LUNES OF HIPPOCRATES History 310. MW: 11:00am Fall 2014 Professor: Dr. Robert Mena

by Noor Shukairy Israel Flores ThuyNguyen Nguyen

LUNES OF HIPPOCRATES

Lunes of Hippocrates states the area of the two lunes outside of the semicircle is the same as the area of the right triangle inscribed inside of that semicircle.

Step 1: Start with a Semi-circle

Step 2: Inscribe a triangle inside of the semi-circle. By Thales’ theorem, any triangle inscribed in a semicircle that shares the hypotenuse with the diameter is a right triangle.

Step 3: Inscribe a semi-circle on each leg of the triangle.

Now that we have the Lunes desired, we can begin to prove it geometrically.

Note: Geometrically, the big semi-circle is equal to the triangle and two slivers.

Also, the two half circles on the legs of the triangle can be broken down into the Lunes and slivers depicted in previous slides.

The smaller semicircles are composed of the two lunes and the two slivers of the large semicircle. The large semicircle is composed of the triangle and the two slivers. Subtract the slivers from both sides of the equation.

We have that the area of the triangle is equal to the area of the two lunes.

ALGEBRA PROOF

Start with a semicircle and inscribe a triangle in the semicircle with the hypotenuse of the triangle being the diameter of the semicircle. Construct a semicircle on each of the two sides of our original triangle. Let us call the sides of the triangle a and b, and let c denote the hypotenuse. By the Pythagorean theorem we know that 𝑎2 + 𝑏2 = 𝑐2 . The radius of the large semicircle is c/2. Thus it’s area is ((π𝑐2)/4)/2=(π𝑐2)/8 . We have that the area of the smaller semicircles are (π𝑎2)/8 and (π𝑏2)/8 . Divide both sides of the following equation by π/8: (π𝑎2)/8+(π𝑏2)/8=(π𝑐2)/8. We get 𝑎2 + 𝑏2 = 𝑐2 Thus, the sum of the areas of the smaller semicircles is equal to the area of the large semicircle. The smaller semicircles are composed of the two lunes and the two slivers of the large semicircle. The large semicircle is composed of the triangle and the two slivers. Subtract the slivers from both sides of the equation. We have that the area of the triangle is equal to the area of the two lunes.

Given a semicircle with a right triangle inscribed on its diameter and a semicircle on each of the sides of the inscribed right triangle, we will show that the area of the two lunes outside of the large semicircle is the same as the area of the inscribed right triangle.