What's down?

The Right Circular Cone unfolds into a Circular Arc of radius $R = 60$

The shortest track $AB$ is a straight line on this arc.

When a perpendicular is dropped from Vertex to $AB$ , the length towards $B$ , $(x)$ is the Downhill part of the track and towards $A$ , $(AB - x)$ is the Uphill part.

Arc lenth = Circumference of base of cone

$60(\theta) = 2π(20) \Rightarrow \theta =\dfrac {2π}{3}$

$\theta=$ Angle of Arc.

Using cosine rule,

$AB^{2} = (50)^{2} + (60)^{2} -2(60)(50)\cos\dfrac{2π}{3}$

$AB^{2} = 9100 \Rightarrow \boxed{AB = 10\sqrt{91}}$

Using Phythagorus theorem,

$x^{2} + h^{2} = 50^{2}$

and $(10\sqrt{91} - x)^{2} + h^{2} = 60^{2}$

Solving, we get, $\boxed{x = \dfrac {400}{\sqrt{91}}}= \dfrac {20^{2}}{\sqrt{10 + 9^{2}}}$

$\boxed{a = 20 , b = 10 ,c = 9} \Rightarrow a + b + c = 20 + 10 + 9 =39$

$ANSWER:\boxed{39}$