Poncelet Figures over Rational Numbers and over Real Quadratic Number Fields

faculty of mathematics and natural sciences

Poncelet Figures over Rational Numbers and over Real Quadratic

Number Fields

Bachelor Project Mathematics

July 2017

Student: T. Mepschen First supervisor: Prof.dr. J. Top Second supervisor: Dr. M.C. Kronberg

Poncelet Figures over Rational Numbers and over Real Quadratic

Number Fields

Tiemar Mepschen July 14, 2017

Abstract

Poncelet figures are polygons whose vertices lie on one conic section and its edges are tangent to a second one. In this thesis we construct examples of Poncelet figures with 7, 8, 9, 10 and 12 vertices over Q and Poncelet figures with 11, 13, 14, 15, 16 and 18 vertices over real quadratic number fields.

2

CONTENTS 3

Contents

1 Introduction 4

2 Preliminaries 4

3 Poncelet Figures over Rational Numbers 5 4 Poncelet Figures over Real Quadratic Number Fields 7

5 General Method 9

6 Other Conic Sections 10

7 Conclusion 12

8 Poncelet Figures 13

Introduction 4

1 Introduction

Poncelet figures are polygons whose vertices lie on one conic section and its edges are tangent to another. Given a field K ⊂ R, a Poncelet figure over K is a Poncelet figure for which both conic sections are given by an equation over K, and all vertices have coordinates in K. In his bachelor’s thesis, J. Los [1]

constructed Poncelet figures over Q with 3, 4, 5, 6 and 7 vertices. Moreover, he shows that no Poncelet figures over Q with n = 11 and n > 13 exist.

In his conclusion he conjectures that the method he used to construct the Poncelet figure with 7 vertices will also construct figures with 8, 9, 10 and 12 vertices, which are all other possibilities over Q. In this bachelor’s thesis, we will check the validity of the conjecture of Los. We also consider the problem of constructing Poncelet figures over real quadratic number fields. The goal is to give an example of Poncelet figures with n vertices for n ∈ {7, . . . , 16, 18}, which are all the possible cases that occur over real quadratic number fields.

2 Preliminaries

An elliptic curve is a smooth projective curve, given by a third degree ho- mogeneous polynomial and equipped with a distinguished point O. On an elliptic curve, we can define a group law, making it into an abelian group, with O as its unit element. Every elliptic curve has a so-called torsion sub- group. This is the group formed by all points of finite order on the elliptic curve. Given an elliptic curve E over Q, its points with coordinates in Q form a subgroup denoted E(Q). A famous theorem of Mazur [2] (see [3] for an amazing historical overview) states that

E(Q)tors ∼= (

Z/nZ, n ∈ {1, . . . , 10, 12}

Z/2Z ⊕ Z/2nZ, n ∈ {1, 2, 3, 4}

Any elliptic curve E where the point P = (0, 0) has order larger than 3 can be put in the special form

E : y²+ uxy − vy = x³− vx².

This has the property that the line X = 0 is tangent to E at P . Using this form, it is quite easy to find relations between u and v that have to hold to get a torsion point of the particular order you want. For example, if we want an elliptic curve with a torsion point of order 8, then we want to have

Poncelet Figures over Rational Numbers 5

4P = −4P ; Using the above form, we find that 4P = −uv + v²+ v

u²− 2u + 1 ,−u²v² + 3uv²− v³− 2v² u³− 3u²+ 3u − 1

 ,

−4P = −uv + v²+ v

u²− 2u + 1 ,u²v − 2uv²− 2uv + v³+ 2v²+ v u³− 3u²+ 3u − 1



Of course, we don’t want P to be a point of order 2 or 4, which means that we should have P 6= −P and 2P 6= −2P . Since −P = (0, −v), we see that v is nonzero. Furthermore, we have 2P = (−v, uv − v) and −2P = (−v, 0).

This implies that u is not 1. So, if we want 4P = −4P , then u and v should satisfy

−u²v² + 3uv²− v³− 2v² = u²v − 2uv²− 2uv + v³+ 2v²+ v or, equivalently,

−u²v + u²− 5uv − 2u + 2v² + 4v + 1 = 0

Using the programming language Magma, a parametrization in one variable of this curve is easy to find.

3 Poncelet Figures over Rational Numbers

We define the dual C^∨ of a conic section C to be the set of pairs (c, d) such that y = cx + d is tangent to C. Now, we will consider the set X = {(P, `) ∈ C<sub>1</sub>×C<sub>2</sub><sup>∨</sup>| P ∈ `}. If C₁∩C₂^∨consists (over some algebraic closure) of 4 distinct points and if a pair (P, `) is given, then X can be viewed as an elliptic curve.

For details on this we refer to [4] and [5]. The process of assigning to a pair (P, `) the “other” intersection point P<sup>0</sup> of ` and C₁, and the “other” line `<sup>0</sup> containing P<sup>0</sup> and tangent to C<sub>2</sub>, we denote by (P, `) 7→ (P⁰, `⁰).

