15 Trisection of an Angle and the Delian Problem

15
15.1
Trisection of an Angle and the Delian Problem
Trisection by Archimedes
Figure 15.1: Trisection of an acute angle
Construction 15.1 (Trisection of an angle by Archimedes). Suppose that on a
straightedge two points C and D of distance d have been marked. Given is the angle
α = ∠AOB to be trisected. We draw a circle C of radius d centered at the vertex O
of the angle. This circle cuts the sides of the angle at points A and B. One places the
marked ruler such that the following three requirements are met:
−→
1. One of the marks on the ruler yields point C on the ray opposite to OA.
2. The other mark yields point D on the circle C.
3. The point B lies on the ruler.
Result: With this construction, the angle γ = ∠OCD is one third of the given angle
α = ∠AOB.
Problem 15.1. Provide a drawing for this construction in the following cases
(a) for a given acute angle α < 90◦ .
(b) for a given angle 90◦ < α < 135◦ .
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Figure 15.2: Trisection of an obtuse angle less than 135◦ .
Figure 15.3: Trisection of an obtuse angle larger than 135◦ .
(c) for a given angle α > 135◦ .
Problem 15.2. In all cases, measure your angles α and γ.
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Answer (Measurements). For my examples, I got
In example (a): α = 54◦ , γ = 18◦
In example (b): α = 108◦ , γ = 36◦
In example (c): α = 153◦ , γ = 51◦
Problem 15.3. Let A be the second end of diameter A A. Since the base angles of
the isosceles triangle COD are congruent, the trisected angle appears both at vertex
C and the vertex O, too. How can one check the accuracy of the construction without
using numbers or a protractor?
Answer. It is easy to use a compass to measure the chord A D of the angle γ = ∠COD =
∠A OD. Then one can check whether this chord fits three times on the arc AB of the
original angle α.
Remark. In general, it is not possible to trisect an angle just with the tradition Euclidean
tools—compass and unmarked straightedge.
But, once the two-marked straightedge is put into the position such that the three
requirements are satisfied, the construction is exact—as shown below.
Problem 15.4. Give the reason, why Archimedes’ trisection construction is exact.
Justification of Archimedes trisection. Since the marked segment CD has a length equal
to the radius of the circle, and all points of a circle have the same distance from the
center, we know that the three segments
OB ∼
= OD ∼
= CD
are congruent. Hence the triangle COD is isosceles. Its congruent base angles
γ = ∠OCD ∼
= ∠COD
appears both at vertex C and at vertex O. The exterior angle at the top of this triangle
is β = ∠ODB. By Euclid I.32, the exterior angle of a triangle is the sum of the two
nonadjacent interior angles. Hence
β = ∠ODB = ∠OCD + ∠COD = 2γ
In the case of an angle α < 135◦ , the bases angle of the second isosceles triangle ODB
are
β = ∠ODB ∼
= ∠OBD
(15.1)
In case (c), the supplements of the base angles of triangle OBD are β = 2γ. Finally,
we use Euclid I.32 for the triangle OBC. Since the given angle α is an exterior angle
for this triangle, we conclude
α = ∠AOB = ∠OCB + ∠CBO = γ + β = γ + 2γ = 3γ
. Hence γ = α3 , as to be shown.
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Problem 15.5. Explain the difference of the drawings for the cases (a), (b) and (c).
Answer (Difference between the cases). In cases (a) and (b), point D in between C and
B. Hence D does not lie in the interior of the given angle, ∠AOB. The base angles of
the isosceles OBD are β = 2γ. This is the same situation as did occur for an acute
angle.
In case (c), point B is between C and D and point D lies in the interior of angle
∠AOB. The supplements of the base angles of triangle OBD are β = 2γ.
15.2
Trisection by Nicomedes
Construction 15.2 (Trisection by Nicomedes, 240 B.C.). Given is the angle α =
∠COB to be trisected. On the ruler is marked a segment P Q with length double as
OB. One drops the perpendicular p from B onto line OC. Next one erects at point B
a second perpendicular q on p. Indeed, line q is the parallel to OC through point B.
Finally, one places the marked ruler such that the following three requirements are met:
1. The ruler line goes through the vertex O of the given angle.
2. The point marked P on the ruler lies on the perpendicular p.
3. The point marked Q on the ruler lies on the parallel q.
Result: We claim that the angle γ = ∠QOC is one third of the given angle α = ∠AOB.
Remark. In general, it is not possible to trisect an angle just with the tradition Euclidean
tools—compass and unmarked straightedge.
But, once the two-marked straightedge is put into the position such that the three
requirements are satisfied, the construction is exact—as we prove below.
Problem 15.6. Provide a drawing for this construction for several different angles:
(a) an angle 90◦ < α < 135◦ .
(b) an angle α > 135◦ .
Tell in a few words, what made you stumble at first, and how this second drawing for
case (b) differs from the drawing in case (a), or the drawing for an acute angle given
above. Measure and report your angles α and the trisected angles γ.
Answer. One needs to turn the ruler such that the point P on the perpendicular lies
below the line OC, on the other side than point B. Too, there exists a second fake
solution with M = O which does not yield angle trisection.
