Lemma 1:

if α = 2π/17

and

ζ

^{k}= cos kα + isin k α

then

ζ

^{k}+ ζ

^{17 - k}= 2 cos kα

Proof:

(1) Using De Moivre's Formula for roots of unity (see Corollary 1.1, here), we note that:

if α = 2π/17, then:

ζ = cos α + isinα is a seventeenth root of unity.

(2) Using De Moivre's Formula (see Theorem 1, here), we note that:

ζ

^{k}= (cos α + isin α)

^{k}= cos kα + isin k α

ζ

^{17-k}= (cos α + isin α)

^{17-k}= cos ([17 - k]α) + isin([17 -k]α) = cos ([17 - k]2π/17) + isin([17 - k]2π/17) = cos(2π - kα) + isin(2π - kα) = cos(-kα) + isin(-kα) = cos(kα) - isin(kα)

(3) So that we have:

ζ

^{k}+ ζ

^{17-k}= cos kα + isin kα + cos kα - isink α = 2cos kα

QED

Lemma 2:

If C is the least positive acute angle such that tan 4C = 4

with

α = 2π/17

ζ

^{k}= cos kα + isin k α

and:

x

_{1}= ζ + ζ

^{9}+ ζ

^{13}+ ζ

^{15}+ ζ

^{16}+ ζ

^{8}+ ζ

^{4}+ ζ

^{2}x

_{2}= ζ

^{3}+ ζ

^{10}+ ζ

^{5}+ ζ

^{11}+ ζ

^{14}+ ζ

^{7}+ ζ

^{12}+ ζ

^{6}

then:

x

_{1}= 2 tan 2C x

_{2}= -2cot 2C

Proof:

(1) Using step #1 thru step #7 in Theorem 1, here, we have:

x

_{1}and x

_{2}are the solutions to:

x

^{2}+ x - 4 = 0

(2) Since tan 4C = 4, it follows that cot 4C = 1/(tan 4C) = 1/4

(3) So 4xcot 4C = 4x(1/4) = x

(4) And we have:

x

^{2}+ 4xcot 4C - 4 = 0

(5) Using the quadratic equation (see Theorem, here), we have:

x = (1/2)[-4cot4c ± √(4cot 4c)

^{2}+ 16]

(6) Using Lemma 1, here, we have:

x = (1/2)[-4(1/2)(cot 2c - tan2c) ± √16(1/4)(cot 2c - tan 2c)

^{2}+ 16) ]= = (tan 2c - cot2c) ± (1/2)√4[cot

^{2}(2c) - 2cot(2c)tan(2c) + tan

^{2}(2c)] + 16 = = (tan 2c - cot 2c) ± √cot

^{2}(2c) - 2 + tan

^{2}(2c) + 4 = = (tan 2c - cot 2c) ± √cot

^{2}(2c) + tan

^{2}(2c) + 2 = = (tan 2c - cot 2c) ± √[cot(2c) + tan(2c)]

^{2}= = (tan 2c - cot 2c) ± [cot(2c) + tan(2c)]

(7) Since x

_{1}is greater than x

_{2}, we have:

x

_{1}= 2 tan 2c x

_{2}= -2 cot 2c

QED

Lemma 4:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17

ζ

^{k}= cos kα + isin k α

and:

y

_{1}= ζ

^{1}+ ζ

^{13}+ ζ

^{16 }+ ζ

^{4}

y

_{2}= ζ

^{9}+ ζ

^{15}+ ζ

^{8}+ ζ

^{2}

y

_{3}= ζ

^{3}+ ζ

^{5}+ ζ

^{14}+ ζ

^{12}

y

_{4}= ζ

^{10}+ ζ

^{11}+ ζ

^{7}+ ζ

^{6}

then:

y

_{1}= tan(C + π/4) y

_{2}= tan(C - π/4) y

_{3}= tan C y

_{4}= -cot C

Proof:

(1) From steps #9 thru #13 in Theorem 1, here, we have:

y

_{1}, y

_{2}are solutions to:

y

^{2}- x

_{1}y - 1 = 0

and

y

_{3}, y

_{4}are solutions to:

y

^{2}- x

_{2}y - 1 = 0

(2) Let's solve first for y

_{1}, y

_{2}

(3) Using the quadratic equation (see Theorem 1, here), we get:

y = (1/2)(x

_{1}± √x

_{1}

^{2}+ 4)

(4) From Lemma 2 above, we know that x

_{1}= 2 tan 2c

(5) So that we have:

y = (1/2)(2 tan 2c ± √4(tan 2c)

^{2}+ 4 = = tan 2c ± √(tan 2c)

^{2}+ 1

(6) Using Lemma 4, here, we have:

tan 2c ± √(tan 2c)

