layout: true class: typo, typo-selection --- count: false class: nord-dark, middle, center # Rational Trigonometry: A New Approach to Geometry 📐 @luk036 👨‍💻 · 2026 📅 --- ### Overview 🌍 In 2005, Norman Wildberger introduced a concept for geometry without transcendental functions in his book *"Divine Proportions"*. Inspired by ancient mathematics, he uses spreads and quadrances instead of angles and lengths to describe geometric relationships. This approach allows most tasks and proofs of Euclidean geometry to be carried out without sine and cosine functions, offering clear didactic advantages. --- ### Key Concepts of Rational Trigonometry 🔑 - **Quadrance** 📏 The squared area over a distance, replacing the concept of length. - **Spread** 📐 The squared ratio of the opposite to the hypotenuse in a right triangle, replacing angles. - **Rational Numbers** ➗ All calculations use only rational numbers, avoiding transcendental functions. --- ### Five Laws Overview .mermaid[ <pre> graph TD RT["Rational\n Trigonometry"] --> PT[Pythagoras' Theorem] RT --> TQF[Triple Quad Formula] RT --> CL[Cross Law] RT --> SL[Spread Law] RT --> TSF[Triple Spread Formula] style RT fill:#9c27b0 style PT fill:#4caf50 style TQF fill:#2196f3 style CL fill:#ff9800 style SL fill:#f44336 style TSF fill:#9c27b0 </pre> ] --- ### Advantages of Rational Trigonometry ✅ - **Simplicity** ✨ Easier to understand and apply, especially for students new to trigonometry. - **Accuracy** 🎯 As accurate as traditional engineering mathematics for practical applications. - **Logical Consistency** 🔗 Avoids problems associated with infinite objects and transcendental functions. --- ### Five Laws of Rational Trigonometry - **Pythagoras' Theorem (PT)** 🔺 Relates quadrances in a right-angled triangle. - **Triple Quad Formula (TQF)** 3️⃣📏 Describes relationships between quadrances in any triangle. - **Cross Law (CL)** ✖️ Replaces the cosine law in traditional trigonometry. - **Spread Law (SL)** ↔️ Analogous to the sine law in traditional trigonometry. - **Triple Spread Formula (TSF)** 3️⃣📐 Replaces the sum of angles rule for triangles. --- class: nord-light, middle, center # 🌐 Hyperbolic/Elliptic Geometry --- ### 🔑 Key Points In Hyperbolic Geometry, two parallel lines meet outside the null circle. ⭕ Given a line $l$, there are more than one parallel lines that pass through a point $p$. Notations: To distinguish with Euclidean geometry, lines are written as capital letters. --- ### 📏 Quadrance and Spread in Hyperbolic/Elliptic Geometry ✨ For efficiency, quadrance and spread can also be written as follows. 📏 The **quadrance** $q(a, b)$ between points $a$ and $b$ is: $$q(a, b) \equiv 1 - (a \cdot b^\perp)^2 / (a \cdot a^\perp)(b \cdot b^\perp)$$ 📐 The **spread** $S(L, M)$ between lines $L$ and $M$ is $$S(L, M ) \equiv 1 - (L \cdot M^\perp)^2 / (L \cdot L^\perp)(M \cdot M^\perp)$$ ⚠️ Note: In Hyperbolic Geometry, the quadrance of two points inside the null circle is negative. --- ### ↔️ Relation with Traditional Distance and Angle 📏 Hyperbolic Distance: $$q(a, b ) = \sinh^2(d(a, b ) )$$ 📐 Hyperbolic Angle: $$S(L, M ) = \sin^2(\theta(L, M))$$ 📏 Elliptic Distance: $$q(a, b ) = \sin^2(d(a, b ))$$ 📐 Elliptic Angle: $$S(L, M ) = \sin^2(\theta(L, M))$$ --- ### 📐 Spread Law ⚖️ Spread Law $$\color{Green}{q_1/S_1 = q_2/S_2 = q_3/S_3}.$$ 📐 (Compare with the sine law in traditional Hyperbolic Geometry): $$\color{red}{\sinh d_1/\sin \theta_1 = \sinh d_2/\sin \theta_2 = \sinh d_3/\sin \theta_3}.$$ 📐 (Compare with the sine law in traditional Elliptic Geometry): $$\color{red}{\sin d_1/\sin \theta_1 = \sin d_2/\sin \theta_2 = \sin d_3/\sin \theta_3}.$$ --- ### 🔢 Triple Formulae 🧮 Theorem (Triple quad formula): If $a_1$, $a_2$ and $a_3$ are collinear points then $$(q_1 + q_2 + q_3)^2 = 2(q_1^2 + q_2^2 + q_3^2) + 4 q_1 q_2 q_3$$ 🧮 Theorem (Triple spread formula): If $L_1$, $L_2$ and $L_3$ are concurrent lines then $$(S_1 + S_2 + S_3)^2 = 2(S_1^2 + S_2^2 + S_3^2) + 4 S_1 S_2 S_3.