To get appropriate angle trisection, the order of the points P, O, M and Q on the
ruler is different in cases (a) and (b). In case (a), points Q and M lie on the same side,
and point P on the other side of point O. In case (b), points P and M lie on the same
side, and point Q on the other side of point O.
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Figure 15.4: Trisection of an acute angle
Figure 15.5: Trisection of an obtuse angle between 90◦ and 135◦ .
Problem 15.7.
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Figure 15.6: Trisection of an obtuse angle larger than 135◦ .
Question. Suppose that M is the midpoint of the segment P Q. Give a reason why the
three segments M P , M Q and M B are congruent.
Answer. The angle ∠P BQ is a right angle by construction. Hence the converse Thales’
Theorem 32 implies that point B lies on the circle with the hypothenuse of triangle
P QB as diameter P Q, and hence the three segments P M, M B and M Q are congruent.
Problem 15.8. Prove that the construction done in part (a) of problem (15.6) produces
an exact trisection.
Validity of the construction. We know that the three segments P M, M B and M Q are
congruent, where M is the midpoint of segment P Q. By construction, segment OB has
half the length of segment P Q. Hence all four segments OB, P M, M B and M Q are
congruent. We get the isosceles triangles M QB and OM B.
Question. Let γ = ∠M BQ ∼
= ∠M QB measure the base angles of the first isosceles
M QB. Find congruent z-angles of size γ. Mark all three angles congruent to γ with
matching color.
Answer. By construction, lines OC and BQ are parallel. With the transversal OQ, they
form z-angles ∠COQ ∼
= ∠OQB = γ, which are congruent by Euclid I.28.
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The base angle ∠OM B of the second triangle is an exterior angle at the top vertex
of the first M QB. By Euclid I.32, the exterior angle of a triangle is the sum of the
two nonadjacent interior angles. Hence
β = ∠OM B = ∠M BQ + ∠M QB = 2γ
The bases angle of the second isosceles triangle OM B are
β = ∠OM B ∼
= ∠BOM
(15.2)
Finally, angle addition at vertex O implies
α = ∠COB = ∠COQ + ∠QOB = 2γ + γ = 3γ
Hence γ = α3 , as to be shown.
Question. The meaning of the angles β from equation 15.2 will change in the case (b)
from problem 15.6. What is the the meaning of the angles β in case one trisects an
angle α > 135◦ . Why is equation 15.2 still true.
Answer. The angles β become the supplements of the base angles of the isosceles triangle
OM B. But, of course, the supplements of the two base angles of an isosceles triangle
are congruent to each other. Hence equation 15.2 is still true.
15.3
Trisection with Nicolson’s angle, and by origami
Construction 15.3 (Trisection with Nicolson’s angle (1883)). Nicolson’s angle
consists of a strip of two parallels with distance AB to which an extra square of congruent
side BC is attached at point B. Thus B becomes the midpoint of segment AC. The
points A, B and C are marked on the Nicolson angle.
−→
To trisect any given angle ∠P OR, we first draw the parallel to side OP at distance
AB which cuts the interior of the given angle. Next we place the Nicolson angle in such
a way that
−→
• on the parallel to side OP lies point A;
−→
• on the other side OR lies point C;
• The vertex O lies on the edge of the Nicolson angle with extension through point
B.
−→
−−→
The rays OA and OB trisect the given angle ∠P OR.
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Figure 15.7: Trisection with Nicolson’s angle.
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Figure 15.8: Trisection by paperfolding
Construction 15.4 (Trisection of an angle by paper folding). Take a rectangular
sheet of paper. One needs the right angle at the lower left corner, which is point A.
Choose an acute angle θ to be trisected. Place the angle you have chosen with its vertex
at A, and one side along the lower edge a of the paper. Draw the other side t of angle
θ in blue across the paper.
The trisection construction is begun by drawing, on the front side of the paper, a
parallel b to the lower edge a. Next one draws a parallel c to b which has the double
distance from the lower edge a. This can conveniently be done by folding the paper along
b. Let B and C be the intersections of the parallels b and c with the vertical left margin
l of the paper. Mark point C on the back side of the paper. Now the paper is folded such
that
(i) the lower left corner A is flipped over to a point A on the line b,
(ii) point C is flipped over onto point C on the side t of angle θ.
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−−→
Claim: The lower edges a and the ray AA form the angle θ/3.
Problem 15.9. (a) Actually do the construction on a separate sheet of paper. Mark
the quantities as explained above. Let α := ∠SAA be the angle between the lower edges
−−→
a and the ray AA . Measure and report your angles θ and α. What about accuracy of
the trisection?
Answer. My choice came out to be θ = 39.5◦ . Measurement yields ∠A Aa = 13◦ . Thus
the relative error is about 1%.
Question (b). Find and mark the trisected angle at five or more places.
Answer. The angle 3θ appears three times with vertex A, and twice with vertex W , where
the paper crest intersects the left paper margin.
Figure 15.9: Reflection at the folding line
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Validity of the construction. The paper folding generates a reflection across the paper
crest s as symmetry axis. Let W be the intersection of the paper crest s with the left
margin. Let S, T and U be the points where the paper crest s intersects lines a, b and
c. Let A , B and C be the reflection images of A, B and C.
