7. Dijana used sticks of length $5 \mathrm{~cm}$ to form a row of equilateral triangles (as shown in the figure). If Dijana used 99 sticks, what is the distance between the two farthest points in the row of triangles formed?

![](https://cdn.mathpix.com/cropped/2024_06_03_acca65fe0c81619e2622g-04.jpg?height=197&width=916&top_left_y=1244&top_left_x=570)

The use of a pocket calculator or any reference materials is not allowed.

## SCHOOL/CITY COMPETITION IN MATHEMATICS
First method:

If we look at the picture:

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-06.jpg?height=145&width=665&top_left_y=1618&top_left_x=724)

it can be noticed that for the first triangle, Diana must take 3 matches,

and for each subsequent triangle in the sequence, she needs to add only 2 more matches.

2 POINTS

Using 99 matches, Diana made the initial triangle and $(99-3) \div 2 = 48$ additional triangles, making a total of 49 triangles.

2 POINTS

If we look at sequences composed of an odd number of triangles,
$\Omega$
![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-06.jpg?height=166&width=1348&top_left_y=2073&top_left_x=455)

we observe that the number of triangles "facing upwards" is 1 more than the number of triangles "facing downwards."

Thus, out of $n$ triangles, $(n+1) \div 2$ are "facing upwards."

The distance between the two farthest points is then $((n+1) \div 2) \cdot a$, where $a$ is the length of the side of the triangle.

2 POINTS

In a sequence of 49 triangles, there are 25 "upward-facing" triangles, and the desired distance is $25 \cdot 5 = 125 \text{ cm}$.