Construction of a Regular Heptadecagon

Moti Ben-Ari http://www.weizmann.ac.il/sci-tea/benari/ Version 1.0

This work is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA. This document presents Gauss’s insight that it is possible to construct a heptadecagon— a regular polygon with 17 sides—with straightedge and compass. The presentation is based on [1]; I have added the detailed calculations leading to Gauss’s formula. An actual construction from [3] is also presented; again I have added the detailed calculations.

1 Construction of regular polygons

History The ancient Greeks knew how to construct some regular polygons with straightedge and compass: a triangle, a square, a pentagon and a regular polygon with 15 sides. Of course, given a regular polygon with n sides, it is easy to construct a polygon with 2n sides by bisecting the sides. No progress was made for two thousand years until in 1796, just before his 19th birthday, Carl Friedrich Gauss awoke one morning and by “concentrated thought” ﬁgured out how to construct a regular heptadecagon, a regular polygon with 17 sides. This achieve- ment inspired him to become a mathematician. The construction of the regular heptadecagon led to the Gauss-Wantzel Theorem which states that a regular polygon with n sides can be constructed with straightedge and compass if and only if n is the product of a power of 2 and zero or more distinct prime Fermat 2k numbers 2 + 1. The known Fermat primes are F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537. A regular polygon with 257 sides was ﬁrst constructed by Magnus Georg Paucker in 1822 and Friedrich Julius Richelot 1832. In 1894 Johann Gustav Hermes claimed to have constructed a regular polygon with 65537 sides and his manuscript is saved at the University of Gottigen¨ should you wish to check it.

The cosine of the central angle To construct a regular polygon, it is sufﬁcient to construct a line segment of length cos θ, where θ is the central angle subtended by a chord that is a side of the polygon inscribed in a unit circle. Given the line segment OB = cos θ, construct a perpendicular at B and label its intersection with the unit circle by C. Then OB cos θ = = OB so θ = cos−1(OB). The chord AC is a side of the polygon. OC C

θ O A cos θ B

Constructible lengths Given a line segment deﬁned to have length 1, the lengths that are constructible are those which can be obtained from line segments of known length √ using the operations {+, −, ×, ÷, }. Appendices A, B present both trigonometric and geometric derivations showing that an equilateral triangle and a regular pentagon are constructible by giving expressions that use only these operations.

2π Construction of a heptadecagon The central angle of a heptadecagon is radians or 17 360◦ ≈ 21.12◦. 17

2π/17≈21.12◦

Gauss showed that [1, 2]:

2π 1 1 √ 1 q √ cos = − + 17 + 34 − 2 17+ 17 16 16 16

r 1 √ q √ q √ 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 . 8 √ This value can be computed using the operations {+, −, ×, ÷, } so it is constructible! Sections 2, 3, 4 present Gauss’s mathematical ideas, together with the detailed calculations. The proof does not use complex numbers explicitly, though I have added some 2π notes on complex numbers. Section 5 shows an efﬁcient construction of cos . 17

2 2 Roots of unity

We use the following theorem without proof: Fundamental Theorem of Algebra Every polynomial of degree n (with complex coefﬁ- cients) has exactly n (complex) roots.

Roots of unity and regular polygons Consider the equation xn − 1 = 0 for any integer n > 1. One root is x = 1. By the Fundamental Theorem of Algebra there are n − 1 other roots. Denote one root by r so that rn = 1. r is called a root of unity.

2π 2π Complex numbers The root r is cos + i sin . n n By de Moivre’s formula:

2π 2π n 2 · nπ 2 · nπ cos + i sin = cos + i sin = 1 . n n n n

Consider now r2. Then: (r2)n = (rn)2 = 12 = 1 .

It follows that: 1, r, r2,..., rn−2, rn−1 are n-th roots of unity. Theorem Let n be a prime and r an n-th root of unity; then {1, r, r2,..., rn−2, rn−1} are distinct, so they are all the n-th roots of unity. Proof Suppose that ri = rj for some 1 ≤ i < j ≤ n, so rj/ri = rj−i = 1. Let m be the smallest positive integer less than n such that rm = 1. Now n = ml + k for some 0 < l < n and 0 ≤ k < m. From 1 = rn = rml+k = (rm)l · rk = 1l · rk = rk, we have 0 ≤ k < m and rk = 1. Since m is the smallest such positive integer, k = 0 and n = ml, so n is not prime. We use the following theorem without proof.

Theorem Let {a1, a2,..., an−1, an} be the roots of an n-th degree polynomial f (x). Then

f (x) = (x − a1)(x − a2) ··· (x − an−1)(x − an) .

Viete’s´ formula [1, p. 28] gives the coefﬁcients of the polynomial in terms of its roots; the formula can be obtained by multiplication. It is easy to see that the coefﬁcient of xn−1 is:

−(a1 + a2 + ··· + an−1 + an) .

Since the coefﬁcient of xn−1 in xn − 1 for n ≥ 2 is obviously zero, we have:

−(1 + r + r2 + ··· + rn−2 + rn−1) = 0 .

3 We will use this in the form:

r + r2 + ··· + rn−2 + rn−1 = −1 .

For the heptadecagon this is:

r + r2 + r3 + r4 + r5 + r6 + r7 + r8 + r9 + r10 + r11 + r12 + r13 + r14 + r15 + r16 = −1 .

3 Gauss’s proof that a heptadecagon is constructable

What Gauss saw is the one need not work with the roots in the natural order r, r2,..., r16. Instead, one can notice that the powers of r3 give all the roots, but in a different order. For k < 17, r17m+k = (r17)m · rk = 1m · rk = rk, so the exponents are reduced modulo 17:

r1, r1·3=3, r3·3=9, r9·3=27=10, r10·3=30=13, r13·3=39=5, r5·3=15, r15·3=45=11, r11·3=33=16, r16·3=48=14, r14·3=42=8, r8·3=24=7, r7·3=21=4, r4·3=12, r12·3=36=2, r2·3=6 .

It is important that you check that this list contains each of the 16 roots exactly once.

The roots of a quadratic equation Consider the monic quadratic equation:

y2 + py + q = 0 ,

and suppose that its roots are a, b. Then:

(y − a)(y − b) = y2 − (a + b)y + ab .

Therefore, p = −(a + b) and q = ab, so that if we are given a + b and ab, we can write down the quadratic equation of which a, b are the roots.1

Let a0 be the sum of the roots in the odd positions in the above list:

9 13 15 16 8 4 2 a0 = r + r + r + r + r + r + r + r , and let a1 be the sum of the roots in the even positions in the above list:

3 10 5 11 14 7 12 6 a1 = r + r + r + r + r + r + r + r .

To obtain a0, a1 as roots of a quadratic equation, we ﬁrst compute their sum:

2 16 a0 + a1 = r + r + ··· + r = −1 .

Now we have to work very hard to compute their product! Figure 1 contains the computation where the values of rirj are written after computing r(i+j) mod 17. Below each root

1Po-Shen Lo used this observation to develop a quick method for solving quadratic equations. See [4] and a document on my website.

4 9 13 15 16 8 4 2 a0a1 = (r + r + r + r + r + r + r + r ) ×

(r3 + r10 + r5 + r11 + r14 + r7 + r12 + r6) r4 r11 r6 r12 r15 r8 r13 r7 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + r12 r2 r14 r3 r6 r16 r4 r15 2 + 1 + 1 + 1 + 2 + 1 + 2 + 2 + r16 r6 r1 r7 r10 r3 r8 r2 2 + 3 + 1 + 2 + 1 + 2 + 2 + 2 + r1 r8 r3 r9 r12 r5 r10 r4 2 + 3 + 3 + 1 + 3 + 1 + 2 + 3 + r2 r9 r4 r10 r13 r6 r11 r5 3 + 2 + 4 + 3 + 2 + 4 + 2 + 2 + r11 r1 r13 r2 r5 r15 r3 r14 3 + 3 + 3 + 4 + 2 + 3 + 4 + 2 + r7 r14 r9 r15 r1 r11 r16 r10 3 + 3 + 3 + 4 + 4 + 4 + 3 + 4 + r5 r12 r7 r13 r16 r9 r14 r8 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4

= −4 .

