[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.)

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 (see figure below), bisect the segment at Then draw the segment perpendicular to . Draw a circle with center and radius Next, arrange the carpenter’s square so that one edge goes through , one edge is tangent to the circle, and the vertex, , sits on . Then .

I made an applet to illustrate this trisection.

Brooks’s proof used trigonometry. My question is: Is there a *geometric proof* that 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 and Then it suffices to show that Let be the point of tangency. Draw segment and extend to on Then draw segments and Because lines and are parallel, and hence is a parallelogram. This implies that is the midpoint of the diagonal so Moreover, Because is the hypotenuse of the right triangle and is the median, We see that so Finally, because is a chord of the circle, that is,

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

In the proof we are referring to the figure below. Let and Then it suffices to show that Let be the point of tangency of the carpenter’s square and the circle. Because Construct Because is a tangent line and is a radius of the circle, Because and are parallel. Thus Construct a point so that is a parallelogram. Notice that and are collinear. Let be the midpoint of . Construct the line segment and the circle with center and radius . Note that the circle passes through and —the first three because and the fourth because and is a diameter of the circle. Because is parallel to Finally, is a central angle and is an inscribed angle and both cut off the same arc of the circle; thus

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*Related*

I think there is no such proof, because there can’t be.

https://en.wikipedia.org/wiki/Angle_trisection

By:

pepe potamoon March 30, 2017at 12:19 pm

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.

By:

Dave Richesonon March 30, 2017at 12:22 pm

@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.

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puntomaupuntoon March 30, 2017at 1:19 pm

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.

By:

chorasimilarityon March 30, 2017at 2:12 pm

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

By:

Dave Richesonon March 30, 2017at 2:33 pm

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.

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chorasimilarityon March 30, 2017at 3:52 pm

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.

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chorasimilarityon March 30, 2017at 2:19 pm

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.

By:

ajakajaon March 30, 2017at 3:29 pm

[…] 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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Trisecare un angolo con riga, squadra e compasso | backup del Poston April 5, 2017at 4:24 pm