Task B-4.1. The lengths of the sides of 5 equilateral triangles form an arithmetic sequence. The sum of the perimeters of these triangles is $120 \mathrm{~cm}$. The sum of the areas of the smallest and the largest triangle is $10 \sqrt{3} \mathrm{~cm}^{2}$ less than the sum of the areas of the remaining three triangles. Determine the lengths of the sides of these triangles. What is the sum of the areas of all the triangles?
## Solution.

Let the sides of the triangles be $\mathrm{s} a_{1}=a-2 d, a_{2}=a-d, a_{3}=a, a_{4}=a+d, a_{5}=a+2 d$, where $d$ is the difference of the sequence.

Since the sum of the perimeters of these triangles is $120 \mathrm{~cm}$, we have

$$
\begin{aligned}
& 3\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)=120 \\
& a-2 d+a-d+a+a+d+a+2 d=40
\end{aligned}
$$

It follows that $a=8$, so the given sequence is $8-2 d, 8-d, 8,8+d, 8+2 d$.

For the areas, we have

$$
\frac{a_{1}^{2} \sqrt{3}}{4}+\frac{a_{5}^{2} \sqrt{3}}{4}+10 \sqrt{3}=\frac{a_{2}^{2} \sqrt{3}}{4}+\frac{a_{3}^{2} \sqrt{3}}{4}+\frac{a_{4}^{2} \sqrt{3}}{4} /: \frac{\sqrt{3}}{4}
$$

Then we have

$$
\begin{aligned}
a_{1}^{2}+a_{5}^{2}+40 & =a_{2}^{2}+a_{3}^{2}+a_{4}^{2} \\
(8-2 d)^{2}+(8+2 d)^{2}+40 & =(8-d)^{2}+8^{2}+(8+d)^{2}
\end{aligned}
$$

It follows that $d^{2}=4, d=2$ or $d=-2$.

If we set $d=2$, we get: $a_{1}=4 \mathrm{~cm}, a_{2}=6 \mathrm{~cm}, a_{3}=8 \mathrm{~cm}, a_{4}=10 \mathrm{~cm}, a_{5}=12 \mathrm{~cm}$. (If we set $\mathrm{d}=-2$, we get the same numbers but in descending order).

The sum of the areas of all the triangles is

$$
P=\frac{\sqrt{3}}{4}(16+36+64+100+144)=90 \sqrt{3} \mathrm{~cm}^{2}
$$