Figure 1: The computation of a0a1 is its number of occurrences so far; check that each root occurs exactly four times so that the value of the product is −4. Since a0 + a1 = −1 and a0, a1 = −4, a0, a1 are the roots of the quadratic equation: y2 + y − 4 = 0 , which are: √ −1 ± 17 a = . 0,1 2 1 3 9 10 Let b0, b1, b2, b3 be the sum of every fourth root starting with r , r , r , r , respectively:

1 13 16 4 b0 = r + r + r + r 3 5 14 12 b1 = r + r + r + r 9 15 8 2 b2 = r + r + r + r 10 11 7 6 b3 = r + r + r + r .

5 Check that b0 + b2 = a0, b1 + b3 = a1. Let us compute the corresponding products:

13 16 4 b0b2 = (r + r + r + r ) × (r9 + r15 + r8 + r2) = r10 + r16 + r9 + r3+ r5 + r11 + r4 + r15+ r8 + r14 + r7 + r1 + r13 + r2 + r12 + r6 = −1 , 3 5 14 12 b1b3 = (r + r + r + r )× (r10 + r11 + r7 + r6) = r13 + r14 + r10 + r9 + r15 + r16 + r12 + r11+ r7 + r8 + r4 + r3 + r5 + r6 + r2 + r1 = −1 . To summarize these computations:

b0 + b2 = a0

b0b2 = −1

b1 + b3 = a1

b1b3 = −1 , so b0, b2 are the solutions of: 2 y − a0y − 1 = 0 . and b1, b3 are the solutions of: 2 y − a1y − 1 = 0 . Using the formula for solving quadratic equations and the values previously computed for a0, a1, we obtain the roots b0, b1 (Figure 2). Finally, let c0, c4 be the sum of every eighth root starting with r1, r13, respectively:2

1 16 c0 = r + r

13 4 c4 = r + r

1 16 13 4 c0 + c4 = r + r + r + r = b0

1 16 13 4 c0c4 = (r + r ) · (r + r )

14 5 12 3 = r + r + r + r = b1 ,

2There are other sums but these two sufﬁce.

6 q 2 a0 + a0 + 4 b = 0 2 v √ u √ !2 (−1 + 17) u (−1 + 17) + t + 4 2 2 = 2 r √ √ 2 (−1 + 17) + −1 + 17 + 16 = 4 √ p √ (−1 + 17) + 34 − 2 17 = , 4 q 2 a1 + a1 + 4 b = 1 2 v √ u √ !2 (−1 − 17) u (−1 − 17) + t + 4 2 2 = 2 r √ √ 2 (−1 − 17) + −1 − 17 + 16 = 4 √ p √ (−1 − 17) + 34 + 2 17 = . 4

Figure 2: Computation of b0, b1 so c0, c4 are the roots of: 2 y − b0y + b1 = 0

1 16 It sufﬁces to compute the root c0 = r + r (Figure 3) since:

2π c = r + r = 2 cos , 0 1 16 17 as shown below (page 9).

