Euclidean geometry

Basic

Line at infinity $l_{\infty} = [0, 0, 1]$
Two special points $I$ and $J$ on $l_{\infty}$ play an important role in Euclidean Geometry:
- $I = [1, - i, 0]$ , $J = [1, i, 0]$
$A = l_{\infty} \cdot l_{\infty}^{T}$
$B = I \cdot J^{T} + J \cdot I^{T}$
If we choose another line $l = M \cdot l_{\infty}$ as line of infinity

Rational Trigonometry in Euclidean geometry

Notations

To distinguish with Euclidean geometry, points are written in capital letters.

Quadrance and Spread in Euclidean geometry

The quadrance $Q$ between points $A_{1}$ and $A_{2}$ is: $Q = (x_{1} / z_{1} - x_{2} / z_{2})^{2} + (y_{1} / z_{1} - y_{2} / z_{2})^{2}$
The spread $s$ between lines $l_{1}$ and $l_{2}$ is: $s = (a_{1} b_{2} - a_{2} b_{1})^{2} / (a_{1}^{2} + b_{1}^{2}) (a_{1}^{2} + b_{1}^{2})$
The cross $c$ between lines $l_{1}$ and $l_{2}$ is: $c = 1 - s = (a_{1} a_{2} + b_{1} b_{2})^{2} / (a_{1}^{2} + b_{1}^{2}) (a_{1}^{2} + b_{1}^{2})$

Triple formulate

Let $A_{1}$ , $A_{2}$ and $A_{3}$ are points with $Q_{1} \equiv Q (A_{2}, A_{3})$ , $Q_{2} \equiv Q (A_{1}, A_{3})$ and $Q_{3} \equiv Q (A_{1}, A_{2})$ . Let $l_{1}$ , $l_{2}$ and $l_{3}$ are lines with $s_{1} \equiv s (l_{2}, l_{3})$ , $s_{2} \equiv s (l_{1}, l_{3})$ and $s_{3} \equiv s (l_{1}, l_{2})$ .
Theorem (Triple quad formula): If $A_{1}$ , $A_{2}$ and $A_{3}$ are collinear points then $(Q_{1} + Q_{2} + Q_{3})^{2} = 2 (Q_{1}^{2} + Q_{2}^{2} + Q_{3}^{2})$
Theorem (Triple spread formula): If $l_{1}$ , $l_{2}$ and $l_{3}$ are concurrent lines then $(s_{1} + s_{2} + s_{3})^{2} = 2 (s_{1}^{2} + s_{2}^{2} + s_{3}^{2}) + 4 s_{1} s_{2} s_{3} .$

Spread Law

Suppose that triangle ${A_{1} A_{2} A_{3}}$ form quadrances $Q_{1} \equiv Q (A_{2}, A_{3})$ , $Q_{2} \equiv Q (A_{1}, A_{3})$ and $Q_{3} \equiv Q (A_{1}, A_{2})$ , and it dual trilateral ${l_{1} l_{2} l_{3}}$ form spreads $s_{1} \equiv s (l_{2}, l_{3})$ , $s_{2} \equiv s (l_{1}, l_{3})$ and $s_{3} \equiv s (l_{1}, l_{2})$ . Then:
Theorem (Spread Law) $Q_{1} / s_{1} = Q_{2} / s_{2} = Q_{3} / s_{3} .$
(Compare with the sine law in Euclidean Geometry): $d_{1} / s i n θ_{1} = d_{2} / s i n θ_{2} = d_{3} / s i n θ_{3} .$

Cross Law

Theorem (Cross law) $(Q_{1} + Q_{2} - Q_{3})^{2} = 4 Q_{1} Q_{2} (1 - s_{3}) .$
(Compare with the Cosine law) $d_{3}^{2} = d_{1}^{2} + d_{2}^{2} - 2 d_{1} d_{2} c o s θ_{3} .$

Right triangles and Pythagoras

Suppose that ${A_{1} A_{2} A_{3}}$ is a right triangle with $s_{3} = 1$ . Then
Theorem (Thales) $s_{1} = Q_{1} / Q_{3} and s_{2} = Q_{2} / Q_{3} .$
Theorem (Pythagoras) $Q_{3} = Q_{1} + Q_{2} .$

Archimedes' function

Archimedes' function $A (Q_{1}, Q_{2}, Q_{3})$

$A r (Q_{1}, Q_{2}, Q_{3}) = (Q_{1} + Q_{2} + Q_{3})^{2} - 2 (Q_{1}^{2} + Q_{2}^{2} + Q_{3}^{2})$
Non-symmetric but more efficient version: $A r (Q_{1}, Q_{2}, Q_{3}) = 4 Q_{1} Q_{2} - (Q_{1} + Q_{2} - Q_{3})^{2}$