Now, we are going to do the opposite. Given an elliptic curve E with equation

E : y²+ uxy + vy = x³+ vx²,

we want to construct C₁ and C₂ such that we can view E as such a set X, and moreover such that the process (P, `) 7→ (P<sup>0</sup>, `⁰) corresponds to the translation over T = (0, 0) on the elliptic curve E. If we do this, then we can get a Poncelet figure by considering the case where T is a point of order n, and the pair (P, `) we start with corresponds to a rational point P₀ on E.

Because T was a point of order n, we will get back where we started after n steps. This means that we obtain a Poncelet figure with 6 n vertices. We

Poncelet Figures over Rational Numbers 6

will require that P₀ is a point of infinite order, because otherwise P₀ might depend on T and we will trace a Poncelet figure of smaller order multiple times.

To view E as a set of the form X, we consider new variables b and c given by

b = x + v

y , c = x + v

−y − ux − v. Rewriting gives

x = −v b + c + bc

b + c + bcu, y = v(u − 1)c b + c + bcu. Substituting this gives the curve

b²c²v + 2v(b²c + bc²) + bc(u²− u + 2v) + v(b²+ c²) + (u − 1)(b + c) = 0.

Now, we set C₁ to be the parabola (b, b²), then we let C₂^∨ be parametrized by (x(c), y(c)) where

x(c) = −2vc² + (u²− u + 2v)c + u − 1 vc²+ 2vc + v

and

y(c) = −vc²+ (u − 1)c vc²+ 2vc + v . In other words, we have b² = x(c)b + y(c).

We now find the general formula for C₂ by substituting y = x(c)x + y(c) into the general formula a₁x²+ a₂xy + a₃y²+ a₄x + a₅y + a₆. Since this line should be a tangent line, we get that the discriminant is zero for all c. This gives us a nonlinear system of equations. Using the programming language Maple, we find that the general formula for C₂ is

(u⁴− 2u³+ 4u²v + u²− 12uv + 4v²+ 8v)x²+ 4v(u²− 2u + 1)xy + 2(u²− 2u + 1)ux + 4v(u − v − 1)y + u²− 2u + 1 = 0.

The process (P, `) 7→ (P<sup>0</sup>, `⁰) corresponds to (b, c) 7→ (b⁰, c⁰), described as follows. The new b⁰ is, for given c, the “other” solution to the equation in b, c. So

b⁰ = −b + x(c).

Next, substituting b⁰into the b, c-equation, c yields one solution and the other one is c⁰. Hence

c⁰ = −c + x(b⁰).

Poncelet Figures over Real Quadratic Number Fields 7

The first involution (b, c) 7→ (b⁰, c) corresponds to the involution P₀ 7→

(−v, 0) − P₀ on the elliptic curve. The second one (b⁰, c) 7→ (b⁰, c⁰) corre- sponds to P₀ 7→ (−v, uv − v) − P₀. Therefore the composition corresponds to translation over (−v, uv − v) − (−v, 0) = [4](0, 0).

Hence if (0, 0) is a torsion point of odd order N we obtain a Poncelet figure consisting of N vertices, and if the order is even we obtain a figure with less vertices.

Now that we have laid the ground work, all that is left is using the methods from the introduction to construct the u and v for the elliptic curve to have a point of order n and positive rank. Using the programming language Magma, we have created the following table of examples of the elliptic curves used to make the Poncelet figures in section 8. We remark that examples of pairs of conic sections defined over Q with the property that a Poncelet figure with N ∈ {5, 8, 10, 12} vertices based on these conic sections exist, were also constructed by Mirman [6, Section 3]. However, he did not insist that also the coordinates of all vertices in the figure are rational numbers.

Moreover Los [1] constructed Poncelet figures over Q with N vertices for each N ∈ {3, 4, 5, 6, 7} and the same was done by Malyshev [7] for N = 5.

Ord (0,0)

u v Point of infinite order

7 -55 -48 (30,198)

8 -⁴³³₅₀ −⁴⁸³₈ ⁷⁵₈ ,³⁷⁵₄

9 13 84 (6,18)

10 -¹⁹₁₁ ¹²⁰₁₂₁ ¹²₁₁,²⁵²₁₂₁
12 ⁴⁶³₈ −¹³¹⁹⁵₄₈ ¹⁴⁵₄₈,²¹⁰²⁵₃₈₄

As discussed above, with T = (0, 0) on the elliptic curve, the function b used here is invariant under the involution P₀ 7→ 2T − P₀ and c is invariant under P₀ 7→ −2T − P₀. As a consequence, the process (P, `) 7→ (P<sup>0</sup>, `⁰) in our case corresponds to translation over 4T on E.

4 Poncelet Figures over Real Quadratic Num- ber Fields

We will now consider the possibilities over a real quadratic number field.

