Problem 3. (4 points) On a table, a cardboard circle with a radius of 5 cm is placed. As long as possible, Peter places cardboard squares with a side of 5 cm outside the circle in such a way that the following conditions are met:

1) one vertex of each square lies on the boundary of the circle;
2) the squares do not overlap;
3) each new square shares a vertex with the one placed before it.

![](https://cdn.mathpix.com/cropped/2024_06_03_336fa2e3a24e7097f2a8g-1.jpg?height=363&width=333&top_left_y=1686&top_left_x=1465)

Find the number of squares Peter can place and prove that the first and the last of them also share a vertex.
Solution. Let the center of the circle be denoted by $O$. If vertex $A$ of the square $ABCD$ lies on the given circle, then points $B, D$, and $O$ lie on a circle with radius 5 cm and center $A$. The inscribed angle $\angle BOD = \frac{1}{2} \angle BAD = 45^{\circ}$. Therefore, each of the placed squares is seen from the center of the circle at an angle of $45^{\circ}$. Since the squares do not overlap and each has a common vertex with the one placed before it, Peter can place $360^{\circ} / 45^{\circ} = 8$ squares.

Let the square "attached" to vertex $D$ be denoted by $EDFG$ (point $E$ lies on the circle). In the rhombus $OADE$, we have $\angle OAD = \angle OED$. Therefore,

$$
\angle OAB = 360^{\circ} - 90^{\circ} - \angle OAD = 360^{\circ} - 90^{\circ} - \angle OED = \angle OEG
$$

Triangles $OAB$ and $OEG$ are congruent by the first criterion, so $OB = OG$.

We have found that in two squares with a common vertex, the opposite vertices are equidistant from $O$. Therefore, one vertex of the first square and one vertex of the eighth lie on the same ray with origin $O$ and are equidistant from $O$, i.e., they coincide.