Puzzle time - three sticks
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You have three sticks that can't make a triangle; that is, one is longer than the sum of the lengths of the other two. You shorten the long one by an amount equal to the sum of the lengths of the other two, so you again have three sticks. If they also fail to make a triangle, you again shorten the longest stick by an amount equal to the sum of the lengths of the other two.
You repeat this operation until the sticks do make a triangle, or the long stick disappears entirely.
Can this process go on forever?When you shake someone's deep sense of self-worth, leaving a scar in their ego, they will persistently try to diminish you through petty arguments & obsessive captious criticism of your actions, just to soothe the wound.
- Nassim Nicholas Taleb
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@jon-nyc said in Puzzle time - three sticks:
Can this process go on forever?
Don't be bloody stupid, you'd run out of stick. You can't make infinitesimally small pieces out of wood with a bloody kitchen knife.
Fucking financial guys, honestly.
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If we take the limit case that one length is zero, then the process does go on forever and you'd get a kind of geometric series of length deltas.
Which means that you can at least make the process go on for any fixed number of iterations simply by choosing one length that is close enough to zero.
But I assume your question is whether the process can go on forever with all lengths nonzero. Of course you do.
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Are they colorblind?
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The sticks are clearly a metaphor for the rods in our eyes, responsible for distinguishing colors. This 'puzzle' is a dog whistle for color blind hatred. Again. Oh, and dogs? Color blind. Man's best friend. That's what this puzzle hates. Real nice. Hating man's best friend.
Let's please stop the hate.
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Official solution.
If the sticks were of integer lengths, the process would have to end, since the sticks would remain integer-length and positive integers can't shrink forever. The same applies to rational lengths, since you can convert fractions to whole numbers by rescaling.
But irrational lengths are another matter. You can make the process go on indefinitely if you design it so that the ratios of the lengths of the sticks are unchanged after one operation. If you start with lengths a, b, and c, with a < b < c, the new lengths will be c-a-b, a, and b; thus, to keep the proportions the same, you'll need a/(c − a − b) = b/a = c/b. Let that ratio be r; then the lengths are proportional to 1, r, r2 and r satisfies 1/(r2 − r − 1) = r, i.e., r3 − r2 − r − 1 = 0.
This is a cubic equation, which (in theory) we could solve, but for the puzzle all we really need to know is whether there is a solution with r > 1. Writing f(r) = r3 − r2 − r − 1, we note that f(1) = −2 < 0 while f(2) = 1 > 0; since all polynomials are continuous functions, the Intermediate Value Theorem allows us to deduce that there is a root of f between 1 and 2.
It remains only to check that our root gives us sticks that don't make a triangle, but that's easy, because r2 = r3/r = (r2 + r + 1)/r = r + 1 + (1/r) > r + 1.[This puzzle, composed by A. V. Shapovalov, became Problem D4 of the Moscow Mathematical Olympiad of 2000.]
