## 24. Walking the Dog

Monsieur and Madame Dubois went for a walk with their dog. Each of them wanted to hold the leash themselves. Therefore, in the end, they attached two leashes, each 1 m long, to the poor animal's collar. Suppose Monsieur and Madame Dubois always walk 1 m apart from each other.

What is the area of the region where the dog can freely run at any moment in time?
24. Let $O$ and $O^{\prime}$ be the points where Monsieur and Madame Dubois are located at a certain moment in time. If the dog were held on a leash by only Monsieur Dubois, the area of the region would be $\pi$ (the area of a circle with a radius of 1 m). The same would be true if the dog were held by only Madame Dubois.

However, in reality, the area where the dog can move coincides with the intersection of these two circles with a radius of 1 m, whose centers are

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-165.jpg?height=557&width=814&top_left_y=470&top_left_x=632)

1 m apart from each other. Let $A$ and $B$ be the points of intersection of the corresponding circles (with centers $O$ and $O^{\prime}$ and radii of 1 m). Then the area of the rhombus $O A O^{\prime} B$ is 2 (the area of $\left.O A O^{\prime}\right)=\sqrt{3} / 2$.

The area of the sector centered at $O$, bounded by points $A$ and $B$, is $\pi / 3$, and the area of the sector centered at $O^{\prime}$, bounded by points $A$ and $B$, is $\pi / 3$.

Therefore, the area of the intersection of our two circles is

$$
\pi / 3 + \pi / 3 - \sqrt{3} / 2 \approx 2.094 - 0.866 = 1.228
$$