These are extensions of degree 2 of the rational numbers, while still being a subfield of the real numbers. In other words, these are fields of the form Q(

√d) = {a + b√

d | a, b ∈ Q}, where d is a positive rational number that is

Poncelet Figures over Real Quadratic Number Fields 8

not a square. According to [8], an elliptic curve E over a quadratic number field K has a torsion subgroup isomorphic to one of the following groups

E(K)_tors ∼=











Z/nZ, n ∈ {1, . . . , 16, 18}

Z/2Z ⊕ Z/2nZ, n ∈ {1, . . . , 6}

Z/3Z ⊕ Z/3nZ, n ∈ {1, 2}

Z/4Z ⊕ Z/4Z.

To obtain an example of such E/K having a point of order n, one starts with y²+ uxy + vy = x³+ vx²

as before. Then (0, 0) ∈ E having order n yields a relation between u and v, and we want a solution u, v ∈ K for this relation. For example, if u and v satisfy

v²+ (u³− u²− 1)v − u²+ u = 0, then we can find constants r and s, given by

s = v + u³ − u

v + u² − u, r = −u² + u

v (1 − s) + s and the elliptic curve given by

y²+ 1 − s(r − 1)
xy − r − rs(r − 1)
y = x³− r − rs(r − 1)
x² has (0,0) as a point of order 13. The curve defined by this equation in u and v is called X1(13). In general, if we want an elliptic curve with torsion subgroup isomorphic to Z/nZ, then this curve is referred to as X1(n). A table of these X₁(n) is given in [9] and [10]. In all the cases that we are interested in, namely n ∈ {11, 13, 14, 15, 16, 18}, these X1(n) are elliptic or hyperelliptic. This means that they can be written in the form of v²₁ = f (u₁), where f is a polynomial.

We want to find elliptic curves with a point of order n over some real quadratic number field. To find the number field, we substitute values for u₁ and define the corresponding elliptic curve over the field Q(pf(u₁)). If this curve has positive rank, then we are done and we can apply the same process as in the previous section to find the corresponding Poncelet figures. The following table gives examples of the elliptic curves used to make the Poncelet figures in section 8. We have taken the example of the elliptic curve with a point of order 18 from [11] and transformed it so that (0, 0) has order 18.

General Method 9

Ord (0,0)

u v Point of infinite order

11 −¹⁴₃−¹₃√

73 −15 − 3√

73 (16 + 2√

73,²⁸₃+⁸₃√ 73) 13 ¹⁵¹₉ −¹⁰₉√

193 ¹³⁰⁶₃ −⁹⁴₃√

193 (125 − 9√

193, −6974 + 502√ 193) 14 ²⁵¹³₇₂₈+¹⁷³₇₂₈√

105 −₁₈₉₂₈⁵⁵⁵ −₁₈₉₂₈⁸⁷ √

105 (₄₇₃₂⁶⁰ +₄₇₃₂³ √

105, −₄₉₂₁₂₈¹³⁵ −₄₉₂₁₂₈¹⁵ √ 105) 15 ⁹⁹³₈₇₅+₈₇₅²√

345 ₁₀₉₃₇₅¹¹⁷⁵⁰+₁₀₉₃₇₅¹⁴⁶ √

345 (₃₁₂₅⁶⁵⁰−₃₁₂₅⁶⁶√

345, −^1299380_2734375+_2734375⁷⁵⁴²⁸√ 345) 16 121 + 39√

10 −3510 − 1107√

10 (−24 − 9√

10, 9402 − 2970√ 10) 18 ¹⁰¹⁰⁰₁₃₆₂₅−₁₃₆₂₅²¹ √

26521 −^13179867_37128125−_37128125⁸¹⁰⁰³ √

26521 (_1703125¹⁸⁸⁵²⁹+_1703125¹¹⁶¹ √

26521,23205078125^6481036062 +23205078125^39796758

√ 26521)

5 General Method

As we saw above, the method described here runs into a problem when applied to finding a Poncelet figure of even order.

A solution to this is to choose more “natural” transformations.

Before, we chose two new variables b and c and we got two involutions X → X,

(P, `) 7→ (P<sup>0</sup>, `), (P, `) 7→ (P, `⁰),

where P⁰ is the other point on ` and `⁰ is the other line that P is on. Now, we want to first choose these involutions and then find corresponding trans- formations. We choose the first involution to be

i₁ : P 7→ −P and we choose the second one to be

i₂ : P 7→ (0, 0) − P

so that i₂ ◦ i₁ maps P to P + (0, 0). We now search for invariants of i₁ and i₂ in the sense that i^∗_kf := f ◦ i_k = id. The explicit formula for i₁ is

i₁(x, y) = (x, −y − ux − v).

We quickly see that x is invariant, because i^∗₁(x) = x. The explicit formula for i₂ is

i₂(x, y) = uvx + vy + v²

x² ,uv²x + v²x²+ v²y + v³ x³

 .

