6. A wolf is surrounded by dogs located at points $M, N, P$ and $Q$ on the sides of a square $A B C D$, $M \in[A ; B], N \in[B ; C], P \in[C ; D], Q \in[D ; A]$ such that $A M: M B=B N: N C=C P: P D=D Q: Q A=1: 3$. The wolf, located inside the square at the intersection of lines $M P$ and $N Q$, can run in a straight line in any direction with a speed of $v_{w}$. The dogs run only along the sides of the square with a speed not exceeding $v_{c}$. The wolf can escape if it meets no more than one dog on the boundary of the square. For what values of the ratio $v_{c} / v_{w}$ does the wolf have a chance to escape?

## Answers and Solutions

Problem 1 Answer: $(x-2 y)_{\max }=\frac{25}{6}$
Solution

If $\varphi \in[0 ; 2 \pi]$, then the numbers $x=\frac{5}{2} \cos \varphi$ and $y=\frac{5}{3} \sin \varphi$ satisfy the equation. Then $x-2 y=\frac{5}{2} \cos \varphi-\frac{10}{3} \sin \varphi=\frac{5}{6}(3 \cos \varphi-4 \sin \varphi)=\frac{25}{6}\left(\frac{3}{5} \cos \varphi-\frac{4}{5} \sin \varphi\right)=$ $=\frac{25}{6}(\cos \theta \cos \varphi-\sin \theta \sin \varphi)=\frac{25}{6} \cos (\varphi+\theta)$
where $\theta=\arccos \frac{3}{5}$. Since the maximum possible value of $\cos (\varphi+\theta), \varphi \in[0 ; 2 \pi]$ is 1, then $(x-2 y)_{\max }=\frac{25}{6}$.