Chapter 2: The Baudhāyana-Pythagoras Theorem
Geometric Dissections, the Nature of √2, Infinite Triples, Līlāvatī Applications & Fermat's Theorem
📐 1. Doubling & Halving Squares
Śulba-Sūtra (c. 800 BCE)Doubling the sides quadruples the area (2 × 2 = 4). Baudhāyana solved how to produce a square of exactly double the area:
Verse 1.9: "The diagonal of a square produces a square of double the area of the original square."
• Halving: Connect side midpoints inward (PQRS has 1/2 area).
• Doubling: Build a square directly on the diagonal.
🌱 2. The Nature of √2
Irrational ContinuumIn a unit square of side 1, diagonal c satisfies c2 = 12 + 12 = 2 → c = √2.
- Tight Decimal Bounds: 1.414 < √2 < 1.415.
- Non-Terminating: Any finite decimal squared ends in a non-zero digit, never exact 2.000...
- Non-Fractional: Cannot be written as m/n (Euclid: 2n2 = m2 violates prime parity count).
- Value: √2 ≈ 1.41421356...
📜 3. The General Theorem
Baudhāyana Verse 1.12Baudhāyana's Geometric Theorem:
Known also as the Baudhāyana-Pythagoras Theorem (studied by Pythagoras ~500 BCE, ~300 years after Baudhāyana).
🔢 4. Baudhāyana Triples
Integer SolutionsInteger sets (a, b, c) where a2 + b2 = c2:
- Primitive Triples (GCD = 1): (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (12, 35, 37), (20, 21, 29).
- Scaled Families: (ka, kb, kc) is always a triple → Infinitely many triples exist!
- Odd-Square Generator: Since 1 + 3 + ... + (2n − 1) = n2, setting the nth odd number (2n − 1) as a square k2 gives (n − 1)2 + k2 = n2!
🏛️ 5. Fermat's Last Theorem
300-Year MilestonePierre de Fermat (17th Century) observed that while sums of squares are infinite, higher powers have no integer solutions:
After 350+ years of attempts, British mathematician Andrew Wiles finally proved it in 1994!
🪷 6. The Lotus Problem
Līlāvatī (1150 CE)Classic Depth Problem: A lotus tip stands 1 unit above water. Swayed by breeze, it submerges 3 units away.
Let depth = x → Stem length = x + 1.
32 + x2 = (x + 1)2
9 + x2 = x2 + 2x + 1
9 = 2x + 1 → x = 4 units depth!
🔍 Master Geometric Applications & Calculation Matrix
Key geometric figures solved directly using the Baudhāyana-Pythagoras relationship:
| Geometric Context | Formula / Relationship | Step-by-Step Working | Calculated Result |
|---|---|---|---|
| Square Diagonal | d2 = s2 + s2 = 2s2 → d = s√2 | For side s = 5 cm → d = 5√2 | ≈ 7.07 cm |
| Rhombus Side | Diagonals bisect at 90°: side2 = (d1/2)2 + (d2/2)2 | Diagonals 24 & 70 → half-lengths 12 & 35 side2 = 122 + 352 = 144 + 1225 = 1369 |
side = 37 units |
| Equilateral Triangle Altitude | Altitude bisects base: h2 + (s/2)2 = s2 | Side s = 6 → h2 + 32 = 62 → h2 = 36 − 9 = 27 h = √27 = 3√3; Area = (1/2) × 6 × 3√3 |
Area = 9√3 ≈ 15.59 sq. units |
| Area Difference Square | Construct right triangle with hypotenuse 7, base 5 | New Side s2 = 72 − 52 = 49 − 25 = 24 | Area = 24 sq. units |
🎯 7. Constructing Grid Squares
Any square drawn on a dot grid with vertices on grid points has sidelength as the hypotenuse of a right triangle with integer steps (dx, dy):
- Area 1: 12 + 02 = 1 sq. unit
- Area 2: 12 + 12 = 2 sq. units (Tilted 1×1 diagonal)
- Area 4: 22 + 02 = 4 sq. units
- Area 5: 22 + 12 = 5 sq. units (Tilted 2×1 diagonal)
- Area 3: IMPOSSIBLE! 3 cannot be written as a sum of two integer squares (a2 + b2 ≠ 3).
🎁 8. Puzzle Time: Find the Colours!
The Challenge: 3 closed boxes contain RED, BLUE, and GREEN balls. ALL three boxes are completely mislabeled. You may draw just 1 ball from 1 box. How do you correctly identify all three?
Deduction Step: Open the box labeled with any chosen color, say RED.
• Since the label is false, if you draw a BLUE ball, this box is definitely BLUE.
• The box labeled GREEN cannot be Green (mislabeled) and cannot be Blue (identified), so it MUST be RED!
• The remaining box labeled BLUE must be GREEN!
⚠️ Common Pitfalls & Conceptual Traps
Hypotenuse Identification
The theorem a2 + b2 = c2 applies only if c is the hypotenuse (side opposite the 90° angle). If a side is unknown, check whether it is a leg or the hypotenuse before adding or subtracting squares!
Linear Area Doubling Fallacy
Doubling side length quadruples the area (2s)2 = 4s2. To double the area, the side must increase by a factor of √2 ≈ 1.414, exactly the diagonal length!
Primitive vs Scaled Triples
(6, 8, 10) is a valid Baudhāyana triple, but it is not primitive because GCD(6, 8, 10) = 2. A primitive triple must have no common divisor other than 1.