Problem 6.7. Eight paper squares $2 \times 2$ were laid out on the table one after another until a large square $4 \times 4$ was formed. The last square placed on the table was $E$. The image shows how the squares are visible: square $E$ is fully visible, and the other squares are partially visible. Which square was placed on the table third in order?

![](https://cdn.mathpix.com/cropped/2024_05_06_a7569897979ba1659de3g-18.jpg?height=366&width=385&top_left_y=899&top_left_x=534)
Answer: $G$.

Solution. It is clear that different squares were placed in different positions (otherwise, the one placed last would completely cover the one placed earlier, but we can see at least one cell from each square).

The positions of squares $A, C, G$, and $F$ can be determined immediately, as they contain corner cells.

Since $F$ does not cover the visible cell of square $H$, $H$ was placed after $F$. (Moreover, the position of square $H$ is now uniquely determined.)

Since $G$ covers the cell to the left of the visible cell of square $H$, $G$ was placed after $H$.

Since $G$ does not cover the visible cell of square $D$, $D$ was placed after $G$.

Since $A$ covers the cell above the visible cell of square $D$, $A$ was placed after $D$. (And the position of square $D$ is now also determined.)

Since $A$ does not cover the visible cell of square $B$, $B$ was placed after $A$.

Since $C$ covers the cell to the right of the visible cell of square $B$, $C$ was placed after $B$.

Square $E$ was placed last according to the condition.

Thus, the squares were placed in the order $F-H-G-D-A-B-C-E$, and the third one was $G$.