Question

In the figure, there are $3$ semicircle touching each other internally and one circle touching two of them externally and third on internally. Radius of the complete circle (in its lowest form) is $\frac{p}{q}(p,q\in N)$, then $p+q$ is _____.

Answer 1

center of bigger semicircle(radius$=3$) is $R1$

,the next bigger one (radius$=2$) is $R2$

and the smaller one(radius$=1$) is $R3$

and complete circle one is $R$ (assume its radius is r )

Now consider the triangle $R2R3R$ and draw a line from $RtoR1$ it acts as a cevian on this triangle lengths of triangle

$R2R3=3$

$R2R=2+R$

$R3R=1+R$

$R1R=3-R$ [bigger circle is normal and complete circle are same so radius of both coincide]

$R2R1=1$

$R3R1=2$

now apply cosine rule at point $R1$

$\frac{(2+r)2-12-(3-r{)}^2}{2\times 1\times (3-r)}=\frac{-\left[\right(1+r)2-22-(3-r{)}^2}{2\times 2\times (3-r)]}$

solving this we get $r=\frac{12}{14}=\frac{6}{7}$

$p+q=13$