Wantzel: Constructibility Criterion

The criterion presented in today's blog was known to Carl Friedrich Gauss in 1796 when he presented his criterion for the constructibility of regular polygons. Even so, the proof of the criteroin was first presented in 1837 by Pierre Laurent Wantzel.

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

To simplify the proof, I offer the following postulates.

Postulate 1:

It is possible to draw a line between any two points that are already determined.

Posulate 2:

It is possible to draw a circle with a center at a point already determined and with the radius equal to the distance between any two points already determined.

Postulate 3:

If two lines are not parallel and not colinear, then they intersect at exactly one point.

Postulate 4:

If a circle's center is at a point on a line then the circle intersects the extended line at two points.

I will now use these postulates to present some lemmas:

Lemma 1: (0,v) is constructible if and only if (v,0) is constructible

Proof:








This follows from Postulate 2 above setting center A at (0,0) and radius = v. If C at (v,0) is constructible, then we can obtain B at (0,v). If B at (0,v) is constructible, then we can obtain C at (v,0).

QED

Lemma 2: (u,0) is constructible and (0,v) is constructible if and only if (u,v) is constructible















Proof:

(1) Assume that B at (u,0) and D at (0,v) are constructible.

(2) If B at (u,0) is constructible and D at (0,v) is constructible, then F at (u,v) is constructible since:

(a) Let BG be a line that intersects B at (u,0) and is parallel to the y-axis. [Euclid: Book I, Prop 12, see here]

(b) Let DE be a line that insersects D at (0,v) and is parallel to the x-axis. [Euclid: Book I, Prop 12, see here]

(c) Then F at (u,v) is constructible since it is the point where BG,DE intersect [See Postulate 3 above]

(4) Assume that F at (u,v) is constructible.

(5) Since F at (u,v) is constructible, we know that B at (u,0) is constructible and D at (0,v) is constructible since:

(a) Let FG be a line parallel to the y-axis that passes through F at (u,v) [Euclid: Book I, Prop 12, see here]

(b) Then B at (u,0) is the point where FG and the x-axis intersect. [See Postulate 3 above]

(c) Let EF be a line parallel to the x-axis that passes through F at (u,v) [Euclid: Book I, Prop 12, see here]

(d) Then D at (0,v) is the point where EF and the y-axis intersect. [See Postulate 3 above]

QED

Lemma 3:

If a line passes through two points (a₁,b₁) and (a₂,b₂), then it has the equation of form:

αX + βY = γ

where

α = b₂ - b₁
β = a₁ - a₂
γ = a₁b₂ - b₁a₂

Proof:

(1) This lemma is proved if it can be shown that both points satisfy the equation given.

That is, both equations, lead to γ = a₁b₂ - b₁a₂

(2) Case I: (a₁,b₁)

α(a₁) + β(b₁) = (b₂ - b₁)(a₁) + (a₁ - a₂)(b₁) = a₁b₂ - b₁a₂

(3) Case II: (a₂,b₂)

α(a₂) + β(b₂) = (b₂ - b₁)(a₂) + (a₁ - a₂)(b₂) = a₁b₂ - b₁a₂

QED

Lemma 4:

If we construct a circle with center (a₁,b₁) and radius equal to the distance between (a₂,b₂) and (a₃,b₃), then it's equation has the form:

X² + Y² = αX + βY + γ

where:

α, β, γ are rational expressions of a₁, a₂, a₃, b₁, b₂, b₃

Proof:

(1) The equation for this circle is (see Theorem 1, here):

(X - a₁)² + (Y - b₁)² = (a₂ - a₃)² + (b₂ - b₃)²

(2) Multiplying out the squares involving X,Y gives us:

X² - 2a₁X + a₁² + Y² - 2b₁Y - b₁²

(3) Moving everything but X²,Y² to the other side gives us:

X² + Y² = 2a₁X + 2b₁Y + [-a₁² -b₁² + (a₂ - a₃)² + (b₂ - b₃)²]

QED

Theorem 5: Criterion of Constructibility

Let O be a circle that with center at (0,0) and radius = 1 unit.

