Task 1. 15 points

Find the minimum value of the expression

$$
\sqrt{x^{2}-\sqrt{3} \cdot|x|+1}+\sqrt{x^{2}+\sqrt{3} \cdot|x|+3}
$$

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

Let $t=|x| \geq 0$. We need to find the minimum value of the function

$$
f(t)=\sqrt{t^{2}-\sqrt{3} \cdot t+1}+\sqrt{t^{2}+\sqrt{3} \cdot t+3}
$$

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

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

Let the points be $\quad M(t, 0), A\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right), B\left(-\frac{\sqrt{3}}{2}, \frac{3}{2}\right) . \quad$ It is not difficult to see that the expression we are considering is nothing but the sum of the lengths of segments $A M$ and $B M$, that is, $f(t)=A M+B M$. 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 $A M+B M=A B$, meaning that 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 $x$-axis.

![](https://cdn.mathpix.com/cropped/2024_05_06_e2fedf2b675f713cdba7g-02.jpg?height=742&width=714&top_left_y=186&top_left_x=274)

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 hence

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

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 $t=|x|=\frac{\sqrt{3}}{4}$ and is equal to:

$$
f\left(\frac{\sqrt{3}}{4}\right)=\sqrt{\frac{3}{16}+\frac{1}{4}}+\sqrt{\frac{27}{16}+\frac{9}{4}}=\frac{\sqrt{7}}{4}+\frac{3 \sqrt{7}}{4}=\sqrt{7}
$$

Answer: $\sqrt{7}$ when $x= \pm \frac{\sqrt{3}}{4}$

## Evaluation Criteria: