Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-32.jpg?height=280&width=582&top_left_y=1198&top_left_x=434)
Answer: 84.

Solution. Rhombuses consisting of eight triangles can be of one of three types:
![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-32.jpg?height=118&width=602&top_left_y=1598&top_left_x=433)

It is clear that the number of rhombuses of each orientation will be the same, so let's consider only the vertical ones. Each of them is uniquely determined by its top triangle. Now it is easy to count the number of such triangles.
![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-33.jpg?height=278&width=572&top_left_y=85&top_left_x=444)

In the first row, there is 1 such triangle, in the second row - 2, in the third row - 3, ..., in the seventh row - 7. In total, there are \(1+2+3+\ldots+7=28\) vertical rhombuses, and therefore, the answer to the problem is \(28 \cdot 3 = 84\).