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© T Madas
© T Madas
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.
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The lunes of Hippocrates
Start with a right angled triangle
How can we circumscribe this triangle?A circle theorem might be useful
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The triangle is now circumscribed
The lunes of Hippocrates
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Find the midpoints of the other two sidesDraw two more circles as shown
The lunes of Hippocrates
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The following result is credited to Hippocrates
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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)
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A
BC
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
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A
BC
What is the area of the two regions enclosed by the lunes and the triangle?
a
b c
p 12ab- =
2
8cp
12´ ( )2
c´2
p 12ab- =1
2´2
4c´
12ab-
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A
BC
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
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-- ( )A
BCa
b c
p´2
8cpp´1
2 ( )2
2a´ 1
2+ ( )2
2b´ 1
2ab- =
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8p ( )
-- ( )p´2
8cpp´1
2 ( )2
2a´ 1
2+ ( )2
2b´ 1
2ab- =
p´2
8cp-p´1
22
4a´ 1
2+2
2b´ 1
2ab+ =
2
8cp-
2
8ap 2
8bp+ 1
2ab+ =
2c-2a 2b+ 12ab+ =
2 2 2a b c+ = Û2 2 2a b c+ - =0
BCa
b c
A
1
2ab
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© T Madas
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 triangleTake 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
CM
N
=
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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 aless 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
CM
N
a
212a
214 ap 21
2a-
214 ap 21
2a-
212a
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To find AN, we use Pythagoras theorem on AMB :
The area of the semi circle, radius AN :
A
B
CM
N
a
214 ap 21
2a-
212a
2a 2a+ 2AB= Û22a2AB = Û
22aAB = Û2AB a=
So AN =
12p
222 a´ ( ) 1
2p= 224a´ 21
4 ap=
The area of the semi circle, radius AN :
22 a
214 ap
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The area of the semi circle, radius AN :
A
B
CM
N
a
214 ap 21
2a-
212a
214 ap
The area of the lune: 21
4 ap -- ( )2 21 14 2a ap -
214 ap= 21
4 ap- 212a+
212a=
212a
=
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© T Madas
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.
A B
CD
O
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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.
A B
CD
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
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If the square has a side length a its area is a 2
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.
A B
CD
O
Pythagoras on AOB
a
x x
2x 2x+ 2a= Û22x 2a= Û
2x = 2
2a Û
x = 2a
Area of circle: p 2
a p=2
2a´
2
2ap=
Area of the 4 segments: 2 2
2a ap -
2 22a ap -
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Area of semicircle: 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.
A B
CD
O
a
x x
p 2a´ 1
2p=2
4a´
2
8ap=
Area of the 4 lunes:
2 22a ap -
12´
Area of 4 semicircles: 2
8ap 4´
2
2ap=
2
2ap 2 2
2a apé ù- -ê úë û
2
2ap=
2
2ap- 2a+ 2a=
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