Question (c). Draw the lines SA , T B and U C and the line W A B C . What do you
observe about these lines?
Answer. The three lines SA , T B and U C are all perpendicular to line W A B C .
Hence they are parallel.
Question (d). Compare the segments A B and B C .
Answer. The two segments A B and B C are congruent. Here is the reason: A segment
is congruent to its reflected segment. Hence AB ∼
= A B and BC ∼
= B C . The two
segments AB and BC are congruent by construction. Hence, because congruence is
transitive A B ∼
= BC .
We need to compare the three angles α := ∠SAA , β := ∠A AT and γ := ∠B AC .
Question (e). How do the two triangles AB A and AB C compare? How do the
two angles β and γ and compare?
Answer. The two triangles AB A and AB C are congruent. Indeed, they have a
common side AB , both have a right angle at vertex B , furthermore the other pair of
legs A B ∼
= B C , is congruent, too. Hence they are congruent by SAS congruence.
Hence, the two angles β and γ are congruent, too.
Question (f ). Give a direct reason that the angles α and β are congruent.
Answer. The angles α := ∠SAA and β := ∠AA T are z-angles (alternate interior
angles) between the parallels a and b. By Euclid I.29, z-angles between parallels are
congruent. The angles β and β are congruent, because they are images by reflection.
By transitivity, we see that α and β are congruent.
Question (g). Draw the diagonals of quadrilateral ASA T . What kind of quadrilateral
is this?
Answer. The two diagonals are perpendicular, because the segment from a point A and
its reflection image A is perpendicular to the reflection axis ST . The four sides are
congruent. The diagonals are perpendicular. Hence the quadrilateral ASA T is a
rhombus.
Question (h). What can you say about the angles that these diagonals form with the
sides of the quadrilateral?
Answer. These two angles are congruent, as shown above. Too, it is well known that
the diagonals in the rhombus bisect its angles.
To conclude the argument recapitulate: we have shown that the three angles α :=
∠SAA , β := ∠A AT and γ := ∠B AC are congruent. Since by angle addition θ =
α + β + γ, we get θ = 3α and α = 3θ , as to be shown.
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Problem 15.10 (Some experimentation). To better understand the trisection construction and the proof of its validity from above, do some paper folding where one or
another of the the assumptions are violated.
Question (a). Choose the three parallel lines a, b and c such that AB < BC. How do
the three angles α, β and γ turn out?
Answer. The two angles α and β are still congruent, but α < γ.
Question (b). Choose the three parallel lines a, b and c such that AB > BC. How do
the three angles α, β and γ turn out?
Answer. The two angles α and β are still congruent, but α > γ.
Question (c). Choose only the two lines a and c parallel, but not line b. How do the
three angles α, β and γ turn out?
Answer.
15.4
Construction of the cubic root by two-marked ruler
The construction given here is a slight simplification of the version reported by Arthur
Baragar in the recent article [5]. Let the ruler have two marks C and D with distance
|CD| = 1. Given is the segment P Q √
of length a < 4. I use the two-marked ruler for the
construction of a segment of length 3 2a.
Construction 15.5 (Nicomedes’ Construction of a Cubic Root). Let T be the
midpoint of the given segment P Q. Construct the perpendicular bisector of the half
segment T Q. Let A be the midpoint of T Q and O be a point on the bisector such that
|OT | = 1. Extend line OT and let B the point such that T is the midpoint of segment
OB. Finally, one places the two-marked ruler. We turn it around point O and put the
two marks on the lines c = BQ and d = P Q, such that the following three requirements
are met:
1. The ruler line goes through point O.
2. The point marked C on the ruler lies on the line c = BQ.
3. The point marked D on the ruler lies on the line d = P Q.
Claim: The segment QD has length
√
3
2a, and the segment OC has the length
3
a2
.
2
Problem 15.11. Provide a drawing and actually do the construction for a = 1. Measure
your results and check for accuracy. What kind of triangle is OT Q?
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Figure 15.10: Nicomedes’ construction of
633
√
3
2a.
Simple modern proof. The proof uses only the two most important facts of Euclidean
geometry: similar triangles, and the Theorem of Pythagoras.
Let u = |QD| and v = |OC| denote the lengths of the segments obtained by the construction. The Theorem of Pythagoras for the two right triangles OAQ and OAD,
and substraction yields
a2
|OA|2 +
= |OA|2 + |AQ|2 = |OQ|2 = 1
16
a 2
2
|OA| + u +
(15.3)
= |OA|2 + |AD|2 = |OD|2 = (1 + v)2
4
a 2 a2
u+
= (1 + v)2 − 1
−
4
16
au
= v 2 + 2v
(15.4)
u2 +
2
To eliminate the quantity v, we use proportions. Point T is the midpoint of both
segments P Q and OB. Hence congruent z angles easily imply that the lines OP and
QB are parallel. With center D, and the parallels CQ = QB OP produce the
equiangular triangles DCQ and DOP . By Euclid (VI.4), the sides of equiangular
triangles are proportional, and hence
v
|OC|
|P Q|
a
=
=
=
1
|CD|
|QD|
u
Now we eliminate of the quantity v from equation (15.4) and get
2a a2
au
=
+ 2
u2 +
2
u
u
a
3
u u+
= a(2u + a)
2
One solution of this equation is u4 = − a2 , corresponding to v4 = −2. In the drawing,
this solution corresponds to putting the ruler marks at C4 = B and D4 = T . Cancelling
the factor (u + a2 ) on both sides yields the equation
We get the solution |QD| = u1 =
have been looking for.
