EPTMath Puzzle [num 17] - Page 2
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GreatFall
United States1061 Posts
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blankspace
United States292 Posts
+ Show Spoiler + A: cot(arctanx) = 1/x B: cos(arctanx) = 1/sqrt(1+x^2) We can get 1 from 0 with cos. We can get 1/sqrt(n) for all n by doing this: B(1) = 1/sqrt2, B(A(1/sqrt2)) = 1/sqrt3 and etc. We can get any sqrt(p/q). Suppose I wanted to get sqrt(23/14), well then if i could get sqrt(9/14) I'd be done (apply B and A). Then if I could get sqrt(14/9) I would be done (by A). Similarly if I could get sqrt(5/9), sqrt (9/5), sqrt (4/5), sqrt (5/4), sqrt(1/4) and then we done. It's pretty easy to see this process has to terminate into form sqrt(1/n), it's euclidean algorithm with 2 relatively prime numbers p,q. the old usamo problem did not have the cotangent key so 1/x was more annoying to get. | ||
]343[
United States10328 Posts
On August 19 2010 08:29 blankspace wrote: The solution idea is fairly simple but kinda annoying to write up (use strong induction) + Show Spoiler + A: cot(arctanx) = 1/x B: cos(arctanx) = 1/sqrt(1+x^2) We can get 1 from 0 with cos. We can get 1/sqrt(n) for all n by doing this: B(1) = 1/sqrt2, B(A(1/sqrt2)) = 1/sqrt3 and etc. We can get any sqrt(p/q). Suppose I wanted to get sqrt(23/14), well then if i could get sqrt(9/14) I'd be done (apply B and A). Then if I could get sqrt(14/9) I would be done (by A). Similarly if I could get sqrt(5/9), sqrt (9/5), sqrt (4/5), sqrt (5/4), sqrt(1/4) and then we done. It's pretty easy to see this process has to terminate into form sqrt(1/n), it's euclidean algorithm with 2 relatively prime numbers p,q. the old usamo problem did not have the cotangent key so 1/x was more annoying to get. rofl euclidean algorithm. i am a retard. (wait i even saw that, kind of, ish... -____-) | ||
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