Download 16 The Heptagon

16
16.1
The Heptagon
The construction
We inscribe a regular 7-gon inside a circle. The construction uses a marked ruler to
trisect an angle. Probably, it was already known to Archimedes.
Construction 16.1 (Construction of a Regular Heptagon). Given is a circle C of
radius 3 with horizontal diameter BOA, and point F on the radius OA with OF = 1.
Let OAC be equilateral. Draw a second circle A around F through point C. Next
use Archimedes’ construction to trisect the angle 3α = ∠AF C. One gets a point G
where the segment HC intersects the circle A just drawn, and three congruent segments
HG ∼
= GF ∼
= F C. The HF C has ∠F HC = α, ∠F CH = 2α. Finally, draw circular
arcs of radius 3 around H. Let I, K be its intersection points with the original circle of
the same radius around O. The isosceles HOI has base angles β = ∠BOI ∼
= ∠IHO.
The assertion is that
360◦
β=
7
Now you can get a regular heptagon with K, B, I three of its seven vertices.
Figure 16.1: The heptagon
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Problem 16.1. (a) Actually do the construction, using an entire sheet. Draw the 7-gon
and check for accuracy.
(b) Use trigonometry and a numerical computation to check the validity of the construction.
Figure 16.2: How I begin
16.2
Trigonometric calculations
Answer. These are the quantities for which exact expressions are needed:
Question (i). Find the radius |F C| of circle A.
Answer. Let√D be the midpoint of segment OA. The altitude of the equilateral OAC
3 3
. Hence the Pythagorean theorem yields |F C|2 = |F D|2 + |DC|2 =
is |DC| =
2
1
+ 27
= 7, and
4
4
√
|F C| = 7
Question (ii). We define angle ∠AF C =: 3α. Calculate cos 3α
Answer.
cos 3α =
1
|F D|
= √
|F C|
2 7
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Figure 16.3: Use Archimedes’ trisection
Question (iii). Calculate
x :=
|F H|
|OA|
in terms of cos α.
Answer. By the sin theorem, the ratio of two sides of a triangle equals the ratio of the
opposite angles. For HF C, the sin theorem and the double angle formula sin 2α =
2 sin α cos α imply
|F H|
sin 2α
=
= 2 cos α
|F C|
sin α
Hence
|F H|
|F H|
=
x=
|OA|
|F C|
Question (iv). Recall that β = ∠BOI ∼
=
HOI. Calculate cos β in terms of x.
√
|F C|
2 7 cos α
·
=
|OA|
3
∠IHO are the base angles of the isosceles
Answer. Let M be the midpoint of segment HO. From the right HIM one gets
cos β =
|HF | − 1
3x − 1
|HO|
=
=
2|HI|
6
6
These are fairly all quantities needed for the construction.
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Proposition 16.1. As the construction is done, we define the two angles 3α := ∠AF C
and β := ∠BOI.
1
cos 3α = √
2 7
√
2 7 cos α
1
|F H|
=
= 2 cos β +
x=
|OA|
3
3
(16.1)
(16.2)
satisfies the third order equation
7
7
x3 − x −
=0
3
27
(16.3)
Proof. Because we have shown that cos 3α =
yields
1
√
,
2 7
the identity cos 3α = 4 cos3 α −3 cos α
1
4 cos3 α − 3 cos α − √ = 0
2 7
3
√
√
7
7 2 7 cos α
2 7 cos α
−
=0
− ·
3
3
3
27
This means that x =
√
2 7 cos α
3
satisfies the third order equation (16.3)
(c) Use formulas (16.1) and (1.11) for a numerical calculation of the quantities 3α, α, x, cos β, β
and 7β.
Answer.
(16.4)
16.3
3α = 79.10660535◦ , α = 26.36886845◦ ,
x = 1.580312937 , cos β = 0.6234898019 ,
β = 51.42857143◦ , 7β = 360◦
Using complex numbers
On giants shoulders. Easier insight is gained from complex numbers. (That is why we
are on ”giants shoulders”.) Define
2π
, z1 = eiγ
7
1
u := 2 cos γ +
3
γ :=
(16.5)
We prove that u is a root of equation (16.3). The complex number z is a root of the
polynomial equation
z7 − 1 = 0
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Division by z − 1, next z 3 , and the binomial formula, next 2 cos γ = z + z −1 , then again
the binomial formula lead to the identities
(16.6)
z7 − 1
z6 + z5 + z4 + z3 + z2 + z + 1
=
(z − 1)z 3
z3
= z 3 + z −3 + z 2 + z −2 + z + z −1 + 1
= (z + z −1 )3 − 3(z + z −1 ) + (z + z −1 )2 − 2 + z + z −1 + 1
= (z + z −1 )3 + (z + z −1 )2 − 2(z + z −1 ) − 1
= 8 cos3 γ + 4 cos2 γ − 4 cos γ − 1
3
1
2 cos γ
1
= 2 cos γ +
−
− 4 cos γ − 1
−
3
3
27
3
1
7
7
1
2 cos γ +
−
−
= 2 cos γ +
3
3
3
27
This means that u := 2 cos γ +
shown.
