How find the $AP+\frac{1}{2}BP$ minmum value An equilateral triangle $ABC$ such $$AB=BC=AC=2a>0$$
 A circle $O$ is inscribed in triangle $ABC$,and the point $P$ on the circle $O$.
Find the minimum $$AP+\dfrac{1}{2}BP$$
My idea: let $$A(-a,0),B(a,0),O(0,\dfrac{\sqrt{3}}{3}a)$$
then the circle equation is $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
let $P(x,y)\;$ , then
$$|PA|+\dfrac{1}{2}|PB|=\sqrt{(x+a)^2+y^2}+\dfrac{1}{2}\sqrt{(x-a)^2+y^2}$$
where $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
then I can't.Thank you 

The solution of Omri Nissan Solan.
Let $K$ and $L$ be common points of the circle with $AB$ and $BC$ respectively
and let $BO\cap KL=\{M\}$.
Thus, our circle it's the Apollonian circle of $\Delta KLB$. 
Id est, $AP+\frac{1}{2}PB=AP+PM\geq AM=\frac{a\sqrt7}{2}$.
