# Problem 1. 15 points

Find the minimum value of the expression

$$
\sqrt{x^{2}-2 \sqrt{3} \cdot|x|+4}+\sqrt{x^{2}+2 \sqrt{3} \cdot|x|+12}
$$

as well as the values of $x$ at which it is achieved.

#
# Solution:

Let $t=\frac{|x|}{2} \geq 0$. Then we need to find the minimum value of the function

$$
f(t)=2\left(\sqrt{t^{2}-\sqrt{3} \cdot t+4}+\sqrt{t^{2}+\sqrt{3} \cdot t+12}\right)
$$

for $t \geq 0$. Transform the function $f(t)$ to the form:

$$
f(t)=2\left(\sqrt{\left(t-\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{4}}+\sqrt{\left(t+\frac{\sqrt{3}}{2}\right)^{2}+\frac{9}{4}}\right)
$$

Let points $M(t, 0), A\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right), B\left(-\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ be given. It is not difficult to understand that the expression we are considering is nothing but twice the sum of the lengths of segments $A M$ and $B M$, that is, $f(t)=2(A M+B M)$. If point $M$ does not lie on segment $A B$, then $A M+B M>A B$. On the other hand, if $M \in A B$, then $A M+B M=A B$, so the sum $A M+B M$ attains its minimum value if and only if point $M$ lies on segment $A B$. Therefore, point $M$ is the intersection of segment $A B$ with the $O x$ axis.

![](https://cdn.mathpix.com/cropped/2024_05_06_e2fedf2b675f713cdba7g-11.jpg?height=746&width=739&top_left_y=181&top_left_x=270)

To find the coordinates of point $M$, consider similar triangles $M D A$ and $M C B$:

$$
\frac{M D}{C M}=\frac{A D}{B C}=\frac{1}{3}
$$

Therefore, if $M D=a$, then $M C=3 a$ and from this

$$
\begin{gathered}
C D=\sqrt{3}=M D+M C=a+3 a=4 a \\
a=\frac{\sqrt{3}}{4} .
\end{gathered}
$$

From this, we find the coordinates of point $M\left(\frac{\sqrt{3}}{4}, 0\right)$. Thus, the minimum value of the function $f(t)$ is achieved when $t=\frac{|x|}{2}=\frac{\sqrt{3}}{4}$ and is equal to:

$$
f\left(\frac{\sqrt{3}}{4}\right)=2\left(\sqrt{\frac{3}{16}+\frac{1}{4}}+\sqrt{\frac{27}{16}+\frac{9}{4}}\right)=
$$

$$
=2\left(\frac{\sqrt{7}}{4}+\frac{3 \sqrt{7}}{4}\right)=2 \sqrt{7}
$$

Answer: $2 \sqrt{7}$ when $x= \pm \frac{\sqrt{3}}{2}$.

## Criteria: