Posted by: Dave Richeson | March 30, 2017

A Geometric Proof of Brooks’s Trisection?

[UPDATE: we have a proof! I included it at the end of the blog post.]

Yesterday I was looking at a few methods of angle trisection.

For instance, I made this applet showing how to use the “cycloid of Ceva” to trisect an angle. (It is based on Archimedes’s neusis [marked straightedge] construction.)

Screen Shot 2017-03-30 at 10.38.54 AM

I also found David Alan Brook’s College Mathematics Journal article “A new method of trisection.” He shows how you can use the squared-off end of a straightedge (or equivalently, a carpenter’s square) to trisect an angle. (This is a different carpenter’s square construction than the one I wrote about recently.)

To perform the trisection of \angle ABC (see figure below), bisect the segment BC at D. Then draw the segment DE perpendicular to AB. Draw a circle with center C and radius CD. Next, arrange the carpenter’s square so that one edge goes through B, one edge is tangent to the circle, and the vertex, F, sits on DE. Then \angle ABF=\frac13\angle ABC.

brookstrisect

I made an applet to illustrate this trisection.

Screen Shot 2017-03-30 at 12.04.47 PM

Brooks’s proof used trigonometry. My question is: Is there a geometric proof that \angle ABF=\frac13\angle ABC? I spent a little while working on it yesterday and couldn’t find one. If you can, let me know!

UPDATE: We have a proof! Thank you Marius Buliga!

In the proof we are referring to the figure below. Let \alpha=\angle ABF and \beta=\angle CBF. Then it suffices to show that \beta=2\alpha. Let G be the point of tangency. Draw segment CG and extend FD to J on CG. Then draw segments DG, BJ, and CF. Because \angle CGF=90^\circ, lines CG and BF are parallel, and hence BFCJ is a parallelogram. This implies that D is the midpoint of the diagonal FJ; so DF=DJ. Moreover, Because FJ is the hypotenuse of the right triangle FJG and DG is the median, \angle DGF=\angle DFG. We see that \angle DFG=\angle ABF so \angle DGF=\alpha. Finally, because DG is a chord of the circle, \angle DCG=2\angle DGF; that is, \beta=2\alpha.

brookstrisectproof

Update 2: Andrew Stacey just sent me an another proof:

In the proof we are referring to the figure below. Let \alpha=\angle ABF and \beta=\angle CBF. Then it suffices to show that \beta=2\alpha. Let G be the point of tangency of the carpenter’s square and the circle. Because \angle BEF=\angle BFG=90^\circ, \angle DFG=\angle EBF=\alpha. Construct CG. Because FG is a tangent line and CG is a radius of the circle, \angle CGF=90^\circ. Because \angle CGF=\angle BFG= 90^\circ, BF and CG are parallel. Thus \angle BCG=\beta. Construct a point H so that BCGH is a parallelogram. Notice that F, B, and H are collinear. Let J be the midpoint of GH. Construct the line segment DJ and the circle with center J and radius GJ. Note that the circle passes through G, H, D, and F—the first three because HJ=GJ=DJ, and the fourth because \angle GFH=90^\circ and GH is a diameter of the circle. Because DJ is parallel to FH, \angle DJG=\beta. Finally, \angle DJG is a central angle and \angle DFG is an inscribed angle and both cut off the same arc of the circle; thus \beta=\angle DJG=2\angle DFG=\alpha.

brookstrisectproof2

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Responses

  1. I think there is no such proof, because there can’t be.
    https://en.wikipedia.org/wiki/Angle_trisection

    • It is impossible to trisect an arbitrary angle with a compass and straightedge, but if you allow certain other tools—like the carpenter’s square in this case—then angle trisection is possible. There is a proof that this works, but it uses trigonometry, not geometry.

    • @pepe: as Dave says, this proof uses a tool different from straightedge and compass. In other words, without the carpenter’s square you cannot construct right angle BFG with the given constraints.

  2. BF || GC and the segments BD and DC have equal lengths. Now denote by J the intersection of the lines FD and GC. Then BFCJ is a parallelogram so the segments FD and DJ have equal length. The triangle FJG is right, therefore DG and FD have equal lengths. So the angles ABF, DFG and DGF are all equal. Use the triangle DCG to prove that the angle DCG is twice the angle DGF.

    • Very nice! Thank you. I’ll update my post in a few minutes!

      • Sorry for the double comment, i thought the computer ate my first one :) Very nice post and construction. From the proof I suspect the construction fails in interesting ways in hyperbolic or elliptic geometry.

  3. Denote by J the intersection of the lines FD and GC. The angles BFG and FGC are right, so BF is parallel with GC. Then BFCJ is a parallelogram, so FD and DJ have equal lengths. The triangle FJG is right so FD and DG have equal lengths, therefore the angles DFG and DGF are equal. Use that HGC is straight and the triangle DCG to prove that the angle DCG is twice the angle DGF.

  4. Hi, random amateur reader here.

    I found it interesting to notice that it’s not the use of a carpenter’s square, specifically, that is needed in order to perform this construction — it’s the ability to find the point ‘F’, subject to the constraint that angle BFJ is a right angle. A carpenter’s square is one tool for doing that, but any method of ‘inverting’ the problem “BFJ is a right angle. solve for F” will do.

    You can do this by binary-searching for the position of F on the line segment ED – but this amounts to constructing an infinite series of points and taking their limit, and that’s equivalent to allowing yourself to compute 1/3 of an angle by using an infinite series of bisections like 1/3 = 1/4+1/16+1/256…, which is a known way of trisecting angles anyway. Infinite series approximations are not allowed, of course, in normal compass+straightedge constructions, so this doesn’t clash with the impossibility proof.

  5. […] Richeson di Division by zero mostra in questo post come sia possibile trisecare un angolo con riga, compasso e squadra: sì, proprio lo strumento […]


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