^{2}+ 1 = (2 tan c)/(1 - tan

^{2}c) ± √(4 tan

^{2}c)/(1 - 2tan

^{2}c + tan

^{4}c) + 1 = = (2 tan c)/(1 - tan

^{2}c) ± √(4 tan

^{2}c + 1 - 2tan

^{2}c + tan

^{4}c )/(1 - 2tan

^{2}c + tan

^{4}c) = = (2 tan c)/(1 - tan

^{2}c) ± √(2tan

^{2}c + 1 + tan

^{4}c )/(1 - 2tan

^{2}c + tan

^{4}c) = = (2 tan c)/(1 - tan

^{2}c) ± √(1 + tan

^{2}c )

^{2}/(1 - tan

^{2}c)

^{2}= = (2 tan c)/(1 - tan

^{2}c) ± (1 + tan

^{2}c)/(1 - tan

^{2}c) = = (2 tan c ± [1 + tan

^{2}c])/(1 - tan

^{2}c)

(7) This then gives us:

(2 tan c + 1 + tan

^{2}c)/(1 - tan

^{2}c) = = (tan c + 1)(tan c + 1)/(1 + tan c)(1 - tan c) = (tan c + 1)/(1 - tan c)

and

(2 tan c - 1 - tan

^{2}c)/(1 - tan

^{2}c) = (tan

^{2}c - 2tan c + 1)/(tan

^{2}c - 1) = = (tan c - 1)(tan c - 1)/(tan c - 1)(tan c + 1) = (tan c - 1)/(tan c + 1)

(8) Since tan(π/4) = 1 (see Lemma 1, here), we can restate this as:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) (tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4))

(9) Using Lemma 3, here, we have:

(1 + tan c)/(1 - tan c) = (tan π/4 + tan c)/(1 - (tan c)(tan π/4)) = tan(c + π/4)

(10) Using Corollary 3.1, here, we have:

(tan c - 1)/(tan c + 1) = (tan c - tan π/4)/(1 + (tan c)(tan π/4)) = tan(c - π/4)

(11) Since y

_{1}is greater than y

_{2},

y

_{1}= tan(c + π/4) y

_{2}= tan(c - π/4)

(11) Now, let's solve for y

_{3}, y

_{4}

(12) Using the quadratic equation (see Theorem 1, here), we get:

y = (1/2)(x

_{2}± √x

_{2}

^{2}+ 4)

(13) From Lemma 2 above, we know that x

_{2}= -2 cot 2c

(14) So that we have:

y = (1/2)(-2 cot 2c ± √4(cot 2c)

^{2}+ 4 = = -cot 2c ± √(cot 2c)

^{2}+ 1

(15) Using Lemma 1, here, we have:

-cot 2c ± √(cot 2c)

^{2}+ 1 = -(1/2)(cot c - tan c) ± √(1/4)(cot c - tan c)

^{2}+ 1 = = (1/2)(tan c - cot c) ± √(1/4)(cot

^{2}c + tan

^{2}c - 2(cot c)(tan c)) + (4/4) = = (1/2)(tan c - cot c) ± √(1/4)(cot

^{2}c + tan

^{2}c - 2 + 4) = = (1/2)(tan c - cot c) ± √(1/4)(cotc + tanc)

^{2}= = (1/2)(tan c - cot c) ± (1/2)(tan c + cot c)

(16) Since y

_{3}is greater than y

_{4}, this gives us:

y

_{3}= tan c y

_{4}= -cot c

QED

Corollary 4.1:

If C is the least positive acute angle such that tan 4C = 4

with:

α = 2π/17 tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α

Proof:

(1) Let us define the following:

y

_{3}= ζ

^{3}+ ζ

^{5}+ ζ

^{14}+ ζ

^{12}

and

y

_{2}= ζ

^{9}+ ζ

^{15}+ ζ

^{8}+ ζ

^{2}

where

ζ

^{k}= cos kα + isin k α

(2) Using Lemma 1 above:

y

_{3}=(ζ

^{3}+ ζ

^{14}) + (ζ

^{5}+ ζ

^{12}) = 2cos 3α + 2cos 5α

y

_{2}= (ζ

^{8}+ ζ

^{9}) + (ζ

^{2}+ ζ

^{15}) = 2 cos 8α + 2 cos 2 α

(3) Using Lemma 5, here, we have:

2 cos(3α) * 2cos(5α) = 2*[2 cos(3α)*cos(5α)] = 2*[cos(5 + 3)α + cos(5 - 3)α] = 2[cos(8α) + cos(2α)] = 2cos(8α) + 2cos(2α)

(4) From Lemma 4 above, we then have:

tan c = 2 cos 3α + 2 cos 5α tan (c - π/4) = 2 cos 2α + 2 cos8α = 2cos(3α)*2cos(5α)

QED

Theorem 5: Constructibility of Heptadecagon using compass and ruler

Proof:

(1) Let CD, CP0 two radii of a circle C1 that are perpendicular to each other.

