Around 250 B.C., Archimedes calculated the ratio of a circle's circumference to its diameter. Today we know that ratio to be PI, or Math.PI in JAVA, but HOW?

• First, he considered the shapes he already understood and that he knew how to calculate the perimeter for, like triangles, squares, and other polygons.
• Next, Archimedes noticed that if you divide a circle into inscribed polygons (square, pentagon, hexagon, heptagon, octagon, ..., etc.) with an increasing number of sides, each side of the polygon decreased in length, and therefore gave a better and better approximation for the circumference of a circle.

• With this observation, and a little more geometry, we can create an algorithm to calculate the perimeter of larger and larger inscribed polygons.
• Remember, the more sides, (n represents the number of sides in the polygon), the more accurate our perimeter estimation is, so let's do the math...

• From the image above, you can see the octagon is divided into isosceles triangles, the bottom of which (labeled s) is the length we need to know.
• If we draw a line from the circle center perpendicular to the base of the triangle s, we divide s in half, so it becomes ${1 \over 2}s$ as depicted.
• From the picture, we can simplify our lives by assuming $h = 1$, and that the circle is a Unit circle.
  • Why can we make this assumption? Remember, we're trying to figure out PI, which we know is the RATIO of $Circumference \over Diameter$, and ratios don't care what values we assign as long as the ratio holds!
• Also, from the diagram we can see 2 angles B and A and because of symmetry $A = {1 \over 2}B$
• Great, so now what!?!
  • WELL, we can calculate B, so let's start there. Remember, a Circle is 360° and we are equally dividing that by n sides ($n = 8$ in the example).
  • So... $B = {360° \over 8}$ or $45°$ and $A = {1 \over 2} * B$ so $A = 22.5°$ in our example.
  • Now we can calculate ${1 \over 2}s$ if we recall $sin(A) = {{1 \over 2}s \over h}$ as shown in the below diagram of the triangle.

• ...continued...
  • Solving for s we see $s = 2 * h * sin(A)$
  • And since we agreed $h = 1$, that simplifies to $s = 2 * sin(A)$
  • Now we have s, so to get the polygon perimeter, we just multiply by the number of sides to get it.
  • Finally, we can calculate PI as $PI = {(n * s) \over 2h}$

OK, so if you've followed the math, feel free to skip ahead and implement calculating PI as an algorithm, but if not lets lay it out a bit more concisely!

• Step 1: Read into an integer the number of sides for our polygon: $n = 8$
• Step 2: Calculate the angle $B = {360.0 \over n}$
• Step 3: Calculate the angle $A = {1 \over 2} * B$
• Step 4: Calculate the length of 1 triangle base $s = 2 * sin(A)$ HINT: YOU NEED TO CONVERT A to RADIANS
  • Math.sin(), and Math.toRadians() will be your friends.
• Step 5: Get the polygon perimeter $p = n * s$
• Step 6: Estimate PI: $PI = {p \over 2}$

Finally, your assignment is to use the book and other resources to put the above algorithm into a program to estimate PI!

• Fork my repo
• Using IntelliJ clone your fork locally
• Use IntelliJ to create a feature branch named Feature1
• Implement the algorithm in ArchimedesPiMethod.java

• Just as you did last week (Reference the Lab video in your Week 1 module), create a Feature1 feature branch of your code if you haven't already
• Commit your working code to your local copy
• Push it to your Remote/origin branch (i.e. GitHub: Feature1 -> origin/Feature1)
• Then issue a Pull request to my instructor branch
• Make sure to save the Pull request URL and submit it for the assignment.