Other Conic Sections 10

Now, we can find that ^y+v_x is invariant. To see this, we calculate

i^∗₁ y + v x



=

uv²x+v²x²+v²y+v³

x³ + v

uvx+vy+v² x²

= uvx + vx²+ vy + v²+ x³ ux²+ xy + vx

= uvx + vy + v²+ y²+ uxy + vy ux²+ xy + vx

= (y + v)(y + ux + v) x(y + ux + v)

= y + v x . Now, if we set

b := y + v

x , c := x

and apply all the previous steps to get a Poncelet figure of order 8, we get Figure 6, which is exactly what we want. In this case, the dual conic C₂^∨ is parametrized as

(x(c), y(c)) = v − uc

c ,c²+ vc + uv c

 . The conic C₂ is given by

u²x²+ y²+ 2uxy + (−2uv − 4v)x − 2vy − v(4u − v) = 0.

6 Other Conic Sections

In the above discussion, we have used the parabola as one of the conic sections by default. Of course, we can use any other conic section. We will go over how to do this in this section.

We want to use the ellipse ^x_α²2 +^y_β²2 = 1 as one of our conics. This means that we want a rational parametrization for this conic. Such a parametrization is given by

 2b

1 + b²α,1 − b² 1 + b²β

 ,

where α, β ∈ Q. We now find functions x(c) and y(c) such that 1 − b²

1 + b²β = x(c) 2b

1 + b²α + y(c).

Other Conic Sections 11

This condition can be rewritten to b² = − 2αx(c)

β + y(c)b + β − y(c) β + y(c). From the previous section, we find the equations

− 2αx(c)

β + y(c) = v − uc

c , β − y(c)

β + y(c) = c²+ vc + uv

c .

Solving these equations, we get x(c) = β

α · uc − v

c²+ (v + 1)c + uv, y(c) = β−c²+ (−v + 1)c − uv c²+ (v + 1)c + uv . Using these functions, we find that the second conic is given by

C2 : − βu²x²+ 2α(uv + u + 2v)xy + α²(4uv − v²− 2v − 1)y²+

2αβ(uv − u + 2v)x + 2α²(4uv − v²+ 1)y + α²β(4uv − v²+ 2v − 1) = 0 We use these methods to construct examples of Poncelet figures of order 12 and order 14 defined over real quadratic fields, given in section 8.

We can do the same thing for the hyperbola _α^x22 − _β^y22 = 1. A rational parametrization for this conic is

 1 + b²

2b α,1 − b² 2b β

 ,

where α, β ∈ Q. We now find functions x(c) and y(c) such that 1 − b²

2b β = x(c)1 + b²

2b α + y(c).

This condition can be rewritten to b² = − 2y(c)

αx(c) + βb + β − αx(c) β + αx(c). From the previous section, we find the equations

− 2y(c)

αx(c) + β = v − uc

c , β − αx(c)

β + αx(c) = c²+ vc + uv

c .

Solving these equations, we get x(c) = β

α · −a²+ (−v + 1)a − uv

a²+ (v + 1)a + uv , y(c) = β ua − v

a²+ (v + 1)a + uv. Using these functions, we find that the second conic is given by

C₂ :β²(4uv − v²+ 2v − 1)x²+ 2αβ(4uv − v²+ 1)xy + α²(4uv − v²− 2v − 1)y²+ 2αβ²(uv − u + 2v)x + 2α²β(uv + u + 2v)y − α²β²u² = 0.

We use these methods to construct an example of a Poncelet figure of order 16 defined over a real quadratic field, given in section 8.

Conclusion 12

7 Conclusion

We have seen how to construct Poncelet figures of a certain order using two methods and have actually constructed examples of these figures in all pos- sible Q-rational cases for the number of vertices, and also for all the possible new cases that arise when we allow the coordinates of the vertices to be de- fined in some real quadratic field.

The general method can likely be extended to number fields of higher degree.

If one substitutes a certain value x₀ for x and find a root y₀ of the corre- sponding polynomial, then the elliptic curve will be defined over Q(y0) and we can apply all the theory in the above sections. We invite the reader to research this.

Poncelet Figures 13

8 Poncelet Figures

Figure 1: A Poncelet figure of order 7 over Q and the conics y = x² and 4569152x²− 5619712xy − 344960x − 702464y + 3136 = 0.

n point line

1 (−¹⁹₉,³⁶¹₈₁) y = −¹²⁹⁴³₆₀₈₄x − ₆₀₈₄²⁰⁹ 2 (−₆₇₆¹¹,₄₅₆₉₇₆¹²¹ ) y = −₃₇₈₅₆³¹⁵¹x − ₁₅₁₄₂₄¹⁶⁵ 3 (−₂₂₄¹⁵,₅₀₁₇₆²²⁵ ) y = −³⁴⁷³₂₂₄x −³⁷⁰⁵₃₅₈₄ 4 (−²⁴⁷₁₆,⁶¹⁰⁰⁹₂₅₆ ) y = −⁶²⁶³₄₀₀x −²⁷¹⁷₈₀₀ 5 (−¹¹₅₀,₂₅₀₀¹²¹) y = −₁₈₀₅₀³⁸²¹x +₁₈₀₅₀³³ 6 (₃₆₁³ ,₁₃₀₃₂₁⁹ ) y = −²³¹⁰²₄₃₆₈₁x +₄₃₆₈₁¹⁹⁵ 7 (−₁₂₁⁶⁵,₁₄₆₄₁⁴²²⁵) y = −²⁸⁸⁴₁₀₈₉x −¹²³⁵₁₀₈₉

