Reading Math & Its History (02)

Greek Geometry

9 min readApr 10, 2020

Reading Math & Its History (01) The Theorem of Pythagoras
Reading Math & Its History (02) Greek Geometry
Reading Math & Its History (03) Greek Number Theory
Reading Math & Its History (04) Infinity in Greek Mathematics

2.1 The Deductive Method

Exercise 2.1.1

Common Notion 1:
Things which are equal to the same thing are also equal to one another.

a ≈ b and b ≈ c => a ≈ c

Common Notion 4:
Things which coincide with one another are equal to one another.

a ≈ a

Exercise 2.1.2

Euclid uses “one another” which means a ≈ b and b ≈ a.

2.2 The Regular Polyhedra

Exercise 2.2.1

The cube is in blue:

Let half of cube side OR = 1, then
RP = √2
OP = √(OR^2 + RP^2) = √3
OP/OR = √3/1 = √3

The octahedron is in blue:

Let OQ = 1, then
the shortest distance to any side OP = 1/√2PQ = √(OP^2 + OQ^2) = √((1/√2)^2 + 1^2) = √3/√2 
OR * PQ = OP * OQ
OR = OP * OQ / PQ = 1/√2 * 1 / (√3/√2) = 1/√3
OQ / OR = √3

Exercise 2.2.2

Icosahedron:

Pacioli’s construction with sides 1 and (1 + √5)/2:

τ = (1 + √5)/2
ττ = (6 + 2√5)/4 = (3 + √5)/2 = τ + 1
BC = 1
AD = τ/2
ED = τ/2 - 1/2
AE^2 = ED^2 + AD^2 = (τ/2 - 1/2)^2 + (τ/2)^2
AC^2 
= AE^2 + EC^2 
= (τ/2 - 1/2)^2 + (τ/2)^2 + (1/2)^2
= (ττ - 2τ + 1)/4 + ττ/4 + 1/4
= ττ/2 - τ/2 + 1/2
= (τ + 1)/2 - τ/2 + 1/2
= 1
AC = 1
AB = AC = BC = 1

Exercise 2.2.3

Let AD = τ, then
A = (-1, 0, τ)
B = (-τ, 1, 0)
C = (-τ, -1, 0)

Each golden rectangle has 3 symmetric points.
(±1, 0, ±τ)
(±τ, ±1, 0)
(0, ±τ, ±1)

Exercise 2.2.4

circumradius = OA = √(ττ + 1) = √(τ + 1)

Exercise 2.2.5

F is the centre of ∆ABC.

OE = AD = τ = (1 + √5)/2
OD = 1
ED = τ - 1
AE 
= √(AD^2 + ED^2) 
= √(2ττ - 2τ + 1)
= √(2(τ + 1) - 2τ + 1)
= √3
AF = (2/3)AE = 2/√3
OA^2 = OD^2 + AD^2 = ττ + 1 = τ + 2
OF = √(OA^2 - AF^2) = √(τ + 2 - 4/3) = √(τ + 2/3) 
OA/OF 
= √(τ + 2)/√(τ + 2/3) 
= 3√(τ + 2)/√(9τ + 6)

Exercise 2.2.6

(3√(τ + 2)/√(9τ + 6))^2
= 9(1/2 + √5/2 + 2)/(9/2 + 9√5/2 + 6)
= 9(5 + √5)/(21 + 9√5)
= 9(5 + √5)(21 - 9√5)/(441 - 405)
= 9(60 - 24√5)/36
= 15 - 6√5
= 3(5 - 2√5)(√(3(7 - 4τ)))^2
= 3(7 - 2 - 2√5)
= 3(5 - 2√5)(√(15/(4τ + 3)))^2
= 15/(2 + 2√5 + 3)
= 15/(5 + 2√5)
= 15(5 - 2√5)/(25 - 20)
= 3(5 - 2√5)

Exercise 2.2.7

Dodecahedron:

Dual dodecahedron:

circumradius of dual dodecahedron
= inradius of icosahedron
= OF 
= √(τ + 2/3)
= (1/3)√(9τ + 6)

A = (-1, 0, τ) 
B = (-τ, 1, 0)
C = (-τ, -1, 0)
D = (0, -τ, 1)
E = (1, 0, τ)
F = (0, τ, 1)

The centre of ∆ABC = (A + B + C)/3
The centre of ∆ACD = (A + C + D)/3
The centre of ∆ADE = (A + D + E)/3
The centre of ∆AEF = (A + E + F)/3
The centre of ∆AFB = (A + F + B)/3