√
3
u3 = 2a
2a and |OC| = v1 =
3
a2
.
2
which is the solution we
Remark. This is another version to derive equation 15.4 from Hartshorn’s book [?],
p. 263. Draw a circle around point O of unit radius. It goes through points Q and
T . Let EF be its diameter on the line OD, with point E between D and O. Since
|DC| = |EO| = 1, we get |DE| = |CO| = v. Now equation 15.4 follows from Euclid’s
theorem of chords III.36 since
a
u u+
= |DQ| · |DT | = |DE| · |DF | = v(v + 2)
2
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15.5
Definition and equations of the conchoid
The conchoid of Nicomedes with focus O, guiding line AB, and marked segment p, is
the location of all points Q away from the guiding line by distance p. as measured on
−→
a radial ray. Thus the radial ray OQ cuts the guiding line AB in point P , and the
segment P Q has length |P Q| = p.
Figure 15.11: The conchoid of Nicomedes
Problem 15.12 (The Conchoid in Cartesian coordinates). From this definition,
find the equation of the conchoid of Nicomedes in Cartesian coordinates, with given
segment length p, the focus at the origin, and vertical guiding line x = d.
Answer. Drop the perpendicular from the generic point Q of the conchoid onto the
x axis, and let F be the foot point. Similar triangles OF Q and OAP yield the
proportion
(15.5)
cos θ =
|OF |
|OA|
|OF | − |OA|
x−d
x
=
=
=
=
r
|OQ|
|OP |
|OQ| − |OP |
p
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Hence multiplying, Pythagoras’ theorem and solving for y yield
x−d
x
=
r
p
r · (x − d) = p · x
(15.6)
2
(x + y 2 )(x − d)2 = p2 x2
x 2
y=±
p − (x − d)2
x−d
Problem 15.13 (The conchoid in polar coordinates). Find the equation of the
conchoid of Nicomedes in polar coordinates.
Answer. Since x = r cos θ and y = r sin θ, the equation (15.6) above implies
x−d
d
x
=
=
r
p
r−p
d
r−p=
cos θ
d
r =p+
cos θ
cos θ =
(15.7)
15.6
The conchoid and the construction of the cube root
We now use the conchoid for a better understanding of Nicomedes’ construction 15.5
of a cubic root. The point O around which the ruler is turned is the focus. We place
this point at the origin of the Cartesian coordinate system. Any one of the two lines
c = BQ and d = P Q can be used as guiding line. I shall use P Q as guiding line. The
guiding line is put parallel to the y-axis. Since A is the foot point of the perpendicular
dropped from the focus
onto the guiding line, the x-axis is the line OA.
2
With |OA| = d = 1 − a16 , the guiding line has again the equation x = d. The
marked segment has length p = |CD| = 1.
Question. Give the equation of the conchoid corresponding to the construction 15.5 in
Cartesian coordinates, implicitly and explicitly.
2
Answer. Since p = 1 and d = |OA| = 1 − a16 , the equation 15.6 becomes
(x2 + y 2 )(x − d)2 = x2
(15.8)
y=±
√
x
32dx + a2
4(x − d)
It is now just an exercise of endurance—left to the reader—to calculate the equation
of the line c = QB, get the coordinates of points C and D,√ and finally the segment
length u = |QD|. A graph with a = |P Q| = 1 and d = 415 is given in the figure
on page 637. One can see that segments OQ and OT are tangent to the loop of this
conchoid at the origin, and the conchoid passes through the two real solutions at points
B and C.
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Figure 15.12: Another gift for Apollo.
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15.7
Duplication of the cube by two-marked ruler
Indeed, the duplication of the cube—also known as the Delian problem—is a special
case of Nicomedes’ more general construction of the third root explained earlier. Nevertheless, I include this section, because of the historic importance of the problem, and
because this special case turns out to be remarkably pretty.
Problem 15.14 (Duplication of the Cube with Two-Marked
Ruler). In the
√
drawing of page 639, Nicomedes’ construction of the third root 3 2a is done in the special
case a = 2. We have assumed that both the radius of the circle as well as the segment
marked on the straightedge have unit length |T B| = |CD| = 1.
(a) Describe the construction.
What kind of quadrilateral is OP BQ? Find two equilateral triangles inside of
it. How is it constructed most easily? How has the the two-marked ruler to be
positioned?
(b) The construction√yields two segments
OC and QD, which we claim to have the
√
3
3
lengths |OC| = 2 and |QD| = 4. To confirm this claim, do the following:
• find two similar triangles and set up a proportion for the segments above.