Poncelet Figures 14

Figure 2: A picture of a Poncelet figure of order 9 over Q and the conics y = x² and 96912x²+ 48384xy + 3744x − 24192y + 144 = 0.

number point line

1 (5, 25) y = ³⁶₇x − ⁵₇ 2 (¹₇,₄₉¹) y = ₁₇₅¹⁸x +₁₇₅¹ 3 (−₂₅¹,₆₂₅¹ ) y = −²⁸₇₅x − ₇₅¹ 4 (−¹₃,¹₉) y = −³⁷₂₁x − ¹⁰₂₁ 5 (−¹⁰₇,¹⁰⁰₄₉) y = −²⁷₁₄x − ⁵₇ 6 (−¹₂,¹₄) y = −₁₆⁹x − ₃₂¹ 7 (−₁₆¹,₂₅₆¹ ) y = ₁₁₂¹ x +₂₂₄¹ 8 (₁₄¹,₁₉₆¹ ) y = ¹⁹₁₄x − ¹₇ 9 (2, 4) y = 7x + 10

Poncelet Figures 15

Figure 3: A Poncelet figure of order 11 over Q(√

73) and the conics y = x² and (⁶²⁰⁸⁶₈₁ −⁻¹⁶¹⁴²₈₁ √

73)x²+(−¹⁷¹⁶⁸₃ −²¹²⁸₃ √

73)xy+(−¹⁵¹⁰⁰₂₇ −¹⁶⁷⁶₂₇ )x+(−2896−

27√

73)y + (³⁶²₉ +³⁴₉√

73) = 0.