7 q 2 b0 + b0 − 4b1 c = 0 2 √ p √ (−1 + 17) + 34 − 2 17 = 4 + 2 v u" √ p √ #2 " √ p √ # u (−1 + 17) + 34 − 2 17 (−1 − 17) + 34 + 2 17 t − 4 4 4 2

1 1√ 1q √ = − + 17 + 34 − 2 17+ 8 8 8 s 1 √ q √ 2 √ q √ (−1 + 17) + 34 − 2 17 − 16 (−1 − 17) + 34 + 2 17 8

1 1√ 1q √ = − + 17 + 34 − 2 17+ 8 8 8 r 1 √ √ q √ √ (−1 + 17)2 + 2(−1 + 17) 34 − 2 17 + (34 − 2 17)− 8

h √ p √ i (−16 − 16 17) + 16 34 + 2 17

1 1√ 1q √ = − + 17 + 34 − 2 17+ 8 8 8 r 1 √ √ q √ q √ 68 + 12 17 + 2(−1 + 17) 34 − 2 17 − 16 34 + 2 17 8

Figure 3: Computation of c0

8 The y-coordinates of r1, r16 are equal but with opposite signs and cancel, while the x- coordinate is counted twice:

θ O 1 θ cos θ

r16

Therefore, the cosine of the central angle of a heptadecagon is constructible with straightedge and compass, since it is composed only of rational numbers and the operations √ {+, −, ×, ÷, }:

2π c cos = 0 17 2 1 1 √ 1 q √ = − + 17 + 34 − 2 17+ 16 16 16 r 1 √ √ q √ q √ 68 + 12 17 + 2(−1 + 17) 34 − 2 17 − 16 34 + 2 17 16

Complex numbers

c0 = r1 + r16 2π 2π 2 · 16π 2 · 16π = cos + i sin + cos + i sin 17 17 17 17 2π 2π −2π −2π = cos + i sin + cos + i sin 17 17 17 17 2π = 2 cos . 17

9 4 Derivation of Gauss’s formula

2π The formula we gave for cos is not the one given by Gauss [2, p. 458], which also 17 appears in [1, p. 68]. I found it only in [5], where Rike gives it as an exercise to transform the formula to the one that Gauss gave. This Section gives that transformation. √ p √ Let us simplify 2(−1 + 17) 34 − 2 17:

√ p √ p √ √ p √ 2(−1 + 17) 34 − 2 17 = −2 34 − 2 17 + 2 17 34 − 2 17+ p √ p √ 4 34 − 2 17 − 4 34 − 2 17 p √ √ p √ = 2 34 − 2 17 + 2 17 34 − 2 17+ p √ −4 34 − 2 17 √ p √ p √ = 2(1 + 17) 34 − 2 17 − 4 34 − 2 17 .

p √ We will remember the term −4 34 − 2 17 for now and simplify the ﬁrst term by squar- ing it and then taking the square root: r √ p √ h √ p √ i2 2(1 + 17) 34 − 2 17 = 2 (1 + 17) 34 − 2 17 q √ √ = 2 (18 + 2 17)(34 − 2 17) q √ = 2 (18 · 34 − 4 · 17) + 17(2 · 34 − 2 · 18) p √ = 2 · 4 34 + 2 17 .

Substituting terms results in Gauss’s formula:

2π 1 1 √ 1 q √ cos = − + 17 + 34 − 2 17+ 17 16 16 16 r 1 √ q √ q √ q √ 68 + 12 17 + 2 · 4 34 + 2 17 − 4 34 − 2 17 − 16 34 + 2 17 16 1 1 √ 1 q √ = − + 17 + 34 − 2 17+ 16 16 16 r 1 √ q √ q √ 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 . 8

10 5 Construction with a straightedgeR and compass

Several constructions are given in [8]. Here I give the construction from [3] because it 2π directly constructs cos . The construction uses only Pythagoras’s Theorem and the 17 Angle Bisector Theorem [7]. Construct a unit circle centered at O, and construct perpendicular diameters QP, SR.3 R ··· A β √ β α α 17 4 1 4 Q P C O B

√ √ 1+ 17 −1+ 17 16 16

··· S 1 Construct A so that OA = OR. By Pythagoras’s Theorem: 4 q √ AP = (1/4)2 + 12 = 17/4 .