• Calculate the lengths of all three sides of the right triangle OAD exactly
and check the theorem of Pythagoras.
Construction 15.6 (Nicomedes’ Solution of the Delian Problem). Let the ruler
have two marks C and D with distance |CD| = 1. Draw a circle with center T and
the same radius 1. Into the circle, we inscribe the regular hexagon OQEBP F by means
of six cords of length 1. Since OT B and QT P are diameters, we get the rectangle
OP BQ with these diameters as its diagonals. Too, the diagonals of this rectangle
form two equilateral triangles OT Q and P T B.
The lines QP and QB need to be extended to the opposite side of Q. Finally, one
places the marked ruler such that the following three requirements are met:
1. The ruler line goes through point O.
2. The point marked C on the ruler lies on the line QB.
3. The point marked D on the ruler lies on the line QP .
√
√
Claim: The segment OC has length 3 2, and the segment QD has length 3 4.
Answer. The quadrilateral OP BQ is indeed a rectangle. Hence the lines QC and
P O are parallel. Thus the figure on page 639 contains the similar triangles CDQ ∼
ODP . We get the proportions
√
3
|CD|
2
|OC|
1
√
=
=
3
|QD|
|P Q|
2
4
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Figure 15.13: Nicomedes’ Solution of the Delian Problem.
639
The right triangle OAD has the sides
√
√
3
1 √
3
3
, |AD| = + 4 , |OD| = 1 + 2
|OA| =
2
2
Hence
2
2
√
√
√
√
√
3
1 √
3
3
3
3
3
3
2
|OA| +|AD| = +
+ 4 = 1+ 4+ 16 , |OD| = 1 + 2 = 1+2 2+ 4
4
2
2
2
which turn out to be the same value.
The lengths of segments OC and QD give both middle proportions between |T P | = 1
and |P Q| = 2.
One can simplify this drawing a bid further, drawing neither the rectangle nor its
circum circle, but adding
a unit circle around Q. This leads to the following construction
√
3
of a segment of length 2.
Construction 15.7 (Simplified Nicomedes’ Construction). Let the ruler have two
marks C and D with distance |CD| = 1. Draw a circle with center Q of radius 1. Let
OQ be the horizontal diameter and let QM be perpendicular to OQ. Next we need on
the circle point T , such that triangle OQT is equilateral.
Finally, one places the marked ruler such that the following three requirements are
met:
1. The ruler line goes through point O.
2. The point marked C on the ruler lies on the perpendicular QM .
3. The point marked D on the ruler lies on the line QT .
√
Claim: The segment OC has length 3 2.
Problem 15.15. Provide a drawing and actually do the construction. Measure your
results and check for accuracy.
Simple modern proof. Being unfaithful to the old Greeks, I use trigonometry. Let α =
∠QOC. (It turns out to be about 37◦ .) Let x = |OC| be the length of the segment
obtained by the construction. From the right OQC, we get
cos α =
|OQ|
1
=
|OC|
x
and
x=
1
cos α
The sin theorem in OQD implies
(15.9)
|QD|
|OC| + |CD|
|OD|
√
=
=
=
◦
1
sin α
sin 120
3
2
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1
+
cos α
√
1
3
2
1
2 + 2 cos α
= √
3 cos α
Figure 15.14: My final gift to Apollo.
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Figure 15.15: The Delian problem, once more
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The sin theorem in QCD implies
|CD|
|QD|
=
◦
sin(90 + α)
sin 30◦
|QD| = |CD| · 2 cos α = 2 cos α
(15.10)
One can eliminate |QD| and get from (15.9) and (9.7)
2 cos α
2 + 2 cos α
= √
sin α
3 cos α
√
2
3 cos α = (1 + cos α) sin α
(15.11)
We use the variable x =
(15.12)
1
cos α
and square
2 1
1
3
1− 2
= 1+
x4
x
x
3 = (x + 1)2 (x2 − 1)
One solution of this equation of forth order is x4 = −2. Hence one can factor
(x + 1)2 (x2 − 1) − 3 = x4 + 2x3 − 2x − 4 = (x + 2)(x3 − 2)
√
Hence another solution of equation 9.9 is x1 = 3 2.
Geometrically, the solution x4 is obtained by putting C4 := B and D4 := T . The
ruler line is AT B, where ∠QOT = −60◦ , and |OT | = |T B| = 1.
The solution x1 is the one we were√
looking for. It corresponds to the ruler line OCD,
where ∠QOC = α ≈ 37◦ , and OC = 3 2.
Duplication of cube using just geometry. To keep the reasoning in the style of the ancient geometers, I need the Theorem of Menelaus 9.1. We use this theorem for the
triangle OCB and transversal T Q. One gets
(15.13)
|DC| |T O| |QB|
·
·
=1
|DO| |T B| |QC|
Now use that by construction |DC| = |T O| = |T B| = 1 and |DO| = |DC|+|CO| = 1+x.
