**
shortest
**
track that starts at point A and ends at point B, the track will start going uphill, but later going downhill. How long is the downhill track?

$\frac{200}{\sqrt{19}}$
$\frac{300}{\sqrt{91}}$
$\frac{400}{\sqrt{91}}$
$\frac{300}{\sqrt{30}}$

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Let the vertex of the cone be $C$ . Cut open the cone along a straight line from $A$ to $C$ via $B$ . Place the cone surface flat. Since $C$ is equidistance, $60$ , from the base circle $AA'$ , the flat surface area is a circle sector of radius $60$ . As the base circle has a circumference of $2\pi \times 20 = 40\pi$ . The angle of the sector $\angle ACA' = \frac {40\pi}{60} = \frac {2\pi}3 \ \rm rad$ or $120^\circ$ .

The shortest track between $A$ and $B$ is the straight line joining $A$ and $B$ on the flat surface. Let any point of $AB$ be $P$ . The shorter the length $CP$ , the higher the point $P$ on the track. Initially the length of $CP$ is decreasing as the track is going uphill, until $CP$ is perpendicular to $AB$ . There, the track has reached its highest point and after that it goes downhill. So we need to find th lenght of $BP$ when $CP$ is perpendicular to $AB$ .

Since $A'C = 60$ and $A'B = 10$ , $BC=50$ . By cosine rule ,

$\begin{aligned} AB^2 & = BC^2 + CA^2 - 2\cdot BC \cdot CA \cdot \cos 120^\circ \\ & = 50^2 + 60^2 - 2 \cdot 50 \cdot 60 \left(-\frac 12\right) \\ \implies AB & = 10 \sqrt{91} \end{aligned}$

By sine rule ,

$\begin{aligned} \frac {\sin B}{CA} & = \frac {\sin C}{AB} \\ \sin B & = \frac {\sin C}{AB} \cdot CA = \frac {60\sqrt 3}{2\cdot 10 \sqrt{91}} = \sqrt{\frac {27}{91}} \\ \implies BP & = BC \cdot \cos B = 50 \sqrt{1-\frac {27}{91}} = \boxed{\frac {400}{\sqrt{91}}} \end{aligned}$