diff --git a/.idea/misc.xml b/.idea/misc.xml
index d15472f..045943f 100644
--- a/.idea/misc.xml
+++ b/.idea/misc.xml
@@ -1,6 +1,6 @@
 <?xml version="1.0" encoding="UTF-8"?>
 <project version="4">
-  <component name="ProjectRootManager" version="2" languageLevel="JDK_18" default="true" project-jdk-name="openjdk-18" project-jdk-type="JavaSDK">
+  <component name="ProjectRootManager" version="2" languageLevel="JDK_20" default="true" project-jdk-name="openjdk-18" project-jdk-type="JavaSDK">
     <output url="file://$PROJECT_DIR$/out" />
   </component>
 </project>
\ No newline at end of file
diff --git a/README.md b/README.md
index 9513c6f..d34df80 100644
--- a/README.md
+++ b/README.md
@@ -11,7 +11,8 @@ Today we know that ratio to be PI, or **Math.PI** in JAVA, but HOW?
 ![Inscribed Polygons](images/geometry003.png)
 * 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...
-![General Equation](./images/geometry001.jpg)
+
+![General Equation](images/geometry001.jpg)
 * 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**.
@@ -21,7 +22,10 @@ Today we know that ratio to be PI, or **Math.PI** in JAVA, but HOW?
     * 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.
-    ![Geometry Reminder](images/geometry002.jpg)
+
+![Geometry Reminder](images/geometry002.jpg)
+
+* ...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.
diff --git a/src/ArchimedesPiMethod.java b/src/ArchimedesPiMethod.java
index 0b6e780..0b4eea6 100644
--- a/src/ArchimedesPiMethod.java
+++ b/src/ArchimedesPiMethod.java
@@ -1,7 +1,22 @@
 import java.util.Scanner;
-
+/*   Chris Shortt's homework Java Assignment-002 */
 public class ArchimedesPiMethod {
     public static void main(String[] args) {
+        while (true) {
+            System.out.println("please type the number of sides");
+            Scanner sc = new Scanner(System.in);
+            int n = sc.nextInt();
+            if (n < 1){
+                break;
+            }
+            double b = 360.0 / n;
+            double a = b / 2;
+            double s = 2 * Math.sin(Math.toRadians(a));
+            double p = n * s;
+            double pi = p / 2;
+            System.out.printf("Our PI estimate is: %.10f %n", pi);
+            System.out.println(pi);
 
+        }
     }
-}
+}
\ No newline at end of file