Furthermore, Pythagoras theorem yields |QB|2 = 22 − 12 = 3 and |QC|2 = |OC|2 −
|QO|2 = x2 − 1. Hence I get
√
3
1
1
· ·√
(15.14)
=1
2
x+1 1
x −1
Now, we go on as in the first version above. Squaring and factoring yields
(x + 1)2 (x2 − 1) − 3 = x4 + 2x3 − 2x − 4 = (x + 2)(x3 − 2)
This is√ an equation of forth order. One solution is x1 = −2. The second solution is
for. It corresponds to the ruler line OCD,
x2 = 3 2, which is the one we are looking
√
3
◦
where ∠OAC = α ≈ 37 , and |OC| = 2.
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15.8
Duplication of the cube and the curve of Agnesi
For the duplication of cube construction, we shall now use the line d = QB as guiding
line of a conchoid. In that special case, the distance from focus to guiding line is exactly
equal to the length of the marked segment. This special conchoid has p = d and is also
known as the curve of Agnesi.
Problem 15.16. Put the simplified Nicomedes’ construction 15.7 in a coordinate system
with origin at focus O = (0, 0), and center of the circle Q = (1, 0). Find the equation of
the conchoid with focus O and guiding line QM ⊥ QO. Find the equation of line QT .
Calculate the coordinates of the points C, D and C4 := B, D4 := T and the distance
|OC|.
Produce a graph and the segments OC and OB and check their lengths.
Answer. The equation of line QT is
(ON)
y=
√
3(x − 1)
The conchoid has parameters p = d = 1, hence its equation is
x 1 − (x − 1)2
(Conchoid)
y=
x−1
Intersection points have the x-coordinate satisfying
√
x 3 (x − 1) =
1 − (x − 1)2
x−1
3(x − 1)4 = x2 (−x2 + 2x)
4x4 − 14x3 + 18x2 − 12x + 3 = 0
(2x − 1)(2x3 − 6x2 + 6x − 3) = 0
(2x − 1)[2(x − 1)3 − 1] = 0
from which we get the coordinates of the two intersection points to be
√ √ 1
3
3
1
,−
D4 = (x4 , y4 ) =
,√
and D = (x, y) = 1 + √
3
3
2
2
2
2
and hence the points on the guiding line are
√ y4
= 1, − 3
and
C4 = 1,
x4
y
=
C = 1,
x
A further exercise in arithmetic yields
√
3
2
2
+
y
2 + 1)2 + 3 √
x
(
3
2
√
=
= 4
|OC| =
3
2
2
x
( 2 + 1)
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√
3
1, √
3
2+1
Figure 15.16: The Delian problem and the curve of Agnesi.
15.9
A close look at Nicomedes’ trisection
Problem 15.17. Repeat Nicomedes’ trisection construction 15.2. Extend the ruler line,
the perpendicular p and the parallel q as long as needed, to both sides. Use a small angle
of about 30◦ . One needs to turn the ruler an entire 360◦ , such that the point P covers the
entire perpendicular line p, and the other marked point slides along the entire parallel q.
Find out how many solutions you get.
Problem 15.18. For all four solutions Pi Qi with i = 1, 2, 3, 4, mark the midpoints Mi
of these segments. What is special about one of these midpoints M4 . What figure do you
get from the segments OM1 , OM2 , OM3 .
Answer. There exists a fake solution with M4 = O which does not yield angle trisection.
The three other segments OM1 , OM2 , OM3 are a star with three congruent angles of
120◦ .
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Figure 15.17: Nicomedes’ trisection construction has four solutions
15.10
Trisection and the conchoid
−→
Let α = ∠AOB be the angle to be trisected, with one side OA on the positive x-axis.
I put Nicomedes’ construction in a Cartesian coordinate system. As above, let
p = |P Q| be the segment marked on the ruler. The origin O is the focus, and as guiding
line, I choose line AB. The point A = (d, 0) is the intersection of the guiding line with
the x-axis. Hence the distance from the focus to the guiding line is
d = |OA| = |OB| cos α =
p cos α
|P Q| cos α
=
2
2
since Nicomedes’ construction requires |OB| = |P2Q| . The Cartesian coordinates of points
A and B are
p cos α p cos α p sin α
, 0 and B =
,
A=
2
2
2
Question. What are the equations of the lines p and q used in the trisection construction?
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Answer. Line p is the parallel to the y-axis through point B, and has the equation
(p)
x=
p cos α
2
Line q is the parallel to the x-axis through point B, and has the equation
(q)
y=
p sin α
2
The ruler in the trisection construction is put onto a radial ray which intersection
guiding line p at point P , and line q at point Q such that |P Q| = p. Hence point Q is
the intersection of the horizontal line q with the conchoid.
Problem 15.19 (Analytic justification of Nicomedes’ angle trisection). Nicomedes
construction can be understood as intersecting the conchoid with a line. Use polar coordinates (r, θ), eliminate distance d. Find the angles θ1 , . . . , θ4 of the four intersection
points of the conchoid with the line q.
