“Neusis” constructions (2): trisections

Here we show how the use of neusis can be used to trisect arbitrary angles. In fact, we shall give four different constructions: two ancient, one medieval, and one modern.


Archimedes of Syracuse 287 BC — c. 212 BC) is not only unquestionably the greatest mathematician of antiquity, but one of the greatest of all time. His writings reveal a mind of unsurpassed inventiveness, as well as extraordinary subtlety and suppleness of thought. His trisection is perhaps the best known, and maybe the simplest. Suppose the acute angle AOB=\theta to be trisected is placed in a circle of radius one and diameter AOC. Extend the diameter of the circle, and from B construct a line which crosses the circle and the diameter extension at P and Q, and for which PQ=1. (This is the neusis construction.) Here is what it looks like:

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Then the angle CQP=\theta/3.

This is easily proved by some angle chasing. Construct the line OP. Since OP=PQ=1 the triangles OPQ and OPB are isosceles. Suppose angle CQP=x, then angle COP=x and so angles OPB=OBP=2x. Then angle BOP=\pi-4x and by adding all the angles at O: \theta + (\pi-4x)+x=\pi , or \theta=3x.


Pappus of Alexandria (c. 290 — c. 350 AD) is still considered an “ancient” mathematician, even though he succeeded Archimedes by nearly half a millennium. He wrote at a time when mathematics (and science generally) was at a low ebb; his achievements are all the more remarkable.

Suppose AOB is the angle to be trisected, and let OA=1. By using appropriate perpendiculars, construct a rectangle OBAC for which OA is a diagonal. Extend CA, and from O construct a line crossing AB at P and CA at Q and for which PQ=2. Then angles AQP=POB=\theta/3.

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The proof is again easy angle chasing: let M be the midpoint of PQ and construct AM. All these new lengths are the same, so that triangles AMQ, AMP and OAM are isosceles. The rest is straightforward.

Jordanus Nemorarius

Jordanus (1225 — 1260) is a shadowy figure about whom nothing is known aside from some books which bear this name. One theory is that “Nemorarius” is simply a copying error by a later scribe; at any rate there is no baptismal or any other record of “Nemorarius”. Although he has been described in Carl Boyer’s history as an “able, but less gifted contemporary” of Leonardo of Pisa (Fibonacci), his works on geometry, statics and mechanics embody the best thought of his time. His trisection is a neat and elegant variant of Archimedes, and again starts with a semicircle with the angle to be trisected. Construct the radius OD perpendicular to AOC, and from B draw a line cutting this radius and the circle at P and Q respectively, and for which PQ=1. Then angle QOC=\theta/3.

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The proof is pretty much the same as for Archimedes’ method: since triangle OPQ is isosceles (by construction) if angle QOC=x then angles OQP=OBP=2x. The rest of the proof is as above.


David Allan Brooks published a delightful trisection method in The College Mathematics Journal in 2007. Brooks is not a professional mathematician, and so his work gives added credence (should any be needed) to excellent work produced by amateurs. This method is in fact not a neusis method, as it uses a right angled ruler (a “carpenter’s square“) instead of two marks, but it very much in the spirit of neusis. Let AOB=\theta be the angle to be trisected, and let M be the midpoint of OB. Construct a circle C with centre B and radius BM and from M construct the perpendicular MD to OA. Now lay the carpenter’s square so that one side goes through O, the right angle lies on MD at E and the other edge is tangent to C. Then the angle

Here is a picture:

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To prove it, suppose the tangent crosses OM at F, and construct the radius BG to the tangent. Let angles DOE and EOF be x and y respectively, so that x+y=\theta. By similar triangles. angle MBG=y. We shall assume that OM=MB=BG=1.

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Since BG=1 we have FB\cos y=1 and so FB=1/\cos y, OF=2-1/\cos y. Then OE=(2-1/\cos y)(\cos y)=2\cos y-1 and so OD=(2\cos y-1)\cos x. But considering triangle ODM we also have OD=\cos(x+y), and so (2\cos y-1)\cos x=\cos(x+y). Expanding both sides produces

    \[ 2\cos y\cos x-\cos x=\cos x\cos y-\sin x\sin y \]

or that

    \[ \cos x\cos y+\sin x\sin y=\cos x \]

and this can be written as

    \[ \cos(y-x)=\cos x \]

so that y-x=x or y=2x. Given that x+y=\theta it follows that x=\theta/3. This is a slightly different proof than that given in Brook’s article, and maybe slightly simpler, in requiring less extra points and lines.

One thought on ““Neusis” constructions (2): trisections

  1. I am the writer of the College Mathematics Journal article referred to.

    After it appeared, I was challenged to develop a method that did not resort to using a second edge of the ruler. In other words, I was to use only a compass and a straightedge with length but no area. And, of course, I was not permitted to place any marks on the straightedge, because Archimedes and others have already been there!

    On the face of it, this sounded exactly like the long-sought method of trisection that we all know is impossible. However, because a real-world straightedge cannot be infinitely long, I did find a way to do it. It is simple, elegant, and mathematically exact.

    No trisection cheat can possibly be brought closer to meeting the terms of the fruitless ancient challenge. Many math teachers, I suspect, would think I was claiming to have done the impossible if I declared without explanation that I use just a compass and one edge of an unmarked straightedge to trisect an angle.

    If you or any other readers are interested in seeing my method and proof for this ultimately challenging trisection cheat, please contact me at davidalanbrooks.uk@gmail.com.

    Congratulations on a great web site, Alasdair, and thank you for your kind words about my earlier trisection.


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