Theorems

Theorem (Archimedes' formula): If $Q_{1} = d_{1}^{2}$ , $Q_{2} = d_{2}^{2}$ and $Q_{3} = d_{3}^{2}$ , then $A r (Q_{1}, Q_{2}, Q_{3})$ = $(d_{1} + d_{2} + d_{3}) (d_{1} + d_{2} - d_{3}) (d_{2} + d_{3} - d_{1}) (d_{3} + d_{1} - d_{2})$

Theorems (cont'd)

Theorem: As a quadratic equation in $Q_{3}$ , the TQF $A r (Q_{1}, Q_{2}, Q_{3}) = 0$ can be rewritten as: $Q_{3}^{2} - 2 (Q_{1} + Q_{2}) Q_{3} + (Q_{1} - Q_{2})^{2} = 0$
Theorem: The quadratic equations $(x - p_{1})^{2} = q_{1}$ and $(x - p_{2})^{2} = q_{2}$ has a common solutions iff $A (q_{1}, q_{2}, (p_{1} - p_{2})^{2}) = 0$

Heron's formula (Hero of Alexandria 60BC)

The area of a triangle with side lengths $a, b, c$ is $area = s (s - a) (s - b) (s - c)$ where $s = (a + b + c) /2$ is the semi-perimeter.

Heron's formula {#fig:heron}

Archimedes' theorem

The area of a planar triangle with quadrances $Q_{1}, Q_{2}, Q_{3}$ is given by $16 (area)^{2} = A r (Q_{1}, Q_{2}, Q_{3})$
Note: Given $Q_{1}, Q_{2}$ . The area is maximum precisely when $Q_{3} = Q_{1} + Q_{2}$ .

Brahmagupta's formula (convex)

Brahmagupta's theorem: $area = (s - a) (s - b) (s - c) (s - d)$ where $s = (a + b + c + d) /2$
Preferred form: $16 area^{2}$ = $(- a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d)$

Quadratic compatibility theorem

Two quadratic equations $(x - p_{1})^{2} = q_{1}, (x - p_{2})^{2} = q_{2}$ are compatible iff $[(p_{1} + p_{2})^{2} - (q_{1} + q_{2})]^{2} = 4 q_{1} q_{2}$ or $A r (q_{1}, q_{2}, (p_{1} - p_{2})^{2}) = 0$
In this case, if $p_{1} \neq = p_{2}$ then there is a unique sol'n: $2 x = (p_{1} + p_{2}) - (q_{1} - q_{2}) / (p_{1} - p_{2})$

Quadruple Quad Formula

Quadruple Quad Formula $Q (a, b, c, d)$ $= [(a + b + c + d)^{2} - 2 (a^{2} + b^{2} + c^{2} + d^{2})]^{2} - 64 ab c d$
Note that $(a + b + c + d)^{2} - 2 (a^{2} + b^{2} + c^{2} + d^{2})$

can be computed efficiently as: $4 (ab + c d) - (a + b - c - d)^{2}$

Brahmagupta's formula

Brahmagupta's formula (convex): $B (a, b, c, d)$ = $(b + c + d - a) (a + c + d - b) (a + b + d - c) (a + b + c - d)$
Robbin's formula (non-convex): $R (a, b, c, d)$ =
$(a + b + c + d) (a + b - c - d) (a - b + c - d) (b + c - a - d)$
Brahmagupta's identity $Q (a^{2}, b^{2}, c^{2}, d^{2}) = B (a, b, c, d) \cdot R (a, b, c, d)$

Cyclic quadrilateral quadrea theorem

$(Area)^{2} - 2 m (Area) + p = 0$

where

$m p = = = = (Q_{12} + Q_{23} + Q_{34} + Q_{14})^{2} - 2 (Q_{12}^{2} + Q_{23}^{2} + Q_{34}^{2} + Q_{14}^{2}) 4 (ab + c d) - (a + b - c - d)^{2} 4 (Q_{12} Q_{23} + Q_{34} Q_{14}) - (Q_{12} + Q_{23} - Q_{34} - Q_{14})^{2}) Q (Q_{12}, Q_{23}, Q_{34}, Q_{14})$

Ptolemy's theorem & generalizations

Claudius Ptolemy: 90-168 A.D. (Alexandria) Astronomer & geographer & mathematician
Ptolemy's theorem If ${A BC D}$ is a cyclic quadrilateral with the lengths $a, b, c, d$ and diagonal lengths $p, q$ , then $a c + b d = p q$ [Actually needs convexity!]

