Q.A man wants to travel from corner A of a square field to the diagonally opposite corner C. The square has side $100\ \text{m}$. A smaller square of side $50\ \text{m}$ sits at the centre of the field (so its edges are $25\ \text{m}$ from each side of the big square) and is filled with sand; its diagonal lies along the diagonal AC. Outside this central sand square the man walks at $1\ \text{m/s}$; inside the sand he can walk only at speed $v\ \text{m/s}$, with $v < 1$. Find the smallest value of $v$ for which travelling along the straight diagonal path from A to C (which cuts through the sand) is faster than the quickest path that stays entirely outside the sand.

Geometry

Put $A=(0,0)$, $B=(100,0)$, $C=(100,100)$, $D=(0,100)$. The central sand square runs from $(25,25)$ to $(75,75)$. The diagonal AC is the line $y=x$; it enters the sand at $(25,25)$ and leaves at $(75,75)$.

Time along the straight diagonal (through the sand)

• Total diagonal length: $AC = 100\sqrt2$ m.
• Portion inside the sand: from $(25,25)$ to $(75,75)$, length $50\sqrt2$ m, walked at $v$.
• Portion outside the sand: the two end segments, $25\sqrt2 + 25\sqrt2 = 50\sqrt2$ m, walked at $1\ \text{m/s}$.

Time along the best path that avoids the sand

The shortest route from A to C that does not cross the central square is a taut path bending around one of its near corners, e.g. $(75,25)$ (the corner $(25,75)$ gives the same length by symmetry):

Each leg has length $\sqrt{75^2+25^2} = \sqrt{6250} = 25\sqrt{10}$ m, so the total is $50\sqrt{10}$ m, all at $1\ \text{m/s}$:

Condition for the straight path to be faster …

Concept: Find the Critical Speed by Solving an Equality First, Then Justify Uniqueness via Monotonicity

Method: Solve $t_{\text{straight}}(v)=t_{\text{around}}$ as a Root-Finding Problem, Then Argue Uniqueness

The direct method manipulates the inequality $t_{\text{straight}}<t_{\text{around}}$ directly to isolate $v$. This method instead treats $t_{\text{straight}}(v)$ as a function of $v$, observes that it is strictly decreasing, solves for the single crossover value $v^\ast$ where the two times are exactly equal, and argues -- from the monotonicity alone -- that $v>v^\ast$ is exactly the condition under which the straight path wins.

Steps

1. Compute both fixed distances, unchanged from the direct method: the diagonal path is $100\sqrt2\text{ m}$ total, split $50\sqrt2\text{ m}$ outside the sand and $50\sqrt2\text{ m}$ inside; the shortest sand-avoiding route is $2\times25\sqrt{10}=50\sqrt{10}\text{ m}$, entirely at $1\text{ m/s}$.

2. Write $t_{\text{straight}}$ explicitly as a function of $v$, and note $t_{\text{around}}$ is a fixed constant (it never touches the sand, so it doesn't depend on $v$ at all):

3. Observe that $t_{\text{straight}}(v)$ is strictly decreasing in $v$ (since $\tfrac{d}{dv}[1/v]<0$ for $v>0$) -- physically obvious, but worth stating explicitly, since it's exactly what guarantees a unique crossover point exists.

4. Solve the equality $t_{\text{straight}}(v^\ast)=t_{\text{around}}$ for the single critical value $v^\ast$:

   $50\sqrt2\left(1+\frac1{v^\ast}\right) = 50\sqrt{10} \;\Rightarrow\; 1+\frac1{v^\ast} = \sqrt5 \;\Rightarrow\; \frac1{v^\ast}=\sqrt5-1 \;\Rightarrow\; v^\ast = \frac1{\sqrt5-1} = \frac{\sqrt5+1}{4}\approx0.81.$ …