# Archimedes found found an interesting property of arbelos, that is the area of the circle whose diameter is AH and has common tangent line to the two smaller semicircles at point A is equal to the shaded area. Prove it?

Nov 1, 2016

see explanation.

#### Explanation:

Let the three respective semicircles be $X , Y , \mathmr{and} Z$, as shown in the diagram.
Let the areas of $X , Y , Z$ be ${A}_{X} , {A}_{Y} , {A}_{Z}$, respectively.
Let the shaded area be ${A}_{S}$
$\implies {A}_{S} = {A}_{X} - \left({A}_{Y} + {A}_{Z}\right)$
${A}_{X} = \frac{1}{2} \frac{\pi {D}^{2}}{4}$, where $D$=diameter$= r + 1 - r = 1$
$\implies {A}_{X} = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$
${A}_{Y} = \frac{1}{2} \cdot \frac{\pi {r}^{2}}{4} = \frac{\pi {r}^{2}}{8}$
${A}_{Z} = \frac{1}{2} \cdot \frac{\pi {\left(1 - r\right)}^{2}}{4} = \frac{\pi \left(1 - 2 r - {r}^{2}\right)}{8}$
$\implies {A}_{S} = \frac{\pi}{8} - \left(\frac{\pi {r}^{2}}{8} + \left(\frac{\pi}{8} \left(1 - 2 r + {r}^{2}\right)\right)\right)$
$\implies {A}_{S} = \frac{\pi}{8} \left(1 - {r}^{2} - 1 + 2 r - {r}^{2}\right)$
$\implies {A}_{S} = \frac{\pi}{4} \left(r - {r}^{2}\right)$

Now let the circle centered at $o$ be $W$, $A H$ is its diameter.
Let $o '$ be the center of $X$
$o ' H =$ radius of$X = \frac{1}{2}$
$a = 1 - r - \frac{1}{2} = \frac{1}{2} - r$
$A {H}^{2} = o ' {H}^{2} - {a}^{2} = {\left(\frac{1}{2}\right)}^{2} - {\left(\frac{1}{2} - r\right)}^{2}$
$A {H}^{2} = \left(\frac{1}{4} - \frac{1}{4} + r - {r}^{2}\right) = \left(r - {r}^{2}\right)$

Another way to find $A {H}^{2}$ is :
$A {H}^{2} = B A \cdot A C = r \cdot \left(1 - r\right) = \left(r - {r}^{2}\right)$

Now let the area of $W$ be ${A}_{W}$
$\implies {A}_{W} = \pi {\left(A H\right)}^{2} / 4 = \frac{\pi}{4} \left(r - {r}^{2}\right)$

Hence, ${A}_{S} = {A}_{W} = \frac{\pi}{4} \left(r - {r}^{2}\right)$ (proved)