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Given an equilateral triangle with side length 100. In each move, you can change the one of the side length of the triangle, but the resulting side length must still be able to form a triangle with the other 2 sides. What is the minimum number of moves needed to form an equilateral triangle of side length 1?

My thoughts and attempt: Clearly, after each move, the triangle inequality should be satisfied: the sum of any two side lengths must be larger than the remaining side length. And to reduce the number of moves, instead of forcibly reducing the length of one of the sides immediately, instead we consider the reduction of length of two side cumulatively for each move. The 3 side lengths were 100, 100, 100. So let's take one of the 100. We change the other two to become 50 and 51, that is 2 moves. Now, consider 51, since $26+26>51$ we will change the largest and smallest length (in order) to 26. Hence another 2 moves. Then I repeat this. However, this seems to fail as the side lengths get smaller. Also, when I changed the side lengths, I strictly changed it into integers. (Although there were no such restrictions)

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    $\begingroup$ If you try to go backwards, we see immediately that we will need to work with more than just the integers because we can't have a triangle with sides length $1,1,2$ $\endgroup$ – WaveX Mar 2 '18 at 6:06
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Going backwards (as per the cute comment above), and supposing that we can stretch one side to the limit $a+b=c$, then the progression will be $(1,1,1),(1,1,2), (1,2,3), (2,3,5), ..(F_{n-2},F_{n-1},F{n})$, till $F_n < 100 < F_{n+1}$.
After that we need $3$ additional moves.

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