339. Three Greek Crosses from One. In the figure shown here, you see an elegant solution to the problem of cutting two smaller Greek crosses of the same shape from a larger symmetrical Greek cross. Part $A$ is cut out as a whole, and it is not difficult to assemble a similar cross from the remaining 4 parts.

However, here is a more challenging question: how can you obtain three crosses of the same shape but smaller size from one large Greek cross, by cutting the large cross into the smallest possible number of pieces?

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-107.jpg?height=360&width=374&top_left_y=331&top_left_x=841)

Note that this problem can be solved using only 13 pieces. I believe many readers who are adept at geometry will enjoy pondering this problem. Of course, all three crosses must be of the same size.
339. Cut off the top and bottom parts of the cross and place them in positions \(A\) and \(B\) (case \(I\)), and cut the remaining larger part into 3 pieces so that the 5 pieces obtained can be used to form the rectangle shown in case II. It can be said that this rectangle is composed of 15 squares - 5 squares for each new cross. The remaining cuts are not difficult to make. From parts \(2,5,8,9\) a cross is clearly formed; from parts 13,6, 10,7, and 11 - the second cross (case \(I I I\)), and from \(1,3,4,12\) the third cross (case \(I V)\). The area of each end of the small cross is \(\frac{1}{3}\) of the area of any end of the large cross.
![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-265.jpg?height=744&width=1194&top_left_y=1595&top_left_x=434)

[The number of parts can be reduced to \(12 .-M \Gamma\).]