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!

Solution 1:

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:

$$\mathscr{S}\approx5.03822\tag{10}$$