Arbelos

Problem

In the Book of Lemmas^[1], Thabit ibn Qurra (c. 826 – 901 AD) describes a shape called the “arbelos”. The “arbelos” in the figure below is the shape bounded by three semicircles with diameters AB, AD, and BD (shaded in blue). Suppose a circle be constructed such that its diameter is CD (shaded in red), determine the ratio of the areas of the arbelos to the circle.

[1] It is uncertain whether the Book of Lemmas was authored by Thabit ibn Qurrah or Archimedes because the earliest edition is written in Arabic and refers to Archimedes in the third person.

Solution

Area of the circle

$$\frac{AD}{CD} = \frac{CD}{BD}$$

$$CD^2 = AD \cdot BD$$

$${A}_{circle} = \frac{\pi}{4} AD \cdot BD$$

Area of the arbelos

$${A}_{arbelos} = \frac{\pi}{2} \left(\frac{AB}{2} \right)^2 - \frac{\pi}{2} \left(\frac{AD}{2} \right)^2 - \frac{\pi}{2} \left(\frac{BD}{2} \right)^2$$

Since

$$AB = AD + BD$$

$$AB^2 = AD^2 + 2AD\cdot BD + BD^2$$

$${A}_{arbelos} = \frac{\pi}{8} \left(AD^2 + 2AD\cdot BD + BD^2 \right) - \frac{\pi}{8} AD^2 - \frac{\pi}{8} BD^2$$

$${A}_{arbelos} = \frac{\pi}{4} AD \cdot BD$$

Therefore,

$${A}_{arbelos} = {A}_{circle}$$

and

$${A}_{arbelos}:{A}_{circle} = 1:1$$