layout: true class: typo, typo-selection --- count: false class: nord-dark, middle, center # Euclidean geometry ⏳ @luk036 👨‍💻 · 2026 📅 --- ## Basic 🔰 - Line at infinity 🧵 $l_\infty = [0, 0, 1]$ - Two special points ⭐ $\mathrm{I}$ and $\mathrm{J}$ on $l_\infty$ play an important role in Euclidean Geometry: - $\mathrm{I} = [1, -i, 0]$, $\mathrm{J} = [1, i, 0]$ - $\mathbf{A} = l_\infty \cdot l_\infty^\mathsf{T}$ ➗ - $\mathbf{B} = \mathrm{I} \cdot \mathrm{J}^\mathsf{T} + \mathrm{J} \cdot \mathrm{I}^\mathsf{T}$ ➕ - If we choose another line 🆕 $l = M \cdot l_\infty$ as line of infinity --- ### Euclidean RT Concepts .mermaid[ <pre> graph LR Q[Quadrance] --> RT["Rational\n Trigonometry"] S[Spread] --> RT C[Cross] --> RT style Q fill:#4caf50 style S fill:#2196f3 style C fill:#ff9800 style RT fill:#9c27b0 </pre> ] --- class: nord-light, middle, center ## 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) 📜 $$\color{Green}{Q_1/s_1 = Q_2/s_2 = Q_3/s_3.}$$ - (Compare with the sine law in Euclidean Geometry) 📊: $$\color{red}{d_1/\sin \theta_1 = d_2/\sin \theta_2 = d_3/\sin \theta_3}.$$ --- ### Cross Law ✖️ - Theorem (Cross law) 📜 $$\color{Green}{(Q_1 + Q_2 - Q_3)^2 = 4 Q_1 Q_2 (1 - s_3)}.$$ - (Compare with the Cosine law) 📊: $$\color{red}{d_3^2 = d_1^2 + d_2^2 - 2 d_1 d_2 \cos \theta_3}.$$ --- ### Right triangles and Pythagoras 📐 - Suppose that $\{A_1 A_2 A_3\}$ is a right triangle with $s_3 = 1$. Then - Theorem (Thales) 📜 $$s_1 = Q_1 / Q_3 \; \text{and} \; s_2 = Q_2 / Q_3.$$ - Theorem (Pythagoras) 📜 $$Q_3 = Q_1 + Q_2.$$ --- ### Archimedes' function ⚙️ - Archimedes' function $A(Q_1, Q_2, Q_3)$ $$ Ar(Q_1, Q_2, Q_3) = (Q_1 + Q_2 + Q_3)^2 - 2(Q_1^2 + Q_2^2 + Q_3^2) $$ - Non-symmetric but more efficient version 🔄: $$ \color{Green}{Ar(Q_1, Q_2, Q_3) = 4 Q_1 Q_2 - (Q_1 + Q_2 - Q_3)^2} $$ --- ### Theorems 📚 - Theorem (Archimedes' formula) 📜: If $Q_1 = d_1^2$, $Q_2 = d_2^2$ and $Q_3 = d_3^2$, then $Ar(Q_1, Q_2, Q_3)$ = $$(d_1 + d_2 + d_3) (d_1 + d_2 - d_3) (d_2 + d_3 - d_1) (d_3 + d_1 - d_2)$$ --- ### Theorems (cont'd) 📚 - Theorem: As a quadratic equation in $Q_3$, the TQF $Ar(Q_1, Q_2, Q_3) = 0$ 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 $$\color{red}{\text{area} = \sqrt{s (s - a)(s - b)(s - c)} }$$ where $s = (a + b + c)/2$ is the semi-perimeter. {#fig:heron} --- ### Archimedes' theorem 📜 - The area of a planar triangle with quadrances $Q_1, Q_2, Q_3$ is given by $$\color{Green}{16(\text{area})^2 = Ar(Q_1, Q_2, Q_3)}$$ - 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 📜: $$\color{red}{\text{area} = \sqrt{(s - a) (s - b) (s - c) (s - d)} }$$ where $s = (a + b + c + d)/2$ - Preferred form: $16\text{area}^2$ = $$ (-a + b + c + d)(a - b + c + d) (a + b - c + d)(a + b + c - d) $$ --- ### Quadratic compatibility theorem 🔄 - Two quadratic equations $$(x - p_1)^2 = q_1, \qquad (x - p_2)^2 = q_2$$ are compatible iff $$[(p_1 + p_2)^2 - (q_1 + q_2)]^2 = 4 q_1 q_2$$ or $$Ar(q_1, q_2, (p_1 - p_2)^2) = 0$$ - 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 4️⃣ - Quadruple Quad Formula $Q(a,b,c,d)$ $$= [(a+b+c+d)^2 - 2(a^2 + b^2 + c^2 + d^2)]^2 - 64 a b c d$$ - Note that $(a+b+c+d)^2 - 2(a^2 + b^2 + c^2 + d^2)$ can be computed efficiently as 🔢: $$4(ab + cd) - (a + b - c - d)^2$$ --- ### 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 🔄 $$(\text{Area})^2 - 2 m (\text{Area}) + p = 0$$ where $$ m = (Q_{12} + Q_{23} + Q_{34} + Q_{14})^2 - 2(Q_{12}^2 + Q_{23}^2 + Q_{34}^2 + Q_{14}^2) \\ = 4(Q_{12}Q_{23} + Q_{34}Q_{14}) - (Q_{12} + Q_{23} - Q_{34} - Q_{14})^2 $$ $$ \begin{array}{lll} p = Q(Q_{12}, Q_{23}, Q_{34}, Q_{14}) \end{array} $$ --- ### Ptolemy's theorem & generalizations 🏛️ - **Claudius Ptolemy**: 90-168 A.D. (Alexandria) Astronomer & geographer & mathematician 🧙‍♂️ - **Ptolemy's theorem** If $\{ABCD\}$ 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 $$\color{Green}{Ar(Q_{12}Q_{34},Q_{23}Q_{14},Q_{13}Q_{24}) = 0}$$ - 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: $$uc = [\lambda^2 - \mu^2, 2 \lambda \mu, \lambda^2 + \mu^2],$$ where $\lambda$ and $\mu$ are not both zero. - Let $A_i = uc(\lambda_i, \mu_i), i = 1, 2, 3, 4$. - Express $Q_{ij} \equiv Q(A_i, A_j), i,j=1,2,3,4$ in terms of $\lambda$'s and $\mu$'s - Express $Ar(Q_{12}Q_{34},Q_{23}Q_{14},Q_{13}Q_{24})$ in terms of $\lambda$'s and $\mu$'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 💻 .font-sm.mb-xs[ ```python 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 💻 .font-sm.mb-xs[ ```python 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 💾 ```terminal 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 ``` --- count: false class: nord-dark, middle, center ## Q&A 🎤