Those Incredible Greeks!
Part 3: Hippocrates Acc.P
The next great hero of Ancient Greek Geometry is Hippocrates of Chios, not to be
confused with the namesake of the Hippocratic Oath. Hippocrates was bent on deriving the area
of a lune - a region bounded by arcs of two circles. The moon, at times, has his shape, which
accounts for his strange name. The lune of figure-10 is obtained by drawing a semicircle with
center at 𝑂 and radius 𝐴𝑂, of length 1. It follows that 𝐴𝐵, the diameter, has length 2. So far, so
good. Now we draw radius 𝑂𝐶 suche that it is perpendicular to 𝐴𝐵. Let’s throw in 𝐴𝐶 for good
measure. Then triangle 𝐴𝑂𝐶 is a right triangle with legs of length 1, in which case hypotenuse 𝐴𝐶
has length √2. (This triangle was a thorn in the backsides of the Pythagoreans, but it returns as a
friend here.)
Figure-10
Notice that we have also drawn a semicircle whose diameter is 𝐴𝐶. This semicircle is
smaller than the original one, whose diameter has length 2. As noted above, 𝐴𝐶 has length √2
which is approximately 1.414. Now the good geometers of that time made a remarkable
observation concerning the areas of similar figures which is highly relevant here since all
semicircles are similar – they have the same shape.
Two similar polygons have a linear magnification ratio – the number that we must use to
multiply the sides of one polygon to get the sides of the other. If we are given two similar
rectangles, the first with sides 2 and 5, and the second with sides 6 and 15, the linear
magnification ratio is 3, that is, each side of the larger rectangle is three times as long as the
corresponding side of the smaller. On the other hand, the areas of the two rectangles are 10 and
90. The ratio of area magnification is 9, not 3. What’s going on here? Why do we gat the square of
the linear magnification ration?
Let’s do this abstractly. Suppose the smaller rectangle has dimensions 𝑎 and 𝑏, and the
larger one has dimensions 𝑘𝑎 and 𝑘𝑏, where we are assuming that 𝑘 is the linear magnification
ratio. Then the area of the larger rectangle is 𝑘 2 𝑎𝑏 which shows us that the area magnification
ratio is the square of the linear magnification ratio. The Greeks extended this observation to
other figures as well including the circle.
In Figure-11, we consider the two semicircles with diameters 𝐴𝐵 and 𝐴𝐶 of the previous
2
2
figure. Since their diameters have lengths 2 and √2, the area magnification ratio is ( 2) , which
√
after squaring the numerator and denominator and simplifying yields 2.
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Lycée Edouard Branly (Amiens)
Those Incredible Greeks!
Part 3: Hippocrates Acc.P
Figure-11
Then the larger semicircle has twice as much area as the smaller one, implying that half
the larger has the same area as the smaller, as indicated in Figure-12, in which we equate the
areas of the semicircle with diameter 𝐴𝐶 and the quarter-circle 𝐴𝑂𝐶.
Figure-12
Now in the original diagram, the semicircle and quarter-circle overlap in the shaded
segment with corners at 𝐴 and 𝐶. If we remove this overlap from the semicircle and quartercircle, the leftovers must have the same area. But the leftovers are the lune with corners at 𝐴 and
𝐶 and the right angle 𝐴𝑂𝐶. Since the base and altitude of this right triangle have length one, its
1
2
area is . Then this is also the exact area of the lune! Wow! Please bear in mind that this
magnificent argument is made even more magnificent when one realizes that Hippocrates did
this in approximately 450 B.C. This is the exact calculation of the area of a curvilinear figure (one
with curved sides). Understand that this is about 200 years prior to the discovery by
Archimedes, that the area of a circle is 𝜋𝑟 2 .
From:
Fermat’s Enigma, by Simon Singh
The Saga of Mathematics by Lewinter-Widulski
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Lycée Edouard Branly (Amiens)