Answer. In polar coordinates, the equations of the conchoid, and of line q are
p cos α
d
=p+
cos θ
2 cos θ
p sin α
r sin θ =
2
r =p+
(15.15)
After division by p, one uses some trigonometric identities and gets:
p cos α
2 cos θ
2 sin θ cos θ + cos α sin θ
2 sin θ cos θ
sin 2θ
p+
p sin α
2 sin θ
= sin α cos θ
= sin α cos θ − cos α sin θ
= sin(α − θ)
=
It is not difficult to see that this equation has the four solutions
θ1 =
α
α − 360◦
α + 360◦
, θ2 =
= θ1 − 120◦ , θ3 =
= θ1 + 120◦ , θ4 = 180◦ − α
3
3
3
The equation q yields the radii corresponding to these intersection points.
ri =
p sin α
2 sin θi
for i = 1 . . . 4
Unfortunately, this formula makes r < 0 if 180◦ < θ < 360◦ . I remedy this by substituting
(r2 , θ2 ) → (−r2 , θ2 + 180◦ ) = (|r2 |, θ1 + 60◦ )
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Figure 15.18: The conchoid gives all four solutions
Question (b). Find the angle τ between the positive x-axis and the tangent to the
conchoid at its double point.
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Answer. On that tangent, the segment from origin O to the intersection point T with
the guiding line AB has length p, too. Hence
cos τ =
d
|OA|
= = 2 cos α
|OT |
p
Question (c). Use the following program LINE on the TI84 to produce some examples
for the trisection. Put the calculator to mode: angle in degrees, graph polar coordinates.
Graph the function
\r1 = P (1+.5cos(A)/cos(θ))
and run program LINE for a few different angles A. Explain your observations.
PROGRAM: LINE
:PROMPT A :2 → P
:Polar
:-P→Xmin: 1.5P→Xmax
:Xmin→Ymin: Xmax→Ymax
:Zsquare
:Horizontal(.5Psin(A))
:Vertical(.5Pcos(A)) :(cos−1 (.5cos(A)))+180→T
:{ 0,A,A/3,A/3+60,A/3+120,180-A,T } → L1
:For(I,1,7)
Line(0,0,3cos(L1 (I)), 3sin(L1 (I)))
:End
15.11
Archimedes trisection yields more solutions
Problem 15.20. Follow the same procedure as given for Archimedes’ trisection, with
one small difference in item 1: Put the point marked C on the ruler anywhere on the
line OA.
Answer. If one follows this modified procedure of construction carefully, one finds six
solutions with Ci Di ∼
= OB for i = 1, . . . , 6. Three of these solution C1 , C2 , C3 are related
to trisection.
−→
For the first solution, C1 lies on the ray opposite to OA and outside the circle. This is
the solution, one usually considers first.
For the second solution C2 lies inside the circle, but in general not at the center O.
−→
For the third solution, C3 lies on the ray OA and outside the circle.
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The other three C4 , C5 , C6 are obvious ”fake” solutions. For the forth solution, the ruler
is a diameter of the circle, hence let C4 = O and BD4 be a diameter. Two further
solutions correspond to D5 = D6 = B, and the two points C5 and C6 on the OA with
distance OA from B.
Proposition 15.1 (A close look at Archimedes’ trisection). A careful analysis
shows that the procedure leads to six solutions.
Problem 15.21. Provide six drawings for this construction with different angles. Use
a different paper, and report your results below—ignore the fake solutions. In each
example, measure and report your chosen angle α, and the three angles γi = ∠OCi Di
for i = 1, 2, 3.
(a) Chose α < 45◦ .
(b) Exactly α = 45◦ .
(c) Chose 45◦ < α < 90◦ .
(d) Exactly α = 90◦ .
(e) Chose 90◦ < α < 135◦ .
(f ) Exactly α = 120◦ .
Answer. (a) Chose α < 45◦ .
Answer. For this example, I got α =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
(b) Exactly α = 45◦ .
Answer. For this example, I got α =
◦
◦
(c) Chose 45 < α < 90 .
Answer. For this example, I got α =
(d) Exactly α = 90◦ .
Answer. For this example, I got α =
(e) Chose 90◦ < α < 135◦ .
Answer. For this example, I got α =
(f) Exactly α = 120◦ .
Answer. For this example, I got α =
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Figure 15.19: Archimedes’ star of Mercedes
Proposition 15.2 (The star of Mercedes in Archimedes’ trisection). The three angles
γi = ∠OCi Di for i = 1, 2, 3 are related to the chosen angle α = ∠AOB via
α
α
α
, γ2 = 60◦ + , γ3 = 60◦ −
3
3
3
−−→ −−→ −−→
The triangle D1 D2 D3 is equilateral and the three rays OD1 , OD2 , OD3 form a star of
Mercedes. The three ruler lines form at vertex B two congruent adjacent angles
γ1 =
∠C1 BC2 ∼
= ∠C2 BC3 ∼
= 60◦
Proof. The claim
α
3
was shown earlier. Recall that the isosceles OC1 D1 has two congruent base angles γ1 ,
and the isosceles OD1 B has two congruent base angles 2γ1 . Finally α = γ1 + 2γ1 ,
because α is an exterior angle in triangle OC1 B, and γ1 , 2γ1 are the two nonadjacent
interior angles of that triangle. The claim
γ1 =
γ3 = 60◦ −
α
3
follows by the same reasoning applied to trisecting the supplementary angle ∠C1 OB ∼
=
180◦ − α. The claim
α
γ2 = 60◦ +
3
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needs a bid more care to be checked. The isosceles triangle OC2 D2 has two congruent
base angles γ2 . The isosceles triangle OD2 B has two congruent exterior base angles
2γ2 . Finally
(15.16)
2γ2 = α + (180◦ − γ2 )
because 2γ2 is an exterior angle in triangle OC2 B, and α, 180◦ − γ2 are the two
nonadjacent interior angles. Solving (15.16) for γ2 yields the claim.