1
3
satisfies the third order equation (16.3), as to be
Question (d). A cubic equation has three real or complex roots, counting multiplicity.
What are the two other roots of equation (16.3)?
Answer. The other two solutions of equation (16.3) are
(16.7)
u2 := 2 cos 2γ +
1
,
3
u3 := 2 cos 3γ +
1
3
Indeed, because z12 = e2iγ and z13 = e3iγ are two further solutions of z 7 − 1 = 0, the
calculation from above shows that u2 and u3 satisfy equation (16.3), too.
Question (e). Calculate u2 and u3 from (16.7) numerically and check that they satisfy
equation (16.3)
Answer.
u2 = −0.1117085346 , u3 = −1.468604402
Question (f ). We have shown above that x = 2 cos β + 13 and u = 2 cos γ + 13 both satisfy
the same equation (16.3). What is the actual value of β? Is the construction working
exactly, or is it only an approximation?
Answer. One can conclude that x = u or x = u2 or x = u3 .
A reason why only the first case is possible, can be obtained from geometry directly.
The second case x = u2 would mean that x = 2 cos β + 13 = 2 cos 2γ + 13 . Hence we
would get β = 2γ! We can see from the construction that this does not happen. Indeed
◦
is obtuse. Similarly, one rules out the
the angle β is acute, but the angle 2γ = 2 · 360
7
third case. Still a different reason for the first case is direct numerical evidence.
◦
Thus we have finally confirmed that x = 2 cos β + 13 = 2 cos γ + 13 and β = γ = 360
.
7
Hence the construction is exactly valid.
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16.4
Geometric proof
The following geometric proof is credited to François Vièta.
Proposition 16.2. The lines QO and IA are parallel.
Proof. By means of Euclid V.6, proportions can be used to show that lines QO and IA
are parallel. To this end, we need to check whether
(16.8)
|HO|
|HQ|
=
|HI|
|HA|
To get (16.8), we use as first step chord in a circle, Euclid III.36. This proposition
implies that |HQ| · |HI| = |HB| · |HA|. Hence
|HQ| |HA|
|HQ| · |HI| |HA|
|HA| · |HB| |HA|
|HB| · |HA|2
·
=
=
=
·
·
|HI| |HO|
|HI|2
|HO|
|OB|2
|HO|
|HO| · |OB|2
That seems
√ stuck, but we need to use the construction yet. From the construction
|HO| = 2 7 cos α − 1 = 3x − 1, |HA| = 3x + 2, |HB| = 3x − 4, |OB| = 3 and hence
|HB| · |HA|2
(3x − 4)(3x + 2)2
=
|HO| · |OB|2
(3x − 1) · 9
Now equation (16.3) mysteriously implies that this last fraction equals one. Hence we
get indeed equation (16.8), Hence by Euclid V.6, the lines QO and IA are parallel.
Proposition 16.3 (The angle sum of a triangle comes in seven parts).
Let Q be the second intersection point of line HI and the main circle around O.
Lines QO and IA are parallel. The angles ε, 2ε = β, 3ε comes up seven times with
vertices H, Q, I, O, A as shown in the drawing. Hence the angle sum of HIA implies
that 7ε = 180◦ and
360◦
β=
7
Question (g). Mark the angles ε, 2ε, 3ε seven times in the drawing below!
Proof. Now we use Euclid I.29 to produce congruent z-angles from the parallel lines,
together with congruence of base angles of isosceles triangles. One gets that
ε = ∠HOQ ∼
= ∠HAI ∼
= ∠AIO ∼
= ∠IOQ
From the congruent base angles of isosceles HIO, one gets
β := ∠OHI ∼
= ∠HOI = 2ε
Form the exterior angle in HQO, and finally the bases angles of isosceles QOI, one
gets
β + ε = ∠OQI ∼
= ∠OIQ = 3ε
Hence the angle sum of HIA comes in seven parts and 7ε = 180◦ , as to be shown.
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Figure 16.4: Angle sum in seven parts
Remark. ε = β2 = 25.71428571 < α. Hence lines OQ and F G are not parallel, but
β
α
intersect in the lower half plane. I conjecture that 360
◦ is irrational, and only 360◦ is
rational.
