Second dart forced farther out than the first

Nicole throws two darts at a circular board of radius $R$. The distance $R_1$ of her first dart from the centre is uniformly distributed on $(0,R)$. Her second dart always lands farther from the centre than the first, and given $R_1$ it is uniformly distributed (by area) over the part of the board at distance greater than $R_1$. The probability that the second dart is at distance at least $\tfrac R2$ from the centre can be written as $\dfrac{a+\ln b}{c}$ in lowest terms with positive integers $a,b,c$. Find $a+b+c$.

$39$. If $R_1>\tfrac R2$ the event is certain. If $R_1=r<\tfrac R2$, the area ratio gives $\tfrac{3/4}{1-(r/R)^2}$. Averaging over $r$ yields $\tfrac12+\tfrac34\int_0^{1/2}\tfrac{du}{1-u^2}=\tfrac12+\tfrac38\ln3=\tfrac{4+\ln 27}{8}$, so $a+b+c=4+27+8=39$.