The angles between the ruler lines at vertex B can now be calculated via the angle
sum in C1 BC2 and triangle C2 BC3 . One gets ∠C1 BC2 ∼
= ∠C2 BC3 ∼
= 60◦ .
−−→
−−→ −−→
To get the angles between the three rays OD1 , OD2 and OD3 , we can use Euclid
III.20 ”the central angle is twice the circum angle”. Hence ∠D1 OD2 = 2∠D1 BD2 =
2∠C1 BC2 = 2 · 60◦ = 120◦ and similarly, ∠D2 OD3 = 2∠D2 BD3 = 2 · 60◦ = 120◦ . As I
imagine, my dear friends, this is all the old guy Archimedes could have told his students,
would he still be alive.
15.12
Archimedes’ trisection and the conchoid
Archimedes construction can be interpreted as intersecting a conchoid and a circle. I
keep the drawing from above. The Cartesian coordinate system get the origin at point
B, the x-axis pointing to the right, and the y-axis upwards.
The marked segment has length p = 1, the guiding line in the horizontal line with
the Cartesian equation y = −d, and the distance from focus to guiding line is d = sin α.
In polar coordinates, this conchoid has the equation
(15.17)
r =p−
d
sin α
=1−
sin θ
sin θ
Problem 15.22. (a) Find the equation of the circle in polar coordinates (r, θ).
(b) Use your TI84 and graph the two curves for α = 40◦ .
(c) Point B is a double intersection point of the self crossing conchoid and the circle. It
occurs at r = 0 and is now discarted. Find the four angles for which the conchoid
and the circle intersect with r = 0.
Answer (a). Let R be a generic point on the circle, and M be the midpoint of segment
|
BR. From the right triangle BOM one gets cos(θ − α) = |BM
= 2r . The circle has
|BO|
the equation
(15.18)
r = −2 cos(θ − α)
Answer (b). To use the TI84 and graph the two curves for α = 40◦ . Choose a graphing
window with Zquare and Ymax = 1.666, Ymin = -1.666.
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Answer (c). To find the four angles θ1 , . . . , θ4 for which the conchoid and the circle
intersect. Putting equal the radial coordinate r from the equations of the conchoid
(15.17) to that of the circle (15.18), and multiplying by sin θ yields the equation
1−
sin α
+ 2 cos(θ − α) = 0
sin θ
for angle θ. One easy solution corresponds to putting the ruler through the midpoint of
the circle. Hence θ4 = α + 180◦ is a solution, as one can check. To cancel that zero, I
factor out cos θ−α
and get, after some calculations
2
(15.19)
sin θ − sin α + 2 cos(θ − α) sin θ = 2 cos
θ−α
3θ − α
sin
2
2
Question (d). Check this formula.
Answer. Put x = θ−α
and y =
2
side of (15.19) gets
3θ−α
.
2
Hence θ = y − x and α = y − 3x, and the left hand
sin θ − sin α + 2 cos(θ − α) sin θ = sin(y − x) − sin(y − 3x) + 2 cos 2x sin(y − x)
= sin(y − x) − sin(y − x) cos 2x + cos(y − x) sin 2x + 2 cos 2x sin(y − x)
= sin(y − x) + sin(y − x) cos 2x + cos(y − x) sin 2x
= sin(y − x) + sin(y − x) cos 2x + cos(y − x) sin 2x
= sin(y − x) + sin(y + x) = 2 sin y cos x
θ−α
3θ − α
= 2 cos
sin
2
2
Question. Find the four angles for which the conchoid and the circle intersect.
Answer. The four zeros are
θ1 =
α
α + 360◦
α − 360◦
, θ2 =
= θ1 + 120◦ , θ3 =
= θ1 − 120◦ , θ4 = 180◦ + α
3
3
3
Unfortunately, the circle equation (15.18) yields r < 0 for some angles. One can again
substitute (r, θ3 ) → (−r, θ + 180◦ ) to remedy that awkward situation. The circle is
passed twice, but the conchoid only once, if the angle takes a 360◦ turn. Hence one gets
four solutions, which are all different.
Question. How does one loose the two fake solutions D5 and D6 ?
Answer. One can loose the origin in polar equations, because it should be checked for
all angles θ— but we did not take this into account. For the conchoid, these two fake
solutions should appear for θ5 = α and θ6 = α − 180◦ . Indeed at these two angles one
get the self crossing loop for the conchoid at r = 0. But the equation (15.18) of the
circle has r = ±2 and yields both times the point D4 .
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Figure 15.20: Archimedes’ trisection and the conchoid
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Figure 15.21: Archimedes’ trisection and the conchoid
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