A point $P$ is at the center of a square with a side length of $10 \mathrm{~cm}$. How many points on the edge of the square are $6 \mathrm{~cm}$ away from $P$?
(a) 1
(b) 2
(c) 4
(d) 6
(e) 8
The correct option is (e).

The points that are $6 \mathrm{~cm}$ away from point $P$ form a circle with center $P$ and radius $R=6$ $\mathrm{cm}$. If $d$ denotes the diagonal of the square, by the Pythagorean Theorem we have

$$
d=\sqrt{10^{2}+10^{2}}=\sqrt{2 \times 10^{2}}=10 \sqrt{2}
$$

The circle with radius $L / 2=5 \mathrm{~cm}$ touches the square at four points.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-048.jpg?height=431&width=440&top_left_y=1769&top_left_x=1393)

The circle with radius $D / 2$ touches the square at four points, specifically at the vertices of the square. We have $L=10, R=6$ and $D=10 \sqrt{2}$, therefore

$$
\underbrace{5}_{L / 2}<\underbrace{6}_{R}<\underbrace{5 \sqrt{2}}_{D / 2}
$$

(Observe that $1.2<\sqrt{2}, 5 \times 1.2<5 \times \sqrt{2}$ and, therefore, $6<5 \sqrt{2}$). Thus, the circle with radius $R=6$ is "between" the two circles with radii 5 and $5 \sqrt{2}$. Therefore, it intersects the square at eight points.