number point line

1 (−¹⁷₃₆ +₃₆¹ √

73,¹⁸¹₆₄₈ −₆₄₈¹⁷√

73) y = (−²⁵₂₄ +₇₂⁷ √

73)x + (−₁₄₄⁵⁹ +₁₄₄⁷ √ 73) 2 (−⁴¹₇₂+ ₇₂⁵√

73,¹⁷⁵³₂₅₉₂ −₂₅₉₂²⁰⁵ √

73) y = (−⁵⁸⁹₅₇₆ +₅₇₆⁴⁹√

73)x + (−²⁵⁹₇₆₈ +₇₆₈³¹√ 73) 3 (−²⁹₆₄+ ₆₄¹√

73,₂₀₄₈⁴⁵⁷ −₂₀₄₈²⁹ √

73) y = (−¹⁷⁵₁₉₂ − ₁₉₂⁵ √

73)x + (−₂₅₆⁴¹ − ₂₅₆³ √ 73) 4 (−¹¹₂₄ −₂₄¹√

73,₂₈₈⁹⁷ +₂₈₈¹¹√

73) y = (−²⁵₂₄ −¹₈√

73)x − (²⁵₄₈ − ₁₆¹√ 73) 5 (−₁₂⁷ − ₁₂¹√

73,⁶¹₇₂ +₇₂⁷ √

73) y = (⁻¹⁹₄₈ −₄₈⁷ √

73)x + (−¹³₄₈ +⁻¹₄₈√ 73) 6 (₁₆³ + ⁻¹₁₆√

73, ₁₂₈⁴¹ +₁₂₈⁻³√

73) y = (²⁸¹₄₈ +⁻³⁵₄₈ √

73)x + (⁻¹⁹⁷₄₈ + ²³₄₈√ 73) 7 (¹⁷₃ + ⁻²₃ √

73,⁵⁸¹₉ +⁻⁶⁸₉ √

73) y = (²⁵₃ +⁻¹₁ √

73)x + (⁻⁹⁴₃ + ¹¹₃√ 73) 8 (⁸₃ + ⁻¹₃ √

73,¹³⁷₉ +⁻¹⁶₉ √

73) y = (⁴⁹₁₈ +⁻⁷₁₈√

73)x + (⁻³₂ +¹₆√ 73) 9 (₁₈¹ + ⁻¹₁₈√

73, ₁₆₂³⁷ +₁₆₂⁻¹√

73) y = (⁻³⁰⁵₁₈ +⁻³⁷₁₈ √

73)x + (⁻⁴³₆ +⁻⁵₆ √ 73) 10 (⁻¹⁷₁ + ⁻²₁ √

73,⁵⁸¹₁ +⁶⁸₁ √

73) y = (⁻⁹³⁵₅₄ +⁻¹⁰⁹₅₄ √

73)x + (⁻¹⁴⁵₁₈ + ⁻¹⁷₁₈ √ 73) 11 (⁻¹⁷₅₄ +⁻¹₅₄√

73,₁₄₅₈¹⁸¹ + ₁₄₅₈¹⁷ √

73) y = (−₁₀₈⁸⁵ +₁₀₈¹ √

73)x − ¹₉

Poncelet Figures 16

Figure 4: A Poncelet figure of order 13 over Q(√

193) and the conics y = x² and (24037458536

6561 −^1730254040₆₅₆₁ √

193)x²+ (^412253056₂₄₃ −^29674624₂₄₃ √

193)xy + (^22880528₇₂₉ −

1646960 729

√193)x + (−^39464320₂₇ + ^2840704₂₇ √

193)y + (³⁹⁴⁶⁴₈₁ − ²⁸⁴⁰₈₁ √

193) = 0.

n point line

1 (¹¹₃₆+₃₆¹√

193,¹⁵⁷₆₄₈) +₆₄₈¹¹√

193) y = (−₁₇₆₄⁵⁵⁹ +₁₇₆₄³¹ )x + (¹²₄₉+₄₉¹√ 193) 2 (−⁶¹₉₈−₉₈¹√

193,¹⁹⁵⁷₄₈₀₂+₄₈₀₂⁶¹√

193) y = (−²⁸⁵⁷₁₃₂₃−₁₃₂₃³⁸√

193)x + (−¹⁴⁶₁₄₇−₁₄₇⁴ √ 193) 3 (−⁸³₅₄−₅₄¹√

193,³⁵⁴¹₁₄₅₈+₁₄₅₈⁸³√

193) y = (²⁸⁸⁷₅₄ +²¹⁵₅₄√

193)x + (⁵⁹³₆ +⁴³₆√ 193) 4 (55 + 4√

193, 6113 + 440√

193) y = (¹⁹⁰⁷₃₆ +¹³⁹₃₆√

193)x + (⁸⁷⁵₄ +⁶³₄√ 193) 5 (−⁷³₃₆−₃₆⁵√

193,⁵⁰⁷⁷₆₄₈ +³⁶⁵₆₄₈√

193) y = (−⁸⁰⁷⁰¹₃₄₅₉₆−₃₄₅₉₆⁵⁵²⁵√

193)x + (−⁴⁵²¹₃₈₄₄−₃₈₄₄³²⁵√ 193) 6 (−²⁹³₉₆₁−₉₆₁²⁰√

193,¹⁶³⁰⁴⁹₉₂₃₅₂₁+₉₂₃₅₂₁¹¹⁷²⁰√

193) y = (−₁₄₁₂₆₇⁶⁷⁰⁹⁶ −₁₄₁₂₆₇⁴⁸⁶² √

193)x + (−₄₇₀₈₉⁵⁰¹⁵ −₄₇₀₈₉³⁶² √ 193) 7 (−₁₄₇²⁵ −₁₄₇² √

193,₂₁₆₀₉¹³⁹⁷ +₂₁₆₀₉¹⁰⁰ √

193) y = (−²⁰¹¹₅₈₈ −¹⁵⁵₅₈₈√

193)x + (⁻²³⁷₁₉₆ −₁₉₆¹⁷√ 193) 8 (−¹³₄ −¹₄√

193,¹⁸¹₈ +¹³₈√

193) y = (²⁹₄ +¹₄√

193)x + (²³³₄ +¹⁷₄√ 193) 9 (²¹₂ +¹₂√

193,³¹⁷₂ +²¹₂√

193) y = (⁵¹₄ +¹₄√

193)x + (¹₂+³₂√ 193) 10 (⁹₄−¹₄√

193,¹³⁷₈ −⁹₈√

193) y = (⁴⁷₂₄−₂₄⁷√

193)x + (−⁶⁵₄₈+₄₈¹√ 193) 11 (−₂₄⁷ −₂₄¹√

193,¹²¹₂₈₈+₂₈₈⁷ √

193) y = (²¹³₁₉₂+₁₉₂¹¹√

193)x + (⁹²⁵₇₆₈+₇₆₈⁶⁷√ 193) 12 (²⁶⁹₁₉₂+₁₉₂¹⁹√

193,⁷¹⁰¹⁷₁₈₄₃₂+₁₈₄₃₂⁵¹¹¹√

193) y = (−⁸¹⁷₁₉₂−₁₉₂⁵⁹√

193)x + (⁴⁰¹⁵₂₅₆ +²⁸⁹₂₅₆√ 193) 13 (−¹⁸¹₃₂ −¹³₃₂√

193,³²⁶⁸⁹₅₁₂ +²³⁵³₅₁₂√

193) y = (−¹⁵⁴¹₂₈₈ −¹⁰⁹₂₈₈√

193)x + (¹²⁵₃₂ +₃₂⁹√ 193)

