Here is my proof that no matter how much the rate of burning varies over the time interval $$ as long as the time to burn when lit at one end is 1 hour then lighting the rope simultaneously at both ends will consume the rope after 1/2 hour.įormally, the amount the rope (lit at one end) that burns in time $a$ is $\int_^1 r(t)dt$ (using substitution rule for definite integrals). When we start burning the remaining rope, rope C from the other end, the same argument for rope A gives the time for complete burning as 0.25 hour. Then it can be seen that integrating g(x) from b to B, which is time needed to burn remaining the rope, which we can call rope C, is half an hour. We know that integrating g(x) from 0 to b gives half an hour, and integrating g(x) from 0 to B gives 1 hour. After burning for 0.5 hour from one end, let position of fire be b.
Next, for rope B, label one end 0 and the other end B and define the corresponding function for derivate of time w.r.t. Hence 2t, which is twice the integral of f(x) from 0 to a = twice the integral of f(x) from a to A = integral of f(x) from 0 to a + integral of f(x) from a to A = integral of f(x) from 0 to A = 1 hour. Since time passed for both processes of burning from each end is the same, integrating f(x) from 0 to a, and integrating f'(x) from A to a (which is the same as integrating f(x) from a to A, as established earlier) gives time t. Now, if rope A burns from both ends, let time needed for complete burning be t hours and meeting point be a. Since rope A takes 1h to burn from either end, integrating f(x) from 0 to A, and integrating f'(x) from A to 0 gives time of an hour. The corresponding function starting from end A is f'(x). Starting from end 0, let derivative of time passed w.r.t. When they're done burning, at time $T/2$, you'll know where the cut would have been.įor the first rope, label one end 0 and the other end A. So you don't need to know where to make the cut you just light both ends. But you don't actually have to make the cut in order to start burning the other portion - you can light it from the other end. If you imagine cutting the rope at exactly that point, the other portion will also burn in time $T/2$. Your mistake is thinking that's the only way - i.e., instead of "one would need to," it's "one could." But here's another way to do it: Suppose you knew, or somehow guessed, exactly how far the rope would burn in time $T/2$ if lit from one end (say the left). That's certainly one way to arrive at a rope that burns in time $T/2$. As soon as it's done burning, you light the next one at both ends, and so forth. Then you light the first one at both ends.
If I understand what you mean here, the idea is that you cut the rope into short segments, each one of (approximately) uniform density. You say "Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. If we look at each $T_i/2$ this is precisely the time it takes to burn the uniform segment $T_i$ if lit at both ends. Now, consider a rope of non-uniform density, the total time T for this rope to burn is the linear combination of the times of the uniform density "chunks" to burn, i.e. Given a rope of uniform density the burn rate at either end is equal so clearly it burns in time $T/2$. I think I have sketched a proof by induction that shows that it's not actually true. Now I usually love brain teasers but this one frustrated me for a while because I could not prove that if a rope of non-uniform density is burned at both ends it burns in time $T/2$. Now you are asked to measure a period of 45 minutes. The rate at which the two ropes burn is not necessarily the same, so the second rope will also take an hour to burn from one end to the other, but may do it at some varying rate, which is not necessarily the same as the one for the first rope. In other words, half the rope may burn in the first five minutes, and then the other half would take 55 minutes. But it doesn't have to burn at a uniform rate. You are told that each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn all the way to the other end. There is a common brain teaser that goes like this:
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