Let B be the intersection of the internal bisector of 6 OAP and OP, and let C be the intersection of the external bisector of 6 OAP and QO. By the angle bisector theorem: S OB AO = BP AP OB 1/4 = √ 1 − OB 17/4 √ 1 1 1 − 17 OB = √ = √ · √ + + − 1 √17 1 17 1 17 −1 + 17 = , 16 and: OC AO = CP AP OC 1/4 = √ 1 + OC 17/4 √ 1 1 1 + 17 OC = √ = √ · √ − + − + + 1 √ 17 1 17 1 17 1 + 17 = . 16 3To save space I have clipped the circle so that R = (0, 1), S = (0, −1) do not appear.

11 R

Construct D on OP such that CD = CA:

A a b Q P f M C O D B b E a f

q 2 2 CD = CA = OA + OC v u √ !2 u12 1 + 17 = t + 4 16

1 q √ = S34 + 2 17 . 16

Construct E on OP such that BE = BA:

q 2 2 BE = BA = OA + OB v u √ !2 u12 1 − 17 = t + 4 16

1 q √ = 34 − 2 17 . 16

Construct M as the midpoint of QD and construct F on OS such that MF = MQ:

1 1 1 MF = MQ = QD = (QC + CD) = ((1 − OC) + CD) 2 2 2 " √ ! p √ # 1 1 + 17 34 + 2 17 = 1 − + 2 16 16 1 √ q √ = 15 − 17 + 34 + 2 17 . 32

12 R

Construct a circle whose diameter is OE. Construct a chord OG = OF. Note that MO = 1 − MQ = 1 − MF:

Q P f M C O D B E e H g g e f G F

q 2 2 q 2 OG = OF = MF − MO = MF − (1 − MF)2

p = 2MF − 1 s 1 √ q √ = 15 − 17 + 34 + 2 17 − 1 16 r S 1 √ q √ = −1 − 17 + 34 + 2 17 . 4 6 OGE is a right angle since it is subtended by a diameter of the circle. Construct H on OP such that EH = EG: q 2 2 q 2 EH = EG = OE − OG = (OB + BE)2 − OG v u √ p √ !2 u −1 + 17 34 − 2 17 1 √ q √ = t + − −1 − 17 + 34 + 2 17 16 16 16 s 1 √ √ q √ √ = (18 − 2 17) + 2(−1 + 17) 34 − 2 17 + (34 − 2 17) + 16 √ p √ 16 + 16 17 − 16 34 + 2 17 r 1 √ q √ √ q √ = 68 + 12 17 − 16 34 + 2 17 − 2(1 − 17) 34 − 2 17 . 16 Let us compute OE: √ −1 + 17 1 q √ OE = OB + BE = + 34 − 2 17 16 16 1 √ q √ = −1 + 17 + 34 − 2 17 . 16 2π Finally, OH = OE + EH which is cos . 17

13 A Constructing an equilateral triangle

Trigonometry The central angle of an equilateral triangle is 360◦/3 = 120◦ and we can compute its cosine from the formula for the cosine of the sum of two angles: √ 3 1 1 cos 120◦ = cos(90◦ + 30◦) = cos 90◦ cos 30◦ − sin 90◦ sin 30◦ = 0 · − 1 · = − . 2 2 2 This value is constructible.

Geometry Consider an equilateral triangle 4ABC whose sides are of length 1. Let AD be the altitude from A to BC. Since AB = AC the line segment bisects BC. Therefore, s √ 12 3 AD = 12 − = . 2 2 √ 1 3 1, 2 , 2 are constructible so we can construct an equilateral triangle by constructing the right triangles 4ADB, 4ADC.

1 1 2 √ 3 2 A D

1 2 1

B Constructing a regular pentagon

Trigonometry The central angle is 360◦/5 = 72◦. Let us compute cos 36◦ using the trigonometric identities for 2θ and θ/2 [9]:

0 = cos 90◦ = cos(72◦ + 18◦) r r 1 + cos 36◦ 1 − cos 36◦ = (2 cos2 36◦ − 1) − 2 sin 36◦ cos 36◦ . 2 2

14 There is now only one angle in the formula; let x = cos 36◦. Then: r r 1 + x √ 1 − x (2x2 − 1) = 2 1 − x2 · x · 2 2 √ √ √ √ (2x2 − 1) 1 + x = 2 1 − x · 1 + x · x · 1 − x

2x2 − 1 = 2x(1 − x)

4x2 − 2x − 1 = 0 .