Poncelet Figures 17

Figure 5: A Poncelet figure of order 15 over Q(√

345) and the conics y = x² and ( 13523377736

586181640625 + 586181640625^419344728

√345)x² + (83740234375^814386560 + 83740234375^31121536

√345)xy + (_669921875^31045104 + _669921875⁹⁹⁸⁶⁰⁸ √

345)x + (_2392578125^32390784 + _2392578125^1328000 √

345)y + (₇₆₅₆₂₅¹⁵³⁰⁴ +

472 765625

√345) = 0.

number point line

1 (⁻⁴⁶⁵⁷⁵₄₄₅₂₁ +₄₄₅₂₁⁻⁸⁵⁰√

345,^2418493125_1982119441+_1982119441^79177500√

345) y = (^640874905_69586323+^41189005_69586323√

345)x + (^342217000_23195441+^58104625_69586323√ 345) 2 (¹⁶⁰³⁰₁₅₆₃+₁₅₆₃⁹⁵⁵√

345,^571609525_2442969 +^30617300_2442969√

345) y = (^376495909_43590507+^27037249_43590507√

345)x + (^212900620_14530169+^12993215_14530169√ 345) 3 (⁻⁴⁵¹⁴⁷₂₇₈₈₉+₂₇₈₈₉²⁵⁸√

345,^2061216189_777796321+^−23295852_777796321√

345) y = (^134421290_23621983+^10983680_23621983√

345)x + (^35021031_3374569+^2261352_3374569√ 345) 4 (⁶¹⁹¹₈₄₇+³⁸⁶₈₄₇√

345,^89732101₇₁₇₄₀₉ +^4779452₇₁₇₄₀₉√

345) y = (^4520816₄₄₁₂₈₇+₄₄₁₂₈₇⁹⁸²⁵⁶√

345)x + (^6702835₄₄₁₂₈₇+¹⁶¹⁴⁶⁰₄₄₁₂₈₇√ 345) 5 (¹⁰⁷⁰⁵₃₆₄₇+⁻⁸⁵⁰₃₆₄₇√

345,^363859525_13300609+^−18198500_13300609√

345) y = (^432668905_154764092+^−27773675_154764092√

345)x + (^730570875_154764092+^−29390125_154764092√ 345) 6 (⁻⁵⁹²⁵₄₂₄₃₆+₄₂₄₃₆²²⁷⁵√

345,^910348125_900407048+^−13479375_900407048√

345) y = (−4046267575 2091415824+_232379536^19088525√

345)x + (−1085271875

1394277216+_4182831648^419196875√ 345) 7 (^−353875₁₉₇₁₃₆ +₂₁₉₀₄⁶²⁵√

345,68071765625

19431301248+^−221171875_2159033472√

345) y = (−1974105450125 759627302976 +84403033664^3897075775

√

345)x + (−1091394921875

675224269312 +331776240625 6077018423808

√ 345) 8 (^−49550625_61653056+_61653056^1087475√

345,1431631042490625

1900549657069568+−53885065921875 1900549657069568

√

345) y = (_1726285568^33492775 +^−147766565_1726285568√

345)x + (10178233125

7891591168+^−769349375_7891591168√ 345) 9 (¹⁴⁷⁵₁₇₉₂+⁻¹⁸⁵₁₇₉₂√

345,^6991625_1605632+^−272875_1605632√

345) y = (⁻¹⁰⁹²⁵₅₄₂₀₈+⁻⁵¹⁹³₅₄₂₀₈√

345)x + (^30761275_27754496+^−3105825_27754496√ 345) 10 (^−222175₂₁₆₈₃₂+₂₁₆₈₃₂¹⁶¹³√

345,25129670465 23508058112+23508058112^−358368275

√

345) y = (^−258221963_112969472+_564847360^16225877√

345)x + (−4868836989

3615023104+_3615023104^112764039√ 345) 11 (⁻⁴²⁰⁵¹₃₃₃₄₄+₁₆₆₇₂₀³⁵⁴⁹√

345,^4855257837_2779555840+^−149238999_2779555840√

345) y = (^−5412739_2300736+_11503680²³⁴⁴⁶¹√

345)x + (^−8404429_6135296 +_6135296¹³⁵⁵⁴³√ 345) 12 (⁻¹²⁰⁵₁₁₀₄+₁₁₀₄⁻¹√

345,⁷²⁶¹⁸⁵₆₀₉₄₀₈+₆₀₉₄₀₈¹²⁰⁵√

345) y = (^−3511445_2041296+_2041296²⁰²³¹√

345)x + (^−929275_1360864+_4082592⁴⁵⁸⁷⁵√ 345) 13 (⁻²³²⁵₃₆₉₈ +₁₈₄₉²⁰√

345,_13675204^5957625+⁻⁴⁶⁵⁰⁰_3418801√

345) y = (−24735962275

12960628366+_6480314183^71343415√

345)x + (^−745578750_925759169+_1851518338^25855625√ 345) 14 (^−4485500_3504767 +_3504767⁶⁷⁵ √