The centre of the pentagon of the above 5 centres:

(1/5)((A + B + C)/3 + (A + C + D)/3 + (A + D + E)/3 + (A + E + F)/3 + (A + F + B)/3)
= (1/15)(5A + 2B + 2C + 2D + 2E + 2F)
= (1/15)(-5 - 4τ + 2, 0, 5τ + 2τ + 4)
= (1/15)(-4τ - 3, 0, 7τ + 4)
= (4τ + 3)/15(-1, 0, τ)

inradius of the dual dodecahedron:

((4τ + 3)/15)√(ττ + 1) = ((4τ + 3)/15)√(τ + 2)

Exercise 2.2.8

circumradius/inradius
= (1/3)√(9τ + 6) / (((4τ + 3)/15)√(τ + 2))
= 5√(9(1 + √5)/2 + 6) / ((4(1 + √5)/2 + 3)(√((1 + √5)/2 + 2)))
= 5√(9 + 9√5 + 12) / ((2 + 2√5 + 3)√(5 + √5))
= 5√(21 + 9√5) / ((5 + 2√5)√(5 + √5))
= 5√(3(7 + 3√5)) / ((5 + 2√5)√(5 + √5))
= 5√(3(7 + 3√5))√(5 - √5) / ((5 + 2√5)√(5 + √5)√(5 - √5))
= 5√(3(7 + 3√5)(5 - √5)) / ((5 + 2√5)√20)
= 5√(3(7 + 3√5)(5 - √5)) / (2(5√5 + 10))
= √(3(7 + 3√5)(5 - √5)) / (2(√5 + 2))
= √(3(7 + 3√5)(5 - √5))(√5 - 2) / (2(√5 + 2)(√5 - 2))
= √(3(7 + 3√5)(5 - √5)(9 - 4√5)) / 2
= √(3(20 + 8√5)(9 - 4√5)) / 2
= √(3(5 + 2√5)(9 - 4√5))
= √(3(5 - 2√5))
which is same as the result of Exercise 2.2.6√(15/(4τ + 3))
= √(15/(2 + 2√5 + 3))
= √(15/(5 + 2√5))
= √(15(5 – 2√5)/(25 – 20))
= √(3(5 - 2√5))

Exercise 2.2.9

Icosahedron I

S = 20 ∆ABC
V = 20 (1/3) ∆ABC * OF
S/V = (1/3)/OF = (1/3)/inradius

Dodecahedron D

Each face is same as pentagon BCDEF.

S = 12 BCDEF
V = 12 BCDEF * (1/3)inradius
S/V = (1/3)/inradius

In Exercise 2.2.8 we already show that circumradius/inradius is same for dodecahedron and icosahedron. So with same circumradius, SofI/VofI = SofD/VofD, and we get

suface area D / surface area I = volume D / volume I

2.3 Ruler and Compass Constructions

Exercise 2.3.1

L / L1 = L2 / 1
L = L1 * L2

L1 / L2 = L / 1
L = L1 / L2

Exercise 2.3.2

∆BAD is similar to ∆ACD
BD / AD = AD / DC
DC = 1
BD = AD * AD
BD = l
AD = √l

Exercise 2.3.3

∆ABP ≈ ∆DCP
AB / DC = BP / DP
AB = x
DC = 1
BP = 1
DP = DB - BP = x - 1
x / 1 = 1 / (x - 1)

Exercise 2.3.4

x(x - 1) = 1
xx - x - 1 = 0
x = (1 ± √(1 + 4))/2
x = (1 + √5)/2

2.4 Conic Sections

Exercise 2.4.1

Exercise 2.4.2

Let OF1 = OF2 = c. Assume PF1 + PF2 = constant. When point P(x, y) moves to X axis, it’s easy to see that that constant = 2a. Similarly when P moves to Y axis, it’s easy to see that each of red lines has length a. So we get a² = b² + c².

Now we just need to check whether x²/a² + y²/b² = 1 and a² = b² + c² are equivalent.

