From the staggers of an athletics track to the infinite series of Mādhava. Discover the laws governing the boundaries and contents of shapes.
In a 4x100m relay, runners in outer lanes start ahead of those in inner lanes. This distance is called the stagger. Without it, the outer runner would travel a much longer distance to reach the same finish line. To calculate this, we must master the geometry of circles.
The total length around a circle. It is directly proportional to its diameter.
The ratio of Circumference (\(C\)) to Diameter (\(D\)) is a universal constant. It connects the straight-edged world with the curves of nature.
Used \(\pi \approx 3 + 1/8 = 3.125\). Realised it must be larger than 3.
"Trapped" \(\pi\) between 96-sided polygons: \( 3\frac{10}{71} < \pi < 3\frac{1}{7} \).
Found the "Close Ratio": \( 355/113 \). Accurate for over 800 years!
Discovered the first exact infinite series for \(\pi/4\). Birthed Calculus.
Using Rotational Symmetry, we can find parts of the circumference easily:
Stagger needed = \( 2\pi(r + w) - 2\pi r = 2\pi w \), where \( w \) is the lane width.
Area = Base \(\times\) Height
Transformation: Cut a right triangle from one side and slide it to the other to form a rectangle.
Area = \(\frac{1}{2} \times \text{Base} \times \text{Height}\)
A median of a triangle divides it into two triangles of equal area (even if they have different shapes!).
Can we find the area if we only know the side lengths? Yes, for triangles and cyclic 4-gons.
For a cyclic 4-gon with sides \( a, b, c, d \):
\( \text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)} \)
Note: If \( d=0 \), this becomes Heron's formula!
In 800 BCE, Indian mathematicians knew how to "square" a rectangle—construct a square with the exact same area as a rectangle of sides \( a \) and \( b \).
Geometric Identity
By constructing a right-angled triangle with hypotenuse \( \frac{a+b}{2} \) and one side \( \frac{a-b}{2} \), the third side automatically becomes \( \sqrt{ab} \)!
Nīlakaṇṭha Somayājī (c. 1500) provided a beautiful visual proof: slice a circle into tiny sectors and rearrange them. They form a parallelogram with base \( \pi r \) and height \( r \).
Subtending angle \( \theta^\circ \):
\( \text{Area} = \pi r^2 \times \frac{\theta^\circ}{360^\circ} \)