Rabbit and Fox Riddle: Maximum Safe Distance for a Rabbit | Solution

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The riddle involves determining the maximum safe distance a rabbit can venture from its hole while being chased by a fox that runs twice as fast. The rabbit can detect the fox at 60 meters and must reach its hole to escape. The initial solution proposed was incorrect because it did not account for the relative positions of the rabbit, fox, and hole. The correct understanding is that the rabbit must remain twice as close to the hole as the fox is at all times. A safe distance of 20 meters from the hole ensures the fox remains at least 40 meters away, allowing the rabbit to escape.
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Homework Statement



Through a combination of sight, sound, and smell a rabbit can detect a fox at 60 meters. When the rabbit detects the fox, it runs immediately straight to its hole. If the fox can run twice as fast as the rabbit, what is the maximum distance from its hole that a rabbit can safely venture? Assume that the rabbit escapes if they arrive at the hole at the same time.

2. The attempt at a solution

2x + 60 = x
x = 60

Apparently this is incorrect
 
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Does the question state where the hole is compared to the fox and rabbit. Your result is correct if the rabbit is directly between the fox and the hole. but what if the hole was between the fox and rabbit?
 
I think I understand now. It would always have to be twice as close to the hole than the fox. A twenty meter radius would ensure the fox is always 40 meters away from the hole.
 
Last edited:
some_letters said:
I think I understand now. It would always have to be twice as close to the hole then the fox. A twenty meter radius would ensure the fox is always 40 meters away from the hole.
Is "twice as close" the same as "half as far" ?
 
I think so. why do you ask?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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