisect an angle? And then the problem would long be solved.
  • If he found a way to trisect an angle, he would have disproved the proofs of those other folks who claim it's not possible. Given that mathematical proofs are usually rigorous, it's hard to envision a fundamental one like this simply falling over backwards.

I'm not a math wiz, but I think I see the fallacy in his thinking: Line AB is perpendicular to line CD, which bisects the angle ACB. So is line IJ. Simply stated, line IJ is "straight across" angle ICJ, it forms the base of an isosceles triangle. AI and JB, on the other hand, do not form isosceles triangles with the vertex at C. They're the same length as IJ, but they're not "straight across" from their respective angles; therefore, they subtend different angles! A geometry-constructing tool or careful measurement of a constructed diagram will reveal that the angles ACI and JCB are different from ICJ.

So yes, the angle is cut into three angles, but they're not equal, so the requirement of trisection is not met.

Next!

Just in case you don't believe that the line AB above is trisected, I can show that it is.

For this, I'm going to rely on Cartesian coordinates. I've chosen not to use standard Euclidian proofs, simply because my geometry is not good enough to do that.

Take point A as the origin of the Cartesian coordinate system, (0,0)

Assume point B lies at (12,0). Call it (12N,0), if you really want to be pernickety.

ABC and ABD are equilateral triangles, so point C is at (6,6√3), while D is at (6,-6√3).

ABDE is a rhombus with side 12*, which means that G is the mid-point of AD. So if D is at (6,-6√3), then G is at (3,-3√3).

* It is easy to prove this using the lengths of the lines and the angles. We know that AE, AD and AB are the same length (they are all radii) while the angles DAE and DAB are both exactly 60°C. This means the two triangles, ABD and ADE are both equilateral, which makes ABDE a rhombus.

The line CG joins C (6,6√3) and G (3,-3√3). The slope of this line is represented by the difference in y coordinates divided by the difference in x coordinates, or (9√3/3), which is 3√3.

So the equation for line CG is y=3√3x + A. Solving for A using the two given points gives A=-12√3.

Thus, CG is represented by y=3√3x - 12√3.

Solving for y=0, or the intercept point I, gives x=4.

By symmetry, J is also 4 units from point B.

Thus we can see that I and J do indeed trisect the line AB.

Congratulations to Major General Panic and his students.


As to whether that also trisects the angle, Elrac is correct: it doesn't.

While the line is accurately trisected, the angle subtended by the line segment IJ at point C are not the same as the angles subtended by AI or BJ.

MGP's suggested technique is to use the trisected line to create a trisected angle. Use the length AB to subtend the relevant angle, and then use the trisections of the line to divide that angle into three parts

While the three angles might look kind of similar, especially where the angle at C is acute, the limitations of this technique emerge when the angle at C is not only obtuse, but close to 180°.

Say you wanted to trisect the angle of 179°, and used MJP's technique. Each of the three angles should be just under 60°.

However, if the point is close to line AB, in order to subtend an angle of 179°, then the middle portion will still subtend an obtuse angle, while the two outer portions will clearly be much less that 60°.

Many budding geometers feel a twinge (or more) of excitement when they learn to trisect an arbitrary line segment. Not because it's necessarily an elegant construction—there are constructions for dividing a segment at any integer ratio that are perfectly formulaic—but because it seems to tiptoe towards the taboo, perhaps to create somehow the so-called impossible: The trisection of an arbitrary angle.

I certainly did. The compass and straightedge afford a lot of leverage; surely they can wrest a trisection from segment to angle. No exclamation was ever sweeter than Ha! And they said it couldn't be done!

Unfortunately, as many as have exclaimed thus over the trisection problem have done so incorrectly. A true solution remains a long way off, for algebraic reasons: As explained by m_turner, trisecting an arbitrary angle requires solving a cubic, and the compass and rule only handle quadratics and linears (respectively). But the segment-trisection approach continues to entice, because it looks so reasonable, so much of the time.

Here's why: People always try it on relatively small angles.

By some quirk of human psychology, we think of acute angles as generic. By another quirk of human psychology, we tend to pick test cases that tell us what we already 'know', providing less advantage than we deserve. And by a quirk of geometry, the smaller the angle the more it looks like we've done it. Here, a triple pitfall for the clever and eager.

Trisecting an angle is equivalent to trisecting the arc between its rays. The arc is approximated by a line segment joining the two rays at the same points. So trisecting the segment approximately trisects the arc and the angle. The smaller the angle is, the nearer the arc falls to the segment, and the better the trisection. If the angle is fairly small, the approximation is better than the accuracy of your compasses, and if the angle is zero, the trisection is exact.

But if the angle is large ... well, let's try it. Empirical counterexamples are fine, it's just empirical proofs you have to be wary of. So rule an arbitrary line, and mark off points ABCD on it at equal intervals:

-------A-----B-----C-----D------

Construct the perpendicular bisector of AD (call it EF):

                E
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-------A-----B--+--C-----D------
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                F

Now we can pick any point G on line EF, and AD will be the line segment that approximates the arc. If we pick a G far from the line, the angle is small, and the trisection looks OK:

                E
                G
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-------A-----B--+--C-----D------
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                F

You'll have to visualize the line segments connecting G to A, B, C, and D. Visualize also the arc AD with center G: Pretty close to the line. But what if G is much closer?

                E
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                G
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-------A-----B--+--C-----D------
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                F

Now it's pretty clear that angle BGC is larger than AGB and CGD are. Keep moving G closer, and BGC continues to increase, approaching 180°, whereas the other two decrease towards 0°. So this is not a trisection. Visualizing the arc shows why: It's substantially longer than the segment, and more of its length is matched to the BC segment than to segments AB or CD.

Thus concludes our foray into geometry. Tune in next week to Scientific European Frontiers, where you'll hear Kepler say,

"I've got epicycles coming out my nose, and I'm not gonna take it anymore!"

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