In the given figure, M is the mid point of AB. Semicircles are drawn on AB with AM and MB as diameters. A circle with centre 'O' touches all three semicircles as shown. Prove that the radius of this circle is $\frac{1}{6}$ AB.

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Solution

Let r be the radius of bigger semi circle

$⟹ r = \frac{A B}{2}$

$r_{1}$ be the radius of smaller semicircles

$⟹ r_{1} = \frac{r}{2}$

And $r_{2}$ be the radius of the circle

Consider $△ O M D$

$O M^{2} + M D^{2} = O D^{2}$

$O M = \sqrt{O D^{2} - M D^{2}}$

$= \sqrt{(r_{2} + \frac{r}{2})^{2} - {\frac{r}{2}}^{2}}$

$M R = M O + O R$

$r = O M + r_{2}$

$O M^{2} = (r - r_{2})^{2}$

$(r_{2} + \frac{r}{2})^{2} - \frac{r^{2}}{4} = r^{2} + r_{2}^{2} = 2 r r_{2}$

$r_{2}^{2} + \frac{r^{2}}{4} + r r_{2} - \frac{r^{2}}{4} = r^{2} + r_{2}^{2} - 2 r r_{2}$

$⟹ 3 r r_{2} = r^{2}$

$3 r_{2} = r$

$r_{2} = \frac{r}{3} = \frac{A B}{6}$

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