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Question Number 43263 by MrW3 last updated on 09/Sep/18

	Answered by MJS last updated on 09/Sep/18

	$A B is a straight line on the mantle$ $in the plain$ $C = (\begin{matrix} 0 \\ 0 \end{matrix})$ $angle A C B = \frac{r}{s} 360 ° = \frac{20}{60} 360 ° = 120 °$ $circle A$ $x^{2} + y^{2} = 60^{2}$ $circle B$ $x^{2} + y^{2} = 50^{2}$ $A = (\begin{matrix} - 60 \\ 0 \end{matrix}) B = (\begin{matrix} 25 \\ 25 \sqrt{3} \end{matrix})$ $A B : y = \frac{5 \sqrt{3}}{17} x + \frac{300 \sqrt{3}}{17}$ $the turning point is T = (\begin{matrix} t \\ \frac{5 \sqrt{3}}{17} t + \frac{300 \sqrt{3}}{17} \end{matrix}) with$ $mimimal ∣ C T ∣$ $\frac{364 t^{2} + 9000 t + 270000}{189} \Rightarrow \min$ $\frac{d}{d t} = \frac{728}{189} t + \frac{9000}{189} = 0 \Rightarrow t = - \frac{1125}{91}$ $T = (\begin{matrix} - \frac{1125}{91} \\ \frac{1275 \sqrt{3}}{91} \end{matrix})$ $∣ A B ∣= 10 \sqrt{91} \approx 95.39$ $∣ A T ∣= \frac{510 \sqrt{91}}{91} \approx 53.46$ $∣ T B ∣= \frac{400 \sqrt{91}}{91} \approx 41.93$
	Commented by MrW3 last updated on 09/Sep/18

	$t h a n k y o u s i r! a n s w e r i s c o r r e c t .$
	Answered by MrW3 last updated on 09/Sep/18

	Commented by MrW3 last updated on 09/Sep/18

	$t h e s h o r t e s t p a t h i s t h e s t r a i g h t A B .$ $θ = \frac{40 π}{60} = \frac{2 π}{3} = 120 °$ $A B = \sqrt{60^{2} + 50^{2} - 2 \times 60 \times 50 \times \cos 120 °} = \sqrt{9100} = 10 \sqrt{91}$ $t h e h i g h e s t p o i n t i n t h e p a t h i s C w i t h$ $O C ⊥ A B . A C i s t h e u p h i l l t r a c k a n d C B$ $i s t h e d o w n h i l l t r a c k .$ $l e t B C = d, t h e n A C = A B - d = 10 \sqrt{91} - d$ $60^{2} - {(10 \sqrt{91} - d)}^{2} = 50^{2} - d^{2}$ $60^{2} - 9100 + 20 \sqrt{91} d = 50^{2}$ $\Rightarrow d = \frac{50^{2} + 9100 - 60^{2}}{20 \sqrt{91}} = \frac{400}{\sqrt{91}} \approx 41.93 m$
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