Consider ropes that take exactly 1 hour to burn. Note that these ropes may burn unevenly (i.e,. It may not take exactly 30 minutes to burn half the rope.) Given two ropes, how can you time $\frac{1}{2} + \frac{1}{2n}$ of an hour, for $n \in \mathbb{Z}^{+}$? Hint: It may help to first consider the widely-known versions of this problem: Given 1 such rope, how can you time 30 minutes? Given two such ropes, how can you time 45 minutes?

Hint Solutions

Given 1 rope, you can light it from both sides to time 30 minutes.

Given 2 ropes, you can light one rope $r_1$ from both ends and light the second rope $r_2$ from one end. Once $r_1$ has completely burned, 30 minutes has elapsed, and $r_2$ only has 30 minutes of burn time left. Light $r_2$ from the opposite end, so that the two remaining flames converge after another 15 minutes.

Solution

Take one rope $r_1$ and light it from both ends. At the same time, light the other rope $r_2$ from one end only. After 30 minutes, light $r_2$ in exactly $n-1$ additional places, chosen at random. Anytime two flames merge, simply re-light $r_2$ in two additional places. In this way, we ensure that $r_2$ is being actively consumed by $n$ flames. This effectively sub-divides $r_2$'s remaining 30 minutes of burn time $n$ ways, giving us $\frac{1}{2}\frac{1}{n}$ of additional time. In sum, we have $\frac{1}{2} + \frac{1}{2n}$ of an hour.

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