$$ --- ### ✖️ Cross Law ✖️ Theorem (Cross law) $$(S_1 S_2 q_3 - (S_1 + S_2 + S_3) + 2)^2 = 4(1 - S_1)(1 - S_2)(1 - S_3).$$ ✖️ Theorem (Cross dual law) $$(q_1 q_2 S_3 - (q_1 + q_2 + q_3) + 2)^2 = 4(1 - q_1)(1 - q_2)(1 - q_3).$$ 🔄 Note: Given three quadrances, three spreads can be uniquely determined. Same as Euclidean Geometry. ⚠️ Note: Given three spreads, three quadrances can be uniquely determined. Not true in Euclidean Geometry. --- ### 📐 Right Triangles and Pythagoras 📏 Theorem (Pythagoras): If $L_1$ and $L_2$ are perpendicular lines ($S(L_1, L_2) = 1$) then $$q_3 = q_1 + q_2 - q_1 q_2.$$ 📐 Theorem (Thales): Suppose that $\\{a_1 a_2 a_3\\}$ is a right triangle with $S_3 = 1$. Then $S_1 = q_1 / q_3$ and $S_2 = q_2 / q_3$. --- ### ➡️ Right Parallax 📐 Theorem (Right parallax): If a right triangle $\\{a_1 a_2 a_3\\}$ has spreads $S_1 = 0$, $S_2 = S$ and $S_3 = 1$, then it will have only one defined quadrance $q_1 = q$ given by $$q = (S - 1)/S.$$ 🔄 We may restate this result in the form: $$S = 1 / (1 - q).$$ --- ### 🔺 Triangle Proportions ⚖️ Theorem (Triangle proportions): Suppose that $d$ is a point lying on the line $a_1 a_2$. Define the quadrances $r_1 \equiv q(a_1, d)$ and $r_1 \equiv q(a_2, d)$, and the spreads $R_1 \equiv S(a_3 a_1, a_3 d)$ and $R_2 \equiv S(a_3 a_2, a_3 d)$. Then $$R_1/R_2 = (S_1/S_2)(r_1/r_2) = (q_1/q_2)(r_1/r_2).$$ --- class: nord-light, middle, center # Euclidean geometry 📏 --- ### 🔑 Key Points - 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 - 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 ``` ] --- ### Practical Application: Measuring a Lighthouse 1. **Measure** 📏 Take spreads *s₁, s₂* and quadrance *Q₃* from two standpoints. 2. **Calculate** 🧮 Use Triple Spread Formula and Spread Law to find quadrance *H*. 3. **Result** Determine height *h* by taking the square root of *H*. --- ### Comparison: Rational vs Traditional ⚖️ - **Rational Method** ➗ Uses spreads and quadrances directly in calculations. Avoids trigonometric functions. Results in quadrance *H = 201.4906705* and height *h ≈ 14.22777860*. - **Traditional Method** 📐 Converts spreads to angles and quadrances to distances. Uses sine law and trigonometric functions. Results in a similar height value with slight differences due to decimal cancellation. --- ### Educational Implications 🎓 1. **Foundation** 🏗️ Rational trigonometry as a base for geometric understanding. 2. **Intermediate** ↔️ Application in problem-solving without transcendental functions. 3. **Advanced** 🚀 Bridge to more complex mathematical concepts. This approach could be useful as a didactic and methodological concept in schools and universities, providing a solid foundation before introducing differential calculus. --- ### Extensions and Applications 🌐 - **3D Geometry** 🧊 Rational trigonometry concepts can be easily extended to three-dimensional space. - **Hyperbolic Geometry** 🌌 The approach can be applied to non-Euclidean geometries, including hyperbolic spaces. - **Special Relativity** ⚡ Potential applications in physics, particularly in the field of special relativity. --- ### Historical Context ⏳ - **Ancient Babylonian Mathematics** 🏺 Used tables of squares and square roots for calculations. - **Ancient Greek Geometry** 🏛️ Focused on ratios and proportions rather than trigonometric functions. - **Modern Rational Trigonometry** 🔄 Wildberger’s approach combines ancient insights with modern algebraic techniques. --- ### Conclusion and Future Directions 🔮 - **Consistent Approach** 🔄 Rational trigonometry offers a systematic method using only rational numbers. - **Practical Applications** 🛠️ Useful in surveying and engineering, avoiding complex trigonometric computations. - **Educational Potential** 🎓 Could revolutionize how geometry and trigonometry are taught in schools and universities. --- ### 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 🎤