Shapovalov A.V.

On the table, 28 coins of the same size are arranged in a triangular shape (see figure). It is known that the total mass of any three coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_7181633d50815d79481eg-07.jpg?height=623&width=692&top_left_y=1644&top_left_x=683)
$$
\begin{aligned}
& 1+2+3= \\
& =2+3+1^{\prime}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_7181633d50815d79481eg-08.jpg?height=354&width=305&top_left_y=0&top_left_x=868)

First solution. Let's take a rhombus of 4 coins. As can be seen from the figure, the masses of two coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the figure, then coins of the same color will have the same mass.

Now it is easy to find the sum of the masses of the coins on the boundary: there are 6 coins of each color there, and the sum of the masses of three differently colored coins is 10 g; therefore, the total mass of the coins on the boundary is $6 \cdot 10=60$ g.
![](https://cdn.mathpix.com/cropped/2024_05_06_7181633d50815d79481eg-08.jpg?height=609&width=688&top_left_y=570&top_left_x=685)

Second solution. All coins except the central one can be divided into 9 triplets (see figure), and all internal coins except the central one can be divided into 3 triplets (see figure). Therefore, the coins on the boundary weigh as much as 9-3=6 triplets, i.e., 60 g.
![](https://cdn.mathpix.com/cropped/2024_05_06_7181633d50815d79481eg-08.jpg?height=1278&width=688&top_left_y=1522&top_left_x=685)

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