FURTHER INFORMAION ON NEUSIS CONSTRUCTIONSThis elaborates on some material in Section III.6 of the course notes. However, the discus-sion assumes familiarity with graduate level material in algebra, including the theory of algebraicextension fields and a little Galois theory.In their study of classical construction problems by means of (unmarked) straightedge andcompass, some Greek geometers studied other means for making such constructions, and one ofthese is the neusis process described in Section III.6. There are actually a few different versions ofthis, but we shall limit ourselves to the following:Suppose that we are given a pair of intersecting lines L and M in the plane, a point x which lieson neither, and a distance a ≥ 0. Then it is possible to construct all pairs of points y ∈ L andz ∈ M such that x, y and z are collinear and the distance from y to z is equal to a.We have three main goals. The first is to show the following result:THEOREM 1. Suppose that we are given L, M, x and a as above, and let F be a Pythagoreansubfield of R such that the lines are given by linear equations with coefficients in F, the coordinatesof x are in F, and a ∈ F. Then the coordinates of y and z lie in a finite extension of F obtained byadjoining the roots of a quartic (degree 4) polynomial with coefficients in F.Note. A subfield F ⊂ R is said to be Pythagorean if for each b ∈ F the square roots of 1 + b2alsolie in F, and the subfield is said to be surd-closed (or Euclidean) if it is closed under taking thesquare roots of all nonnegative numbers. It follows that if F is Pythagorean and (a, b) ∈ F2, then√a2+ b2∈ F.In the course notes, there are references to proofs that angles can be trisected and cubes canbe doubled by a finite sequence of straightedge, compass and neusis constructions. Here is anotheronline reference in the course directory:http://math.ucr.edu/∼math133/neusis-classical.pdfSignificant portions of the cited document are devoted to proving existence proofs for the pointsin question. Usually it seems “obvious” that one can find points on a pair of lines such that thedistance is some predetermined number, and by Theorem 1 it is possible to write down equations forthe coordinates for such lines, but from a mathematical viewpoint it is also necessary to prove thatthese equations have solutions over the real numbers. We shall do this in the following situation,which turns out to include the given constructions for angle trisection and cube duplication:THEOREM 2. Suppose that we are given three noncollinear points A, B, C in the coordinateplane, and let X be a point which lies on the same side of BC as A, but not in the interior of6ABC or on this angle. Then for each r > 0 there are points Y ∈ (BA and Z ∈ (BC such thatX ∗ Y ∗ Z and d(Y, Z) = r.The final objective here is to show that one cannot square the circle using a finite sequenceof such constructions. Since π is not the root of a nontrivial polynomial with rational coefficients,this will follow immediately from the following immediate consequence of Theorem 1:PROPOSITION 3. Suppose that x ∈ R2is obtained from points in Q2by a finite sequence ofsequence of straightedge, compass and neusis constructions. Then x lies in a subfield E ⊂ R such1that E is a vector space over Q whose (finite) dimension has the form 2b3cfor some nonnegativeintegers b and c.Proof of Proposition 3. By an inductive argument it will suffice to prove that if we are given acollection of lines and points such that (i) the coordinates of the points lie in some subfield F, (ii)the lines are defined by linear equations whose coefficients lie in F, and x is obtained from thesepoints and lines by a classical or neusis construction, then the coordinates of x lie in an extentionE of F such that the dimension of E over F divides 24 = 23· 3. By Theorem 1 we know that thecoordinates of x lie in an extension obtained by adjoining roots of some polynomial of degree (atmost) 4, and therefore it follows that the dimension divides the order of the symmetric group on 4letters, which is 24.There is one further consequence that we shall mention here. Namely, it is not possible toconstruct a regular 11-gon using straightedge, compass and neusis constructions. This followsfrom the corollary because cos(2π/11) satisfies a minimal irreducible polynomial of degree 5 overthe rationals; by the corollary, the coordinates of a point obtained by a sequence of straightedge,compass and neusis constructions must satisfy a minimal irreducible polynomial over the rationalswhose degree is given by 2u3vfor some nonnegative integers u and v.Change of coordinatesIn nearly all uses of coordinates to prove geometrical statements, it is convenient to choosecoordinates so that the computations become as simple as possible, and Theorem 1 is no exceptionto this principle. Specifically, it will be useful to have the following:CLAIM. It will suffice to prove the result when the line L is the x−axis, the line M meets L at theorigin, and x is a point whose second coordinate is positive.Verification of Claim. Let w be the point where L and M meet. It follows immediately (forexample, by Cramer’s Rule) that the coordinates of w must lie in F. Therefore, if A is the affinetransformation of R2which translates a vector v to v − w, then A sends F2into itself. If we canprove Theorem 1 when the two lines meet at the origin, then in the general case it will follow thatthe result is true if we translate everything using A, so that we have points y0and z0. But now wecan take the translates of these points under A−1(in other words, translate by w) to obtain thedesired points y and z. Thus we see that it suffices to prove the result when L and M meet at theorigin, and we shall assume this holds for the rest of this argument.Next, we need to show that it is enough to consider the case where L is the x−axis. SinceL is defined by a linear equation with coefficients in F, it follows that there is a nonzero point(p1, p2) ∈ L with p1, p2∈ F. The length of this vector is in F because the latter is Pythagorean,so we may divide by its length to obtain a unit vector (q1, q2) ∈ L ∩ F2. If B is the rotation of R2given by the matrixq1−q2q2q1then B maps L to the x−axis, and as before if we can prove the result when L is the x−axis (andthe two lines meet at the origin), then we can recover the result in the general case.Finally, assume the conditions in the