Solving the quadratic equation gives: √ 1 + 5 cos 36◦ = , 4 √ which can be computed using {+, −, ×, ÷, } so it is constructible. The following ﬁgure shows how to construct a regular pentagon from cos 36◦. From D at distance cos 36◦ from O construct a perpendicular to OA that intersects the unit circle centered at O at C. Construct OC. Construct a perpendicular from A to OC. Its intersection with the unit circle at B deﬁnes AB, the side of the inscribed pentagon.

O 36◦ A cos 36◦ D

Geometry Here is a solution to Exercises 2.3.3–2.3.4 of [6, page 28], showing that a regular pentagon is constructible.

1 1

x C A θ θ

x x 1 ψ 1 F ψ φ φ D 1 E

15 Let ABCDE be a regular pentagon. By deﬁnition all the sides and all the interior angles are equal. It is easy to show that all diagonals are equal.4 Let the length of the sides be 1 and the length of the diagonals be x. 4ACE =∼ 4CAD so 6 ACE = 6 CAD = θ. 4AED =∼ 4CDE so 6 ADE = 6 CED = φ. 6 AFC = 6 EFD = ψ are vertical angles. ψ + 2θ = ψ + 2φ = 180◦ so θ = φ. By alternate interior angles, we conclude that AC k DE. Construct a line through E parallel to DC and let F be its intersection with AC. 4ACE is an isoceles triangle with base angles α. 4AEF is also isoceles so 6 AFE = 6 FAE = α.

1 1

1 F x − 1 C A α α

1 1 1 x α

D 1 E

Therefore, 4ACE ∼ 4AEF: x 1 = . 1 x − 1 Multiplying out gives the quadratic equation:

x2 − x − 1 = 0 , whose positive root is: √ 1 + 5 . 2

This length is constructible using rational numbers√ and square roots. The regular pen- 1+ 5 tagon is constructible: construct AC of length 2 and then construct the isoceles triangle 4ABC with sides of length 1 intersecting at B. AB and BC are two sides of the regular pentagon with interior angle 6 ABC.

4For example, 4ABC =∼ 4AED by SAS, so AC = AD.

16 References

[1]J org¨ Bewersdorff. Galois Theory for Beginners: A Historical Perspective. American Math- ematical Society, 2006.

[2] Todd W. Bressi and Paul Groth, editors. Disquisitiones Arithmeticae. Yale University Press, 2006.

[3] James J. Callagy. The central angle of the regular 17-gon. The Mathematical Gazette, 67(442):290–292, 1983. https://wwww.jstor.org/stable/3617271.

[4] Po-Shen Lo. A different way to solve quadratic equations, 2019. https://www. poshenloh.com/quadratic/.

[5] Tom Rike. Fermat numbers and the heptadecagon, 2005. https://mathcircle. berkeley.edu/sites/default/files/BMC6/ps0506/Heptadecagon.pdf.

[6] John Stillwell. Mathematics and Its History (Third Edition). Springer, 2010.

[7] Wikipedia contributors. Angle bisector theorem — Wikipedia, the free encyclopedia. https://en.wikipedia.org/w/index.php?title=Angle_bisector_theorem& oldid=984147660, 2020.

[8] Wikipedia contributors. Heptadecagon — Wikipedia, the free encyclopedia. https: //en.wikipedia.org/w/index.php?title=Heptadecagon&oldid=975964212, 2020.

[9] Wikipedia contributors. Pentagon — Wikipedia, the free encyclopedia. https://en. wikipedia.org/w/index.php?title=Pentagon&oldid=983136827, 2020.