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The Constructability of the Regular Heptadecagon

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The Constructability of the Regular Heptadecagon The Constructability of the Regular Heptadecagon Daniel Humphreys June 2020 1 Introduction The construcability of various geometric figures using only a compass and unmarked straightedge is an area of investigation that dates back to the time of Euclid in ancient Greece. It was known at that time that regular polygons with sides lengths 3; 4; 5; 6; 8; 10; 12; 15::: could be constructed, but no further progress was made on this problem for a couple thousand years. In fact no progress was made for such a long time that it was thought that the Greeks had found all the regular polygons that could be constructed. However, on March 30th, 1796, a 19 year old Carl Gauss rose from bed and was struck by an idea regarding how to prove that the regular 17-gon was constructable. It is said that Gauss was so pleased by this result that it not only made him decide to peruse a career in mathematics (lucky for us) but that he also requested that a 17 sided polygon be inscribed on his tombstone. This paper will follow the proof that the regular 17-gon is constructable that Gauss wrote in his diary, given in [2]. Later on, we will briefly discuss the broader theory on constructable n-gons including a theorem given by Gauss that gives a necessary and sufficient condition for a regular n-gon to be construable. 2 Definitions and Proofs Definition 2.1. (Constructable Number) A number is said to be constructable if it has a representation that consists of only finitely many additions, subtrac- tions, multiplications, divisions, and square roots of positive integers. Note that this definition implies that the sum, difference, product, quo- tient and square root of a finite number of constructable numbers is also a constructable number. Definition 2.2. (Constructable n-gon) A regular n-gon is said to be con- structable if the number cos(2π=n) is construable. The reasoning behind this definition is that the problem of constructing a regular n-gon is equivalent to the problem of dividing the unit circle (or any 1 circle really) into n equal arcs. Once this is done, it becomes easy to construct the regular n-gon by simply connecting the points that divide the circle. The number cos(2π=n) represents the x-coordinate of the first of these points (note that the cosine of any integer multiple of the angle 2π=n will be the x-coordinate of one the vertices of the n-gon), so if it is a constructable number then the first vertex of the n-gon is constructable. The rest of the vertices can then be added using the compass and the first vertex. We now prove two lemmas which will be needed in our main proof, the first of which will turn out to be as useful as its proof is straightforward. c Lemma 2.1. If two non zero real numbers r1 and r2 satisfy r1r2 = a and b r1 + r2 = − a for real numbers a; b; c, with a non zero, then r1 and r2 are the 2 b c roots of the quadratic x + a x + a . Proof. We will only prove the case for r1, as the case for r2 is essentially identical. c From the first condition, we have r2 = , which, when substituted into ar1 b r1 + r2 = − a , yields c b r1 + = − ar1 a Multiplying by r1 and rearranging we see that b c r2 + r + = 0 1 a 1 a 2 b c and hence r1 is a root of the quadratic x + a x + a as desired. Pn Lemma 2.2. Denote the sum k=1 cos(kθ) by Sn. Then 2n+1 θ sin( 2 θ) − sin( 2 ) Sn = θ 2 sin( 2 ) whenever θ 6= 2nπ for some integer n. Proof. Using the product to sum formula for cos(α) sin(β) (that is cos(α) sin(β) = 1 2 (sin(α + β) − sin(α − β))), we have 2k + 1 2k − 1 2 cos(kθ) sin(θ=2) = sin θ − sin θ 2 2 Summing over k we have n n X X 2k + 1 2k − 1 2 sin(θ=2) cos(kθ) = sin θ − sin θ 2 2 k=1 k=1 3 1 5 3 = (sin θ − sin θ ) + (sin θ − sin θ ) + ::: 2 2 2 2 2n − 1 2n − 3 2n + 1 2n − 1 + (sin θ − sin θ ) + (sin θ − sin θ ) 2 2 2 2 2 This sum clearly telescopes, and so we are left with n X 2n + 1 θ 2 sin(θ=2) cos(kθ) = sin θ − sin 2 2 k=1 Since θ 6= 2nπ for some integer n, sin(θ=2) is non zero, and so we can divide to obtain n 2n+1 θ X sin( 2 θ) − sin( 2 ) Sn = cos(kθ) = 2 sin( θ ) k=1 2 as desired. We can now use these results to prove the main theorem, that the regular heptadecagon is constructable. We will follow the proof that Gauss wrote in his diary given in [2]. Theorem 2.3. The regular heptadecagon is constructable. 