A player on a computer controls a fox hunting two rabbits. At vertex $A$ of the square $A B C D$ there is a burrow: if at least one rabbit gets into it in the absence of the fox, the game is lost. The fox catches a rabbit as soon as it is in the same point (possibly at point $A$). Initially, the fox is at point $C$, and the rabbits are at points

$B$ and $D$. The fox runs everywhere with a speed not exceeding $v$, while the rabbits run along the rays $A B$ and $A D$ with a speed not exceeding 1. For what values of $v$ can the fox catch both rabbits?
Let's introduce a rectangular coordinate system, the origin of which coincides with point $A$, and points $B, C$, and $D$ have coordinates $(0,1), (1,1)$, and $(1,0)$ respectively.

We will specify such behavior of the rabbits that for certain values of $v$ it will certainly allow one of them to escape, and then we will determine these values of $v$. The distance between two lines given in our coordinate system by the equations $y=x+a$ and $y=x+b$ is $\frac{|a-b|}{\sqrt{2}}$. Therefore, the time it takes for the fox to move from one of these lines to another, each rabbit will be able to run a distance of $\sqrt[4-\sqrt{2}]{2}$. By the condition and due to the choice of the coordinate system, at the initial moment of time, the fox is on the line $y=x$. Therefore, at the moment when the fox is on the line $y=x+a$, the rabbits starting from points $B$ and $D$ will always be able to be at points with coordinates $\left(0,1+\frac{a}{v \sqrt{2}}\right)$ and $\left(1-\frac{a}{v \sqrt{2}}, 0\right)$ respectively (as long as these coordinates are non-negative). Let the rabbits follow this strategy. We will call these rabbits the first and the second. Without loss of generality, we can assume that the fox will catch the first rabbit first. This will happen at the point $\left(0,1+\frac{a}{2}\right)$, which belongs to the line $y=x+a$, i.e., when $1+\frac{a}{v \sqrt{2}}=a$. Finding $a$, we get that this point has an ordinate $v \sqrt{2} a=\frac{v \sqrt{2}}{\sqrt{2}-1}$. Acting according to the specified strategy, the second rabbit at this moment will be at a point with abscissa $1-\frac{a}{2 \sqrt{2}}=\frac{v \sqrt{2}-2}{\sqrt{2}-1}$ (if this abscissa turns out to be negative, the second rabbit will escape before the fox catches the first rabbit). If at the moment of catching the first rabbit, the ratio of the ordinate of the fox to the abscissa of the second rabbit is greater than $v$, then the second rabbit will escape by running to point $A$. This means that for $\frac{v \sqrt{2}}{\sqrt{2}-1} > v \cdot \frac{v \sqrt{2}-2}{\sqrt{2}-1}$, i.e., for $v < 1+\sqrt{2}$, the proposed behavior of the rabbits will certainly allow one of them to escape.

Now let $v \geqslant 1+\sqrt{2}$. We will specify the behavior of the fox, with which she will be able to catch both rabbits. Let her first run along the line $y=x$ to point $A$ at maximum speed. Since the fox will be able to reach point $A$ before each of the rabbits, there will certainly come a moment when the distance of the fox from this point will be equal to the smallest of the distances of each of the rabbits to point $A$. Without loss of generality, we will assume that the smallest will be the distance of the first rabbit from $A$, and the coordinates of the fox and the first rabbit at this moment will be $(a, a)$ and $(0, a \sqrt{2})$ respectively. Then, moving in a straight line at maximum speed first to the point with coordinates $(0,2a)$, and then to point $A$, the fox will catch the first rabbit and will be at point $A$ no later than the second (see fig.). Indeed, the distance from the point with coordinates $(0, a \sqrt{2})$ to the point with coordinates $(0,2a)$, equal to $a(2-\sqrt{2})$, is not less than $\frac{a \sqrt{2}}{v}$ - the distance from the point with coordinates $(a, a)$ to the point with coordinates $(0,2a)$ divided by $v$. This means the first rabbit will certainly be caught. The second rabbit will also not escape, as its distance to $A$ is not

![](https://cdn.mathpix.com/cropped/2024_05_06_7041536e9c9bacd5befeg-34.jpg?height=89&width=1004&top_left_y=1236&top_left_x=1)

## Answer

$v \geqslant 1+\sqrt{2}$.

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