# A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m.

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A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m x 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

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Consider the straight line path APQC through the sand. Time taken to go from A to C via this path

Tsand = $\frac{AP+QC}{1}+\frac{PQ}{v}$

= $\frac{25\sqrt2+25\sqrt2}{1}+\frac{50\sqrt2}{v}$

= $50\sqrt2[\frac{1}{v}+1]$

The shortest path outside the sand will be ARC.

Tsand = $\frac{AR+AC}{1}S$

AR = $\sqrt{75^2+25^2}=\sqrt{75\times75\times25\times25}$

= 5 × $5\sqrt{9+1}=25\sqrt10m$

RC = AR =$\sqrt{75^2+25^2}=25\sqrt10m$

Toutside = 2AR = $2\times25\sqrt10=50\sqrt10 s$

For Tsand < Toutside

$50\sqrt2(\frac{1}{v}+1)<2\times25\sqrt10$

$\frac{2\sqrt2}{2}(\frac{1}{v}+1)=\sqrt10$

$(\frac{1}{v}+1)<\frac{2\sqrt10}{2\sqrt2}=\frac{\sqrt5}{2}\times2=\sqrt5$

$\frac{1}{v}<\frac{\sqrt5}{2}\times2-1$

$\frac{1}{v}<\sqrt5-1$

⇒ v> $\frac{1}{\sqrt5-1}$ ≈ 0.81 m/s

⇒ v> 0.81 m/s.