What is the minimum of $BP+\frac{1}{2}CP$? 
In $Rt\triangle ABC,\ \angle A=90^{\circ},\ AB=4,\ AC=6.$ The radius of $\odot A$ is $2$. What is the minimum of $BP+\frac{1}{2}CP$?


Actually, the question should be "What is the minimum of $BP+\frac{1}{3}CP$", and then it will be simple by using similarity. But I still wonder how to deal with this question.
One possible way I've got is to put it in Rectangular Coordinates and assume $P(p\ ,\ \sqrt{4-p^2})$ . By using $d=\sqrt{(x_1-x_2)^2+(y_1+y_2)^2\ }$ I can write $BP+\frac{1}{2}CP$ with $p$ . And then it can be worked out by calculating the minimum of the function $f(p)=BP+\frac{1}{2}CP$.
However , it is clear that it is very hard to work out. I tried to work it out with better and easier methods but failed. Are there other possible approaches? Thank you for your ideas in advance!
 A: Well, I made the following picture of the situation:

And we can write the following equations:

*

*$$\left|\text{BP}\right|=\sqrt{\left(\left|\text{AB}\right|-x_1\right)^2+\left(0-\text{y}_1\right)^2}=\sqrt{\left(\left|\text{AB}\right|-x_1\right)^2+\text{y}_1^2}\tag1$$

*$$\left|\text{CP}\right|=\sqrt{\left(0-x_1\right)^2+\left(\left|\text{AC}\right|-\text{y}_1\right)^2}=\sqrt{x_1^2+\left(\left|\text{AC}\right|-\text{y}_1\right)^2}\tag2$$

*$$\text{y}_1=\sqrt{\text{r}^2-x_1^2}\tag3$$
Combining, gives:

*

*$$\left|\text{BP}\right|=\sqrt{\left(\left|\text{AB}\right|-x_1\right)^2+\text{r}^2-x_1^2}\tag4$$

*$$\left|\text{CP}\right|=\sqrt{x_1^2+\left(\left|\text{AC}\right|-\sqrt{\text{r}^2-x_1^2}\right)^2}\tag5$$
So, we get:
\begin{equation}
\begin{split}
\mathscr{S}&=\left|\text{BP}\right|+\frac{1}{2}\cdot\left|\text{CP}\right|\\
\\
&=\sqrt{\left(\left|\text{AB}\right|-x_1\right)^2+\text{r}^2-x_1^2}+\frac{1}{2}\cdot\sqrt{x_1^2+\left(\left|\text{AC}\right|-\sqrt{\text{r}^2-x_1^2}\right)^2}\\
\\
&=\sqrt{\left|\text{AB}\right|^2+\text{r}^2-2\left|\text{AB}\right|x_1}+\frac{1}{2}\cdot\sqrt{\left|\text{AC}\right|^2+\text{r}^2-2\left|\text{AC}\right|\sqrt{\text{r}^2-x_1^2}}
\end{split}\tag6
\end{equation}
Now, we can solve:
$$\frac{\partial\mathscr{S}}{\partial x_1}=0\space\Longrightarrow\space x_1=\dots\tag7$$

Using your values, we get:
$$x_1\approx1.92745\tag8$$
The exact solution 'can' be found by solving:
$$3x_1\sqrt{5-2x_1}=4\sqrt{4-x_1^2}\sqrt{10-3\sqrt{4-x_1^2}}\tag9$$
So, we get:
$$\mathscr{S}\approx5.03822\tag{10}$$
A: 
Here is a solution using angles. Our aim was to investigate whether we could find an answer to OP’s question without much ado. But, we could not. Our answer is as laborious as that of J.W.L. Jan Eerland.
For brevity, we denote $BP = a$, $CP = b$, and $\measuredangle PAB = \omega$. Furthermore, let $L$ be the length, which we expect to find the minimum value of. We apply cosine rule to the triangles $PAB$ and $CAP$ to find expressions for $a$ and $b$ in terms of $\omega$.
$$a^2 = 16 + 4 – 16\cos\left(\omega\right)\qquad\Longrightarrow\qquad a = 2\sqrt{5-4\cos\left(\omega\right)}\tag{1}$$
$$b^2 = 36 + 4 – 24\sin\left(\omega\right)\qquad\Longrightarrow\qquad b = 2\sqrt{10-6\sin\left(\omega\right)}\tag{2}$$
Using (1) and (2), let us express $L$ in terms of $\omega$.
$$L = a + \dfrac{b}{2} = 2\sqrt{5-4\cos\left(\omega\right)} + \sqrt{10-6\sin\left(\omega\right)}$$
The equation  $\dfrac{dL}{d\omega} =0= \dfrac{4\sin\left(\omega\right)}{\sqrt{5-4\cos\left(\omega\right)}} - \dfrac{3\cos\left(\omega\right)}{ \sqrt{10-6\sin\left(\omega\right)}}$ must give us the value of $\omega$, for which the length $L$ is a minimum or maximum.
When simplified, it becomes,
$$f\left(\omega\right) = 160\sin^2\left(\omega\right)-96\sin^3\left(\omega\right)-45\cos^2\left(\omega\right)+36\cos^3\left(\omega\right) = 0. \tag{3}$$
The easiest way to solve equation (3) is to apply the numerical method of Newton-Raphson. You need the following Newton-Raphson formula to do so. By the way, $f’\left(\omega\right)$ is the first derivative of $f\left(\omega\right)$. As an educated guess for $\omega_{\text{0}}$, we chose $\dfrac{\pi}{4}$ after some spot checks revealed that the minimum of $L$ occurs for $\dfrac{\pi}{4} \gt \omega \gt 0.$
$\omega_{\text{n+1}} = \omega_{\text{n}} - \dfrac{ f\left(\omega_{\text{n}}\right) }{ f’\left(\omega_{\text{n}}\right) }= \dfrac{160\sin^2\left(\omega\right)-96\sin^3\left(\omega\right)-45\cos^2\left(\omega\right)+36\cos^3\left(\omega\right) }{2\sin\left(\omega\right) \cos\left(\omega\right)\Bigl\{205-144\sin\left(\omega\right) - 54\cos\left(\omega\right)\Bigr\}},\text{ where }\space\omega_{\text{0}} = \dfrac{\pi}{4}$.
When this numerical procedure is executed in an EXCEL worksheet, it took only five iteration steps for $\omega$ to converge to the following values.
$\omega = 0.2701631649520730\enspace \text{rad}= 15.4792091316505000^0$
$L_{\text{min}} = 5.0382236048810800$
