There are two ants having same crawling speed. Ant 1 follows route 1 to reach finish point while the ant 2 follows path 2 (zig zag route 2) to reach finish point. Which ant will reach the finish point first?
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Answer:
Although solving the problem appears to require analytical techniques, it can actually be answered by a combinatorial argument by considering a variation in which the rope stretches suddenly and instantaneously each second rather than stretching continuously. Indeed, the problem is sometimes stated in these terms, and the following argument is a generalisation of one set out by Martin Gardner, originally in Scientific American and later reprinted.[1]
Consider a variation in which the rope stretches suddenly and instantaneously before each second, so that the target-point moves from {\displaystyle x=c}x=c to {\displaystyle x=c+v}{\displaystyle x=c+v} at time {\displaystyle t=0}t=0, and from {\displaystyle x=c+v}{\displaystyle x=c+v} to {\displaystyle x=c+2v}{\displaystyle x=c+2v} at time {\displaystyle t=1}t=1, etc. Many versions of the problem have the rope stretch at the end of each second, but by having the rope stretch before each second we have disadvantaged the ant in its goal, so we can be sure that if the ant can reach the target-point in this variation then it certainly can in the original problem or indeed in variants where the rope stretches at the end of each second.
Let {\displaystyle \theta (t)}\theta (t) be the proportion of the distance from the starting-point to the target-point which the ant has covered at time t. So {\displaystyle \theta (0)=0}\theta(0)=0. In the first second the ant travels distance {\displaystyle \alpha }\alpha , which is {\displaystyle {\frac {\alpha }{c+v}}}{\displaystyle {\frac {\alpha }{c+v}}} of the distance from the starting-point to the target-point (which is {\displaystyle c+v}c+v throughout the first second). When the rope stretches suddenly and instantaneously, {\displaystyle \theta (t)}\theta (t) remains unchanged, because the ant moves along with the rubber where it is at that moment. So {\displaystyle \theta (1)={\frac {\alpha }{c+v}}}{\displaystyle \theta (1)={\frac {\alpha }{c+v}}}. In the next second the ant travels distance {\displaystyle \alpha }\alpha again, which is {\displaystyle {\frac {\alpha }{c+2v}}}{\displaystyle {\frac {\alpha }{c+2v}}} of the distance from the starting-point to the target-point (which is {\displaystyle c+2v}{\displaystyle c+2v} throughout that second). So {\displaystyle \theta (2)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}}{\displaystyle \theta (2)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}}. Similarly, for any {\displaystyle n\in \mathbb {N} }n\in \mathbb {N} , {\displaystyle \theta (n)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}+\cdots +{\frac {\alpha }{c+nv}}}{\displaystyle \theta (n)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}+\cdots +{\frac {\alpha }{c+nv}}}.
Notice that for any {\displaystyle k\in \mathbb {N} }{\displaystyle k\in \mathbb {N} }, {\displaystyle {\frac {\alpha }{c+kv}}\geqslant {\frac {\alpha }{kc+kv}}=\left({\frac {\alpha }{c+v}}\right)\left({\frac {1}{k}}\right)}{\displaystyle {\frac {\alpha }{c+kv}}\geqslant {\frac {\alpha }{kc+kv}}=\left({\frac {\alpha }{c+v}}\right)\left({\frac {1}{k}}\right)}, so we can write
{\displaystyle \theta (n)\geqslant \left({\frac {\alpha }{c+v}}\right)\left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)}{\displaystyle \theta (n)\geqslant \left({\frac {\alpha }{c+v}}\right)\left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)}.
The term {\displaystyle \left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)}{\displaystyle \left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)} is a partial Harmonic series, which diverges, so we can find {\displaystyle N\in \mathbb {N} }N\in {\mathbb {N}} such that {\displaystyle 1+{\frac {1}{2}}+\cdots +{\frac {1}{N}}\geqslant {\frac {c+v}{\alpha }}}{\displaystyle 1+{\frac {1}{2}}+\cdots +{\frac {1}{N}}\geqslant {\frac {c+v}{\alpha }}}, which means that {\displaystyle \theta (N)\geqslant 1}{\displaystyle \theta (N)\geqslant 1}.
Therefore, given sufficient time, the ant will complete the journey to the target-point. This solution could be used to obtain an upper-bound for the time required, but does not give an exact answer for the time it will take.
Explanation:
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