Ptolemy's theorem

{#fig:Ptolemy}

Exercise

Ex. $A_{1} = (1, 0)$ , $A_{2} = (3/5, 4/5)$ , $A_{3} = (- 12/13, 5/13)$ , $A_{4} = (15/17, - 8/17)$
Then the quadrances are: $Q_{12} = 4/5, Q_{23} = 162/65, Q_{34} = 882/221, Q_{14} = 4/17$
The diagonal quadrances are: $Q_{24} = 144/85, Q_{13} = 50/13$

Ptolemy's theorem (rational version)

Ptolemy's theorem (rational version): If $A_{1} A_{2} A_{3} A_{4}$ _ is a cyclic quadrilateral with quadrances $Q_{ij} \equiv Q (A_{i}, A_{j}), i, j = 1, 2, 3, 4$ then

$A r (Q_{12} Q_{34}, Q_{23} Q_{14}, Q_{13} Q_{24}) = 0$
Ex. For $A_{1} = (1, 0)$ , $A_{2} = (3/5, 4/5)$ , $A_{3} = (- 12/13, 5/13)$ , $A_{4} = (15/17, - 8/17)$ with

$Q_{12} = 4/5, Q_{23} = 162/65$

$Q_{34} = 882/221, Q_{14} = 4/17$

$Q_{24} = 144/85, Q_{13} = 50/13$
we can verify directly that $A (Q_{12} Q_{34}, Q_{23} Q_{14}, Q_{13} Q_{24}) = 0$ .
Note that with the rational form of Ptolemy's theorem, the three quantities appear symmetrically: so convexity of the cyclic quadrilateral is no longer required!

Proof of Ptolemy's theorem

Sketch of the proof:

Without loss of generality, we can assume that the circle is a unit circle.
Recall that a point on a unit circle can be parameterized as: $u c = [λ^{2} - μ^{2}, 2 λ μ, λ^{2} + μ^{2}],$ where $λ$ and $μ$ are not both zero.
Let $A_{i} = u c (λ_{i}, μ_{i}), i = 1, 2, 3, 4$ .
Express $Q_{ij} \equiv Q (A_{i}, A_{j}), i, j = 1, 2, 3, 4$ in terms of $λ$ 's and $μ$ 's
Express $A r (Q_{12} Q_{34}, Q_{23} Q_{14}, Q_{13} Q_{24})$ in terms of $λ$ 's and $μ$ 's.
Simplify the expression and derive that it is equal to 0. (we may use the Python's symbolic package for the calculation. It took about 8 minutes on my computer :-)

🐍 Python Code

import numpy as np
from fractions import *
from proj_geom import *

def quad1(x1, z1, x2, z2):
    if isinstance(x1, int):
        return (Fraction(x1,z1) - Fraction(x2,z2))**2
    else:
        return (x1/z1 - x2/z2)**2

def quadrance(a1, a2):
    return quad1(a1[0], a1[2], a2[0], a2[2]) + \
            quad1(a1[1], a1[2], a2[1], a2[2])

def uc_point(lambda1, mu1):
    return pg_point([lambda1**2 - mu1**2,
                2*lambda1*mu1, lambda1**2 + mu1**2])

def Ar(a, b, c):
    ''' Archimedes's function '''
    return (4*a*b) - (a + b - c)**2

🐍 Python Code

if __name__ == "__main__":
    import sympy
    sympy.init_printing()

    lambda1, mu1 = sympy.symbols("lambda1 mu1", integer=True)
    lambda2, mu2 = sympy.symbols("lambda2 mu2", integer=True)
    lambda3, mu3 = sympy.symbols("lambda3 mu3", integer=True)
    lambda4, mu4 = sympy.symbols("lambda4 mu4", integer=True)
    a1 = uc_point(lambda1, mu1)
    a2 = uc_point(lambda2, mu2)
    a3 = uc_point(lambda3, mu3)
    a4 = uc_point(lambda4, mu4)
    q12 = quadrance(a1, a2)
    q23 = quadrance(a2, a3)
    q34 = quadrance(a3, a4)
    q14 = quadrance(a1, a4)
    q24 = quadrance(a2, a4)
    q13 = quadrance(a1, a3)
    t = Ar(q12*q34, q23*q14, q13*q24)
    t = sympy.simplify(t)
    print(t) # get 0

Backup

>  pandoc -t latex -F pandoc-crossref -o temp2.svg .\01proj_geom.md .\02ck_geom.md .\03RT.md .\04RT_2.md latex.yaml .\crossref.yaml
>  pandoc -t beamer -F pandoc-crossref -o temp2.svg .\01proj_geom.md .\02ck_geom.md .\03RT.md .\04RT_2.md beamer.yaml .\crossref_2.yaml

Projective Geometry