(2) Let CG be 1/4 the length of CD (bisector of a bisector).

(3) Let ∠ CGL be 1/4 ∠ CGP0 with L on CP0 (bisector of a bisector).

(4) Let ∠ LGP = π/4 (bisector a right angle since right angle = π/2, see here for review of radians) where P is on CP0.

(5) Let Q be the midpoint of PP0 and draw a circle C2 with center Q and radius QP0.

(6) Let R be the point where the circle C1 intersects with CD.

(7) Draw a circle C3 with center L and radius equal to RL.

(8) Let S,Q be the points where the circle C3 intersects with CP0.

(9) Let SP5, QP3 be lines parallel to CP0 that intersect with circle C1.

(10) Let C = measurement ∠ CGL

(11) Then 4C = measurement ∠ CGP0

(12) cos P0CP3 = CQ/CP3 = CQ/CP0 [See here for definition of cosine]

(13) cos P0CP5 = cos (π/2 + π/2 - SCP5) = cos (π - SCP5)

(14) Using cos(a + b) = (cos a)(cos b) - (sin a)(sin b) (see Theorem 2, here) and cos(-x) = cos(x) (see Property 9, here), we have:

cos(π - SCP5) = (cos π)(cos -SCP5) - (sin π)(sin -SCP5) = (-1)cos(SCP5) - (0)(-SCP5) = -cos(SCP5)

(15) So, cos P0CP5 = -cos(SCP5) = -SC/CP5 = -SC/CP0

(16) This gives us that:

2cos P0CP3 + 2cos P0CP5 = 2(CQ - SC)/CP0

(17) Since CQ = CL + LQ and SC = SL - CL and SL=LQ:

SC = SL - CL = LQ - CL

CQ - SC = (CL + LQ) - (LQ - CL) = CL + LQ - LQ + CL = 2CL

(18) This gives us that:

2(CQ - SC)/CP0 = 2(2CL)/CP0 = 4CL/CP0

(19) Since CG is 1/4 of CD and CD=CP0, we have:

4CL/CP0 = 4CL/CD = 4CL/(4CG) = CL/CG

(20) Since GCL is a right angle and C = measurement ∠ CGL, we have:

tan CGL = tan c = sin CGL/cos CGL = (LC/GL)/(GC/GL) = LC/GC = 2cos P0CP3 + 2cos P0CP5

(21) Since cos P0CP3 =CQ/CP0 and cos P0CP5 = -SC/CP0, we have:

2 cos(P0CP3)*2cos(P0CP5) = -4*(SC)(CQ)/(CP0)(CP0)

(22) Using the Pythagorean Theorem (see Theorem, here):

CR

^{2}= CL

^{2}+ RL

^{2}= (CL - RL)(CL + RL)

(23) Since LQ=LR and LS=RL:

CL + RL = CL + LQ = CQ

CL - RL = CL - LS = SC

(24) So we have:

CR

^{2}= (CL - LR)(CL + LR) = (CQ)(SC)

(25) Using the Pythagorean Theorem:

(RC)

^{2}= RQ

^{2}- CQ

^{2}= (RQ - CQ)(RQ + CQ)

(26) Now RQ = QP and RQ=QP0 so that:

(RQ - CQ) = (QP - CQ) = PC

(RQ + CQ) = (QP1 + CQ) = CP1

(27) This gives us that (RC)

^{2}= (PC)(CP0) so that:

-4(RC)

^{2}/(CP0)

^{2}= -4(PC)(CP0)/(CP0)(CP0) = -4(PC)/(CP0)

(28) Since CG = (1/4)CD = (1/4)CP0, we have:

-4(PC)/(CP0) = -4(PC)/(4CG) = -PC/CG

(29) Also:

tan ∠ CGP = sin ∠ CGP/cos ∠ CGP = (PC/GP)/(CG/GP) = PC/GC

(30) Since ∠ CGP = ∠ LGP - ∠ CGL, we have:

tan ∠ CGP = tan (∠ LGP - ∠ CGL) = tan(π/4 - C)

and this means:

-PC/GC = -tan(π/4 - C) = tan(C - π/4)

(31) Using Corollary 4.1 above, we have:

2cos ∠ P0CP3 + 2cos ∠ P0CP5 = 2 cos 3α + 2 cos 5α = tan c where α = 2π/17

2 cos(∠P0CP3)*2cos(∠P0CP5) = 2cos(3α)*2cos(5α) = tan (c - π/4)

(32) Now, it is clear that ∠ P0CP3 and ∠ P0CP5 map onto 3α and 5α.

(33) Since ∠ P0CP5 is greater than ∠P0CP3, it is clear that:

∠ P0CP3 = 3α

∠ P0CP5 = 5α

QED

References

- G. H. Hardy & E. M. Wright, An Introduction to the Theory of Numbers, Oxford Science Publications, 1979