PF1 + PF2 = 2a
√((x + c)^2 + yy) + √((x - c)^2 + yy) ?= 2a
√((x + c)^2 + yy) ?= 2a - √((x - c)^2 + yy)
(x + c)^2 + yy ?= 4aa + (x - c)^2 + yy - 4a√((x - c)^2 + yy)
2cx ?= 4aa - 2cx - 4a√((x - c)^2 + yy)
4a√((x - c)^2 + yy) ?= 4aa - 4cx
a√((x - c)^2 + yy) ?= aa - cx
aa((x - c)^2 + yy) ?= (aa - cx)^2
aa(xx - 2cx + cc + yy) ?= a^4 - 2aacx + ccxx
aaxx + aacc + aayy ?= a^4 + ccxx

Cool, all terms with c are gone, leaving c² terms. Let’s continue.

cc = aa - bb
aaxx + aa(aa - bb) + aayy ?= a^4 + (aa - bb)xx
aaxx + a^4 - aabb + aayy ?= a^4 + aaxx - bbxx
-aabb + aayy ?= -bbxx
bbxx + aayy ?= aabb
xx/aa + yy/bb = 1

Exercise 2.4.3

The sum of 2 sides of a triangle is greater than the 3rd side. So FPF2' < F1P’F2'.

Exercise 2.4.4

F1PF2 = F1PF2' < F1P’F2. Angles 1, 2 and 3 are same.

2.5 Higher-Degree Curves

Exercise 2.5.1

Cissoid

OM / QN = OR / QR
OR = 1
XX + YY = 1
QN = √(1 - xx)
OM = QN * OR / QR = √(1 - xx) / (1 + x)
Y = (OM/OR)X - OM
Y = (OM)X - OM = OM(X - 1)
Y = (√(1 - xx) / (1 + x))(X - 1)

Exercise 2.5.2

P is the intersection of line NR and line X = x. So y is Y when X = x.

y = (√(1 - xx) / (1 + x))(x - 1)
yy = ((1 - xx)/(1 + x)^2)(x - 1)^2
yy = ((1 - x)/(1 + x))(x - 1)^2
yy(1 + x) = (1 - x)^3
y^2(1 + x) = (1 - x)^3

Exercise 2.5.3

cardioid

Exercise 2.5.4

Review some trigonometrical identities first using a unit circle.

sin2θ = CD = ACsinθ = (ABcosθ)sinθ = 2cosθsinθ
cos2θ = OD = AD - AO = ACcosθ - 1 = (ABcosθ)cosθ - 1 = 2(cosθ)^2 - 1

Q is a point on cardioid. a = 1, φ = θ.

x = OR + RQcosθ
y = RQsinθ
RQ = 2RN = 2SM = 2(OM - OS) = 2(1 - ORcosθ) = 2(1 - cosθ)
x = 1 + 2(1 - cosθ)cosθ = 1 + 2cosθ - 2(cosθ)^2 = 2cosθ - cos2θ
y = 2(1 - cosθ)sinθ = 2sinθ - 2cosθsinθ = 2sinθ - sin2θ

Exercise 2.5.5

x = 1 + 2(1 - cosθ)cosθ = 1 + 2cosθ - 2(cosθ)^2
y = 2(1 - cosθ)sinθ = 2sinθ - 2cosθsinθ = 2sinθ(1 - cosθ)(xx + yy - 1)^2
= ((x + 1)(x - 1) + yy)^2
= ((2 + 2cosθ - 2(cosθ)^2)(2cosθ - 2(cosθ)^2) + yy)^2
= (4(cosθ + (sinθ)^2)(cosθ - (cosθ)^2) + yy)^2
= (4(cosθ + (sinθ)^2)cosθ(1 - cosθ) + yy)^2
= (4(cosθ + (sinθ)^2)cosθ(1 - cosθ) + 4(sinθ)^2(1 - cosθ)^2)^2
= 16(((cosθ)^2 + cosθ(sinθ)^2)(1 - cosθ) + (sinθ)^2(1 - cosθ)^2)^2
= 16((1 - cosθ)((cosθ)^2 + cosθ(sinθ)^2 + (sinθ)^2 -cosθ(sinθ)^2))^2
= 16((1 - cosθ)((cosθ)^2 + (sinθ)^2))^2
= 16(1 - cosθ)^24((x - 1)^2 + yy)
= 4((2cosθ - 2(cosθ)^2)^2 + 4(sinθ)^2(1 - cosθ)^2)
= 4(4(cosθ)^2(1 - cosθ)^2 + 4(sinθ)^2(1 - cosθ)^2)
= 16((cosθ)^2(1 - cosθ)^2 + (sinθ)^2(1 - cosθ)^2)
= 16(1 - cosθ)^2(xx + yy - 1)^2 = 4((x - 1)^2 + yy)