Construct a figure with six adjacent equilateral triangles, the first with a side length of $1 \mathrm{~cm}$ and the following triangles with a side equal to half the side length of the previous triangle, as indicated in the given figure. What is the perimeter of this figure?
![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-099.jpg?height=160&width=262&top_left_y=1042&top_left_x=1231)

I

The perimeter of the figure is formed by thirteen segments, in the sequence of the formation of the triangles, which can be described as follows.

- 2 segments of $1 \mathrm{~cm}$ and 1 segment of $\frac{1}{2}$ cm in triangle I,
- 1 segment of $\frac{1}{2} \mathrm{~cm}$ and 1 segment of $\frac{1}{4}$ cm in triangle II,
- 1 segment of $\frac{1}{4} \mathrm{~cm}$ and 1 segment of $\frac{1}{8}$ cm in triangle III,
- 1 segment of $\frac{1}{8} \mathrm{~cm}$ and 1 segment of $\frac{1}{16}$ cm in triangle IV,
- 1 segment of $\frac{1}{16} \mathrm{~cm}$ and 1 segment of $\frac{1}{32}$ cm in triangle V, and
- 2 segments of $\frac{1}{32}$ cm in triangle VI.

Solution 1: By counting the lengths of the segments, we can see that the perimeter measures

$$
\begin{aligned}
2 \times 1 & +2 \times \frac{1}{2}+2 \times \frac{1}{4}+2 \times \frac{1}{8}+2 \times \frac{1}{16}+3 \times \frac{1}{32} \\
& =2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{3}{32}=3+\frac{16+8+4+3}{32} \\
& =3+\frac{31}{32}=\frac{127}{32} \mathrm{~cm}
\end{aligned}
$$

Solution 2: The contour of the figure, starting from the left corner and following in a counterclockwise direction, measures

$$
1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{32}+\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1
$$

centimeters. The sum of the geometric progression with the first term 1, common ratio $\frac{1}{2}$, and the last term $\frac{1}{32}$ is given by

$$
1-\frac{1}{64} / 1-\frac{1}{2}=2-\frac{1}{32}
$$

Therefore, the perimeter of the figure measures

$$
\left[2-\frac{1}{32}\right]+\frac{1}{32}+\left[2-\frac{1}{32}\right]=4-\frac{1}{32}=\frac{127}{32} \mathrm{~cm}
$$

Solution 3: Observe that each time we add a triangle with side length $a$, we replace one segment of length $a$ in the perimeter with two segments of length $a$, so the perimeter increases by $a$.

Since the first triangle has a perimeter of $3 \mathrm{~cm}$, adding a triangle with side length $\frac{1}{2} \mathrm{~cm}$, the new figure has a perimeter of $3+\frac{1}{2} \mathrm{~cm}$; if we add another triangle with side length $\frac{1}{4} \mathrm{~cm}$, the new figure has a perimeter of $3+\frac{1}{2}+\frac{1}{4} \mathrm{~cm}$. Following this process, after the sixth triangle, the figure has a perimeter of

$$
3+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=3+1-\frac{1}{32}=\frac{127}{32} \mathrm{~cm}
$$

where we use the sum of the geometric progression with the first term $\frac{1}{2}$, common ratio $\frac{1}{2}$, and the last term $\frac{1}{32}$, given by

$$
\frac{1}{2}-\frac{1}{64} / 1-\frac{1}{2}=1-\frac{1}{32}
$$