鸭子与狐狸

能。

策略要点：

• 先让鸭子在离中心不远、半径略小于 $R/4$ 的圆上绕圈游，使得狐狸不得不沿岸追着“同步绕圈”。
• 由于狐狸的线速度是 4 倍，只有当鸭子在半径 $r<R/4$ 的圆上绕圈时，鸭子的角速度 $v/r$ 才能超过狐狸的角速度 $4v/R$，从而逐渐在角度上拉开相位差。
• 当鸭子与狐狸在湖岸的方位相差接近 $\pi$（对径）时，鸭子立刻改为朝最近岸边直线冲刺并上岸起飞。

关键不等式：选择 $r<R/4$ 使得 $\frac{v}{r} > \frac{4v}{R}$，从而“转得比狐狸快”。

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Original Explanation

At a radius of slightly less than $R/4$, the duck can swim in circles, forcing the fox to run around. Once the duck is at a phase of $\pi$ from the fox it starts swimming towards the shore and flies away.

Solution

This is a classic problem involving geometry and relative speed, often referred to as the "Duck and Fox Problem". Here's the approach:

1. Swim in Circles: The duck begins by swimming in a circle with a radius of $r < R/4$ where $R$ is the radius of the lake. The center of this circle is the same as that of the lake. The fox is four times as fast as the duck. By swimming in a small circle, the duck can start creating an angular separation between itself and the fox.

2. Force the Fox to Run: Since the fox is trying to chase the duck, it will continue to run around, trying to move to the point nearest to the duck but on the shore. As the duck goes around slightly faster, the fox lags behind. Eventually, the duck will be at a point where the fox is directly across the lake from it.

3. Wait for the Right Moment: The duck continues swimming in its smaller circle until the fox is straight across the lake from it. At this point, the duck and fox are separated by half the circumference of the lake, a distance that the fox has to cover to reach the duck if it heads straight for the shore.

4. Head for the Shore: While standing directly opposite, the duck can now swim straight to the shore. Duck has to travel around $R \cdot 3/4$ distance, while the fox has to travel $\pi R$. Given the ratio of their speeds, it will take $3 R$ and $3.14 \cdot R$ units of time respectively. Hence the duck will have a few extra moments to start the flight.

5. Escape: Reaching moments before the fox, the duck can now fly away to safety.

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Limits of $r$

Let us also formalize the limits of the choice of $r$.

It is slightly less than $R/4$, to create a phase lag

$r < R/4$ (to create a phase lag)

Also, in the end, the remaining time to reach the shore should be less for the duck, than the fox.

$\dfrac{(R-r)}{1v} < \dfrac{\pi R}{4v}$ (where $v$ is the speed.)

Solving these two equations gives:

${R} \cdot (1 - \dfrac{\pi}{4}) < r < \dfrac{R}{4}$

Hence, the inner circle can have a radius between $21.5\%$ to $25\%$ of $R$