2π Proof. In order to prove construcability, we must show that cos( 17 ) is a con- 2π structable number. To do so, we set θ = 17 and make the following definitions: a = cos(θ) + cos(4θ) b = cos(2θ) + cos(8θ) c = cos(3θ) + cos(5θ) d = cos(6θ) + cos(7θ) e = a + b f = c + d By lemma 2.2, we have 8 X sin( 17 2π ) − sin( π ) a + b + c + d = e + f = cos(kθ) = 2 17 17 2 sin( π ) k=1 17 sin(π) − sin(π=17) = 2 sin(π=17) 1 = − 2 We now must form several products. 2ab = 2(cos(θ) + cos(4θ))(cos(2θ) + cos(8θ)) = 2(cos(θ) cos(2θ) + cos(θ) cos(8θ) + cos(4θ) cos(2θ) + cos(4θ) cos(8θ)) We again use the product to sum formula to obtain 2 cos(θ) cos(2θ) = cos(θ) + cos(3θ) 2 cos(θ) cos(8θ) = cos(7θ) + cos(9θ) 2 cos(4θ) cos(2θ) = cos(6θ) + cos(2θ) 2 cos(4θ) cos(8θ) = cos(12θ) + cos(4θ) 3 We now note that, for any integer n 2π πn cos (17 − n) = cos 2π − 17 17 = cos(2π) cos(2πn=17) + sin(2π) sin(2πn=17) 2πn = cos 17 and so we have cos(12θ) = cos(5θ) and cos(9θ) = cos(8θ). Combining all this, we have 2ab = cos(θ) + cos(2θ) + cos(3θ) + cos(4θ) + cos(5θ) + cos(6θ) + cos(7θ) + cos(8θ) = e + f 1 = − 2 Similarly, we have 2ac = 2(cos(θ) + cos(4θ)(cos(3θ) + cos(5θ)) = 2(cos(θ) cos(3θ) + cos(θ) cos(5θ) + cos(3θ) cos(4θ) + cos(4θ) cos(5θ)) = (cos(4θ) + cos(2θ)) + (cos(6θ) + cos(4θ)) + (cos(7θ) + cos(θ)) + (cos(9θ) + cos(θ)) = 2(cos(4θ) + cos(θ)) + (cos(2θ) + cos(8θ)) + (cos(6θ) + cos(7θ)) = 2a + b + d Continuing on in this fashion, we see that 2ad = b + c + 2d 2bc = a + 2c + d 2bd = a + 2b + c 1 2cd = − 2 We then see that 2ac + 2ad + 2bc + 2bd = (2a + b + d) + (b + c + 2d) + (a + 2c + d) + (a + 2b + c) = 4(a + b + c + d) = 4(e + f) = −2 Factoring the LHS we have 2(ac + ad + bc + bd) = 2(a(c + d) + b(c + d)) = 2(c + d)(a + b) = 2ef 4 and so 2ef = −2 ) ef = −1 1 Thus we have found that ef = −1 and e + f = − 2 , and so by lemma 2.1 we 2 1 know that e and f are the roots pf the quadratic x + 2 x − 1. We can find these roots explicitly using the quadratic formula, yielding 1 r17 r = − + 1 4 16 1 r17 r = − − 2 4 16 Using some numerical estimates, we have e = cos(2π=17) + cos(4π=17) + cos(8π=17) + cos(16π=17) ≈ 0:93247 + 0:73901 + 0:09226 − 0:98297 ≈ 0:78077 f = cos(6π=17) + cos(10π=17) + cos(12π=17) + cos(14π=17) ≈ 0:44573 − 0:27366 − 0:60263 − 0:85021 ≈ −1:28077 1 and so we must have that r1 = e and r2 = f. Next, we saw before that ab = − 4 , and by definition a+b = e, so by lemma 2.1 a and b are the roots of the quadratic 2 1 x − ex − 4 . These roots are given explicitly by r e 1 e2 r = + + 1 2 4 4 r e 1 e2 r = − + 2 2 4 4 We again use some numerical estimates to determine which root is a and which is b. We have a = cos(2π=17) + cos(8π=17) ≈ 0:93247 + 0:09226 ≈ 1:0247 b = cos(4π=17) + cos(16π=17) ≈ 0:73900 − 0:98297 ≈ −0:24397 5 and so we must have 1 1p 1q p r = a = − + 17 + 34 − 2 17 1 8 8 8 1 1p 1q p r = b = − + 17 − 34 − 2 17 2 8 8 8 1 We can determine c and d in a similar fashion. Recall from earlier that 2cd = − 2 1 and so cd = − 4 , and by definition c + d = f. Thus, by lemma 2.1, c and d are 2 1 the roots of the quadratic x − fx − 4 , which are given explicitly by r f 1 f 2 r = + + 1 2 4 4 r f 1 f 2 r = − + 2 2 4 4 Estimating c and d numerically we have c = cos(6π=17) + cos(10π=17) ≈ 0:44573 − 0:27366 ≈ 0:17206 d = cos(12π=17) + cos(14π=17) ≈ −0:60263 − 0:85021 ≈ −1:45284 and so we must have 1 1p 1q p r = c = − − 17 + 34 + 2 17 1 8 8 8 1 1p 1q p r = d = − − 17 − 34 + 2 17 2 8 8 8 Finally, we have, by the product to sum formula, 1 1 cos(θ) cos(4θ) = (cos(3θ) + cos(5θ)) = c 2 2 and since cos(θ) + cos(4θ) = a by definition, we have, by lemma 2.1, that cos(θ) 2 1 and cos(4θ) are the roots of the quadratic x − ax + 2 c.
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