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 
 A: Sorry I can't manage English well, so I might just use the google translate.
Take $B$'s inverse points about the circle, and call it $B'$.
By conclusions concerning Apollonian circles, we knew that $PB'=\dfrac{1}{2}PB$.
Hence we only need to determine the minimum of $PA+PB'$, thus it's easy to observe that when $P$ lies on the line $AB$ we get the minimum value, and that is quite easy to calculate.
The answer is $\dfrac{\sqrt{7}}{2}a$, if I'm not mistaken.（which happens a lot)
A: Some ideas:
Taking it from where you left it and correcting the coordinates of $\;O\;$ , we get the circle's equation is
$$x^2+\left(y-\frac a{2\sqrt3}\right)^2=\frac{a^2}{12}\;\;\;\;(I)$$
Observe that since the $\;x$-axis is tangent to the circle, the absolute value of the center's $\;y$-coordinate  equals the circle's radius...
Now, with your notation
$$|PA|+\frac12|PB|=\sqrt{(x+a)^2+y^2}+\frac12\sqrt{(x-a)^2+y^2}\;\;\;\;(II)$$
Thus, we want to minimize (II) contiioned to (I) , and thus Lagrange's Multipliers may help here...or
$$(I)\implies x=\pm\sqrt{\frac{a^2}{12}-\left(y-\frac a{2\sqrt3}\right)^2}$$
and substitute in (II) to have a one-variable extrema problem (first derivative and stuff)
A: 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, $AP+\frac{1}{2}PB=AP+PM\geq AM=\frac{a\sqrt7}{2}$.