The point (u,v) is constructible from O if and only if (u,v) can be obtained from (1,0) by a sequence of operations of the following types:

(i) rational operations
(ii) extraction of square roots

Proof:

(1) I will prove the first part of the theorem using induction.

(2) For the base case, we assume that u=a or 0 and v=a or 0.

(3) To prove the base case, we have four cases to prove:

(a) (a,0),(0,0) [Constructible from the given]

(b) (0,a) [Constructible from (a,0) and Proposition 2 above if we set center at (0,0) and radius=a].

(c) (a,a) [Constructible from (a,0) and Lemma 1 above]

(4) So we assume that all points (u',v') are constructible.

(5) To complete the first half of the proof, we need to show that if (u',v') is constructible and (u,v) is obtained in a single operation from (u',v'), then (u,v) is constructible.

(6) Using Lemma 2 above, step #5 is achieved if we show that the following points are constructible:

(u'+v',0)
(u' -v',0)
(u'v',0)
(u'v'^-1,0) [assuming v ≠ 0 ]
(√u',0) [assuming u ≥ 0]

(7) (u'+v',0) and (u' -v',0) are constructible [From Postulate 2 above where the center is (u',0) and the radius=v and Postulate 4 above where (u'-v',0) and (u'+v',0) are the intersection of the circle and the x-axis.]

(8) (u'v',0) is constructible since:















(a) We are given that F at (0,v') is constructible.

(b) We are given that C at (u',0) is constructible.

(c) Let E at (0,b) be a point where b is less than v' and b is greater than 0 such that we consider b our unit of measure (i.e. b is length = 1 unit)

(d) Draw a line EC connecting E at (0,b) and C at (u',0) [See Posulate 1 above]

(e) Draw a line called FB parallel to EC and passing through F at (0,v') [Euclid: Book I, Prop 12, see here]

(f) Let B at (z,0) be the point where FB intersects with the x-axis. [See Postulate 3 above]

(g) Triangles EAC and FAB are similar since:

∠ EAC and ∠ FAB are the same.

∠ AEC is congruent to ∠ AFB and ∠ ACE is congruent to ∠ ABF since FB and EC are parallel [Euclid: Book I, Prop 29]

(h) So, we know that:

AC/AE = AB/AF [see Lemma 3, here]

(i) Since in step #8c above we assumed that AE = 1 unit, we have:

AC = AB/AF

or equivalently:

AB = AF*AC

(j) Since the length of AB is z [since A is at (0,0) and B is at (z,0)], the length of AF is v' [since A is at (0,0) and F is at (0,v')] and the length of AC is u' [since A is at (0,0) and C is at (u',0)], it follows that:

z = u'*v'

(9) (u'v'^-1,0) is constructible since:

(a) Using the same diagram as step #8 above, we are given that F at (0,v') is constructible.

(b) Let us assume that we are given B at (u',0) is constructible.

(c) Let E at (0,b) be a point where b is less than v' and b is greater than 0 such that we consider b our unit of measure (i.e. b is length = 1 unit)

(d) Draw a line EC connecting E at (0,b) with a point C at (c,0) on the x-axis. [Euclid: Book I, Prop 12, see here]

(e) We can use the same arguments as #8g above to establish that Triangles EAC and FAB are similar

(f) Using the same argument as #8i, we have:

AC = AB/AF

(g) Since the length of AB is u' [since A is at (0,0) and B is at (u',0)], the length of AF is v' [since A is at (0,0) and F is at (0,v')] and the length of AC is c [since A is at (0,0) and C is at (c,0)], it follows that:

c = u'/v' = u'*v'^-1

(10) Assuming u ≥ 0, then ( √u , 0) is constructible since:













(a) Assume that point B is at (0,0)

(b) Let D be a point on the x-axis at (d,0) such that BD = 1 unit.