16.5
A false but almost regular heptagon
Problem 16.2 (A false heptagon). A false construction of a regular heptagon is
shown in the figure on page 16.5. We use a circum circle of radius |OA| = 2.
1. Calculate the false side |AM |.
2. Calculate the true side of the regular heptagon.
3. How much too short is the false side.
4. Calculate the accumulated center angle ∠W OA by which the false heptagon does
not close.
Answer.
1. |OA| = 2 is assumed, hence |OM | = 1. Pythagoras for the triangle
OAM gives
|AM |2 = |OA|2 − |OM |2 = 3
√
|AS| = |AM | = 3 ≈ 1.732050808
for the constructed side.
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Figure 16.5: A false heptagon.
2. The central angle for the regular heptagon is 360◦ /7. The true side is
2|OA| sin
360◦
≈ 1.735534956
14
3. The true side is about 0.0035 longer.
4. More instructive is to know the angle ∠AOW by which the false 7-gon fails to
close. Let α = ∠SOA.
√
3
α
|AS|
=
hence
sin =
2
2|OA|
4
α ≈ 51.3178◦ and ∠AOW = 7α − 360◦ ≈ −0.775◦
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16.6
Plemelj’s construction
Figure 16.6: A regular heptagon.
Construction 16.2 (Plemelj’s construction of a regular heptagon). Draw the
circle with center O passing through A. Find on this circle the point M so that AM ∼
=
OA. Bisect segment OM at N , and trisect it at P . Find point T on the segment N P
so that ∠N AT = ∠N AP/3. The segment AT is the needed side of the regular heptagon
with the circum circle drawn in the beginning.
For the justification of the construction, we check that both the constructed side,
and the true side of the regular heptagon inscribed into the unit circle satisfy the cubic
equation
√
√
y 3 + 7y 2 − 7 = 0
But this cubic equation has a unique positive solution. Hence the construction yields
the regular heptagon.
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Calculation of the constructed side. The tangent and cosine of the angle α = ∠N AP to
be trisected are
√
√
27
1
√
=
tan α = 1/(3 3) and cos α = √
28
1 + tan2 α
For the trisected angle, u = 2 cos(α/3) is a root of the polynomial
√
27
3
3
u − 3u − 2 cos α = u − 3u − √ = 0
7
For the constructed side length
√
√
3
3
=
y := |N A| =
2 cos(α/3)
u
we get the polynomial equation
√ 3
√
√
3
3
27
− √ =0
−3
y
y
7
√
√
y 3 + 7y 2 − 7 = 0
This equation has exactly one positive solution. You can use Descartes’ rule of signs
and elementary calculus to check. Hence it really determines the constructed piece
uniquely.
Reduction of the true side to a cubic root. We have seen in equation (16.6) that
8 cos3 γ + 4 cos2 γ − 4 cos γ − 1 = 0
has the solutions γ = 2πk/7 with k = 1, 2, 3. Hence the polynomial R(v) := v 3 + v 2 −
2v − 1 has the zeros 2 cos(2πk/7). The actual side is
πk
7
with k = 1 and the diagonals are obtained for k = 2, 3. They are related to half the
angle. We use
πk
2πk
2 − x2k = 2 − 4 sin2
= 2 cos
7
7
and see that these are zeros of the polynomial R7 (v). Now we substitute:
xk = 2 sin
R(2 − x2 ) = (2 − x2 )3 + (2 − x2 )2 − 2(2 − x2 ) − 1 = −x6 + 7x4 − 14x2 + 7
√
The polynomial is still reducible over Q( 7) and hence over the constructible field.
Indeed
√
√
√
√
x6 − 7x4 + 14x2 − 7 = (x3 + 7x2 − 7)(x3 − 7x2 + 7)
From Descartes rule of signs, and by direct calculation, we see that the first fac, −2 sin 4πk
, whereas the second factor has zeros
tor has the zeros 2 sin π7 , −2 sin 2πk
7
7
π
2πk
4πk
−2 sin 7 , 2 sin 7 , 2 sin 7 .
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Question. How can we now see that the regular 7-gon is not constructible with ruler
and compass?
Answer. The polynomial x6 − 7x4 + 14x2 − 7 has as zeros the side and diagonals of the
regular heptagon inscribed into the unit circle. By the Eisenstein criterium with p = 7,
we see that this polynomial is irreducible over the integers. Hence any of its roots r
generate a field extension [Q(r) : Q] = 6 of dimension six. But this is not a power of
two. Hence r is not in the constructible field.
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