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2. The mathematician Hippocrates of Chios (470 BC- 380 BC) He is not to be confused with the famous physician Hippocrates of Cos (author of the Hippocratic Oath). Hippocrates of Chios taught in Athens and worked on the classical problems of squaring the circle and duplicating the cube. Little is known of his life but he is reported to have been an excellent geometer.

3. The lunes of Hippocrates Start with a right angled triangle How can we circumscribe this triangle? A circle theorem might be useful

4. © T Madas The triangle is now circumscribed The lunes of Hippocrates

5. © T Madas Find the midpoints of the other two sides Draw two more circles as shown The lunes of Hippocrates

6. © T Madas The following result is credited to Hippocrates

7. © T Madas The area of the right angled triangle is equal to the sum of the lunes’ area, off its perpendicular sides Lune = Moon shaped (Μηνίσκος in Greek)

8. © T Madas A B C What is the area of the two regions enclosed by the lunes and the triangle? Together they must equal the area of the “circumscribing semicircle” less the area of the triangle a b c

9. © T Madas A B C What is the area of the two regions enclosed by the lunes and the triangle? a b c

10. © T Madas A B C What is the area of the two lunes? a b c Together they must equal the area of the two semicircles (off the perpendicular sides of the triangle), less the area we just found

11. © T Madas A B C a b c

12. B C a b c A

14. Consider an isosceles right angled triangle. Circumscribe the triangle. [centre the midpoint of the hypotenuse and radius half the hypotenuse] Draw the height of the triangle Take one of the 2 resulting triangles Circumscribe it The arcs of the two circumscribing circles form a lune. Prove that the area of the lune is equal to area of ABM A B C M N =

15. Suppose that AM = a The area of AMB = Let’s find the area of the green segment Its area must be a semi circle, radius a less the area of the triangle The area of the lune must be: The area of a semi circle radius AN less the segment A B C M N a

16. To find AN, we use Pythagoras theorem on AMB : The area of the semi circle, radius AN : A B C M N a So AN = The area of the semi circle, radius AN :

17. © T Madas The area of the semi circle, radius AN : A B C M N a The area of the lune: =

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19. A square is circumscribed by a circle. 4 semicircles are drawn outside the square, each having as a diameter the 4 sides of the square. Prove that the area of the 4 lunes equals the area of the square. AB C D O

20. A square is circumscribed by a circle. 4 semicircles are drawn outside the square, each having as a diameter the 4 sides of the square. Prove that the area of the 4 lunes equals the area of the square. AB C D O The area of one lune: Area of a semicircle Less the area of a segment Area of a segment: area of the circle less area of the square divide by 4 as there are 4 identical segments The area of 4 lunes: area of 4 semicircles less area of 4 segments

21. © T Madas If the square has a side length a its area is a 2 A square is circumscribed by a circle. 4 semicircles are drawn outside the square, each having as a diameter the 4 sides of the square. Prove that the area of the 4 lunes equals the area of the square. AB C D O Pythagoras on AOB a x x Area of circle: Area of the 4 segments:

22. © T Madas Area of semicircle: A square is circumscribed by a circle. 4 semicircles are drawn outside the square, each having as a diameter the 4 sides of the square. Prove that the area of the 4 lunes equals the area of the square. AB C D O a x x Area of the 4 lunes: Area of 4 semicircles:

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