(c) Let C be a point on the x-axis at (1+u',0) where 1 = BD. [See step #7 above]

(d) Let us a draw a circle O whose center is at the midpoint M of BC and whose radius is MC. [See Postulate 2 above]

(e) Let A at (1,a) be the intersection of O and a line drawn from D that is perpendicular to BC. [Euclid: Book I, Prop 11, see here]

(f) We can see that triangle BDA and triangle ADC are similar since:

Both ∠ BDA and ∠ ADC are right angles

And, ∠ BAC is a right angle [see Euclid: Book III, Prop 31]

So, ∠ BAD + ∠ DAC = 90 °

Since triangle ADC is a right triangle, we also have ∠ DAC + ∠ DCA = 90 ° [See Lemma 4, here]

So, we have shown that ∠ BAD ≅ ∠ DCA.

(g) Since the two triangles are equiangular, we have [see Lemma 3, here]:

DA/BD = DC/DA

(h) Since the length of DA is c [since A is at (1,c) and D is at (1,0)], the length of BD is 1 [since B is at (0,0) and D is at (1,0)] and the length of DC is u' [since D is at (1,0) and C is at (1+u,0)], it follows that:

c/1 = u'/c

Further, this gives us that:

c² = u'

which means that:

c = √u'

(11) To complete the second half of the proof, I will show that if a point (u,v) can be constructed from (1,0) then it can be obtained from (1,0) through a finite sequence of operations.

(12) To accomplish, I need to show that the following imply points obtained from (1,0):

(a) two lines intersecting
(b) two circles intersecting
(c) line intersecting with a circle

(13) A point at two lines intersecting implies that this new point is obtained from the previous points since:

(a) By Lemma 3 above, the two lines can be represented by:

α₁X + β₁Y = γ₁

α₂X + β₂Y = γ₂

(b) Assuming that α₁ ≠ 0, we multiply the top equation by (-α₂/α₁) and add the two equations together to get:

(-α₂/α₁)β₁Y + β₂Y = γ₁ + γ₂

which gives us that:

Y[(-α₂/α₁)β₁ + β₂] = γ₁ + γ₂

which shows that Y can be obtained from the given points.

Y = [γ₁ + γ₂ ]/[(-α₂/α₁)β₁ + β₂]

(c) Now, all we need to do is show that X can be obtained from the given points which can be shown using the first equation:

α₁X + β₁Y = γ₁

which then shows:

X = (γ₁ - β₁Y)/(α₁)

(14) A point at two circles intersecting implies that this new point is obtained from the previous points since:

(a) By Lemma 4 above, the two circles can be represented by:

X² + Y² = α₁X + β₁Y + γ₁

X² + Y² = α₂X + β₂Y + γ₂

which gives us:

α₁X + β₁Y + γ₁ = α₂X + β₂Y + γ₂

and moving X to one side and Y to the other gives us:

α₁X - α₂X = β₂Y - β₁Y + γ₂ - γ₁

(b) Now, assuming that (α₁ - α₂) ≠ 0, we can state X in terms of Y

X = [(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)

(c) Now, we can state Y as a quadratic equation since:

Using:

X² + Y² = α₁X + β₁Y + γ₁

We get:

{[(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)}² + Y² = α₁{[(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)} + β₁Y + γ₁

(d) Since a quadratic equation is solvable in terms of the coefficients (see Theorem, here if needed), we have shown that in this case, the points are obtained from the previous points.

(15) A point at a circle intersecting with a line is obtained from the previous points since:

(a) By Lemma 3 above, the line can be represented by:

α₁X + β₁Y = γ₁

(b) By Lemma 4 above, the circle can be represented by:

X² + Y² = α₂X + β₂Y + γ₂

(c) Assuming α₁ ≠ 0, we can restate X in terms of Y using step #15a since:

X = (γ₁ - β₁Y)/α₁

(d) Substituting into the equation in #15b gives us the following quadratic equation:

([γ₁ - β₁Y]/α₁)² + Y² = α₂([γ₁ - β₁Y]/α₁) + β₂Y + γ₂

(e) Since a quadratic equation is solvable in terms of the coefficients (see Theorem, here if needed), we have shown that in this case, the points are obtained from the previous points.

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001