Two Fuses Solution

Light both ends of one fuse and one end of the second. When the one you lit at both ends burns out, light the other end of the second one. When the second fuse burns out 45 minutes will have gone by.

In order to understand the solution to this problem, one must be able to prove that it takes half the time for a fuse to burn when lit at both ends. This is what I will show.
Show that the burning time for a fuse with an unknown position dependent rate is halved when it is lit from both ends:
Consider a fuse of length L with a burn time T.
After lighting one end and waiting a time T the whole length L will have burned.
Call one end of this fuse LHS and the other RHS
(for right and left hand sides . . .)
If I light the LHS and RHS at the same time the fuse will burn until the whole fuse is burned up. Let the spot where the burning stops (the point where the left hand side burn meets the right hand side burn) be some point X. This point X is determined by the burning rate of the fuse, for different fuses it will be in different locations. (ex. For a linear fuse the point X would be the middle) Define the burn time for the LHS and RHS as the time from when the fuse’s ends are lit, to the time the fuse stops burning. Obviously these times are equal for both ends.
Call the LHS burn time TL and the RHS burn time TR.
In the time TL = TR the whole fuse burns. If I were to have only lit the LHS the fuse would have kept burning past X and gone right to the RHS. It is not hard to se that the total burn time for the fuse is TL + TR = T. But since TL = TR, 2TL(or R) = T, so TL(or R) = 1/2 T. This is what I wanted to show.

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