345,20119867440625

12283391724289+ −6055425000 12283391724289

√

345) y = (−617461769125 451697875727+64528267961^4035757275

√

345)x + (−52253093750 451697875727+36051778125

451697875727

√ 345) 15 (⁻⁷⁸⁶²⁵₉₀₂₁₆₇+₉₀₂₁₆₇⁵⁶²⁵⁰√

345,1097783453125

813905295889+−8845312500 813905295889

√

345) y = (−45518891650 40165377007+40165377007^1737464300

√

345)x + (12833353125 40165377007+40165377007^2553012500

√ 345)

Poncelet Figures 18

Figure 6: A Poncelet figure of order 8 over Q and y = x² and −11434224x²+ 37746552xy − 101494849y + 532800x + 5442800y − 40000 = 0.

number point line

1 ⁸⁹₂₅,⁷⁹²¹₆₂₅

y = ¹¹¹₅₀x + ⁵⁹⁶³₁₂₅₀ 2 ⁻⁶⁷₅₀ ,⁴⁴⁸⁹₂₅₀₀

y = ³⁰⁸⁹₄₀₀x +¹⁹⁴³₁₆₀ 3 ¹⁴⁵₁₆,²¹⁰²⁵₂₅₆

y = ⁸⁴³³₅₀ x +^−1851331₁₂₈₀ 4 ⁶³⁸³⁹₄₀₀ ,^4075417921₁₆₀₀₀₀

y = ³⁴⁶³₄₀₀x +^481792933₂₀₀₀₀ 5 ⁻⁷⁵⁴⁷₅₀ ,^56957209₂₅₀₀

y = ^−135523₉₅₀ x + ^5939489₄₇₅₀ 6 ⁷⁸⁷₉₅,⁶¹⁹³⁶⁹₉₀₂₅

y = ¹²⁹¹⁷₁₇₀₀x + ^17437559_3068500 7 ⁻²²¹⁵⁷₃₂₃₀₀ ,_1043290000^490932649

y = ²⁴⁴⁷₉₅₀x +^24572113_10982000 8 ¹¹⁰⁹₃₄₀,^1229881₁₁₅₆₀₀

y = ¹¹⁵⁹⁷₁₇₀₀x + ⁻⁹⁸⁷⁰¹₈₅₀₀

Poncelet Figures 19

Figure 7: A Poncelet figure of order 10 over Q and y = x² and ³⁶¹₁₂₁x²−³⁸₁₁xy −

720

1331x + y²− ²⁴⁰₁₂₁y +¹¹⁴⁷²⁰₁₄₆₄₁ = 0.

number point line 1 ³¹₁₁,⁹⁶¹₁₂₁

y = ²⁹₁₁x + ₁₂₁⁶² 2 ⁻²₁₁,₁₂₁⁴

y = ⁸⁹₇₇x + ²⁰⁶₈₄₇ 3 ¹⁰³₇₇,¹⁰⁶⁰⁹₅₉₂₉

y = ⁶⁸₁₁x + ⁻³⁸⁴¹⁹₅₉₂₉ 4 ³⁷³₇₇,¹³⁹¹²⁹₅₉₂₉

y = ²⁷⁷₁₅₄x +²⁴⁹⁹¹₁₆₉₄ 5 ⁻⁶⁷₂₂ ,⁴⁴⁸⁹₄₈₄

y = ⁻¹³₁₁ x +²⁷⁴⁷₄₈₄ 6 ⁴¹₂₂,¹⁶⁸¹₄₈₄

y = ⁻¹⁷₂₂ x +¹¹⁸⁹₂₄₂ 7 ⁻²⁹₁₁ ,⁸⁴¹₁₂₁

y = ²⁰₁₁x + ¹⁴²¹₁₂₁ 8 ⁴⁹₁₁,²⁴⁰¹₁₂₁

y = ⁶³₁₁x + ⁻⁶⁸⁶₁₂₁ 9 ¹⁴₁₁,¹⁹⁶₁₂₁

y = ¹³₁₁x + ₁₂₁¹⁴ 10 ⁻¹₁₁,₁₂₁¹

y = ³⁰₁₁x + ₁₂₁³¹

Poncelet Figures 20

Figure 8: A Poncelet figure of order 12 over Q, using the conics x²+ y² = 1 and −²¹⁴³⁶⁹₆₄ x²−^6298181₁₉₂ xy −^6342629₁₉₂ x −^319466449₂₃₀₄ y²− ^320728561₁₁₅₂ y −^321999889₂₃₀₄ = 0.

number point line

1

,

y =

x +

2

,

y =

x +

3

,

y =

x +

4

,

y =

x +

5

,

y =

x +

6

,

y =

x +

7

,

y =

x +

8

,

y =

x +

9

,

y =

x +

10

,

y =

x +

11

31868578513

,

y =

x +

12

,

y =

x +