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        riddles >> medium >> hard
        
(Message started by: tony123 on Jul 13^th, 2007, 2:59pm)

Title: hard
Post by tony123 on Jul 13^th, 2007, 2:59pm

http://www.mathramz.com/phpbb/latexrender/pictures/a5ffb80bf98c5c3833ad324d67f24c8d.png

Title: Re: hard
Post by pex on Aug 23^rd, 2007, 9:26am

Hard? Not at all.

More generally, let us prove for any positive integer n that
sqrt(2)sin[(1/2)arcsin{sqrt(1/n² + (n-1)/n³ + (n-1)/n⁴)}] = 1/n.

We divide by sqrt(2), take the arcsin, multiply by 2, and simplify the expressions of which arcsins are taken:

arcsin(sqrt((2n²-1)/n⁴)) = 2*arcsin(sqrt(1/(2n²))).

Recall sin(2x) = 2sin(x)cos(x). Thus,

sqrt((2n²-1)/n⁴) = 2 * sqrt(1/(2n²)) * sqrt(1-1/(2n²)),

since we are working in the first quadrant. Squaring,

(2n²-1)/n⁴ = 4 * 1/(2n²) * (2n²-1)/(2n²),

which is easily verified. Plugging in n = 2006 gives the desired result.

Title: Re: hard
Post by Sameer on Aug 23^rd, 2007, 9:38am

tony is incorrigible :-/

Title: Re: hard
Post by pex on Aug 23^rd, 2007, 9:48am

on 08/23/07 at 09:38:18, Sameer wrote:

tony is incorrigible :-/

Well, this is an old thread. The first request for more descriptive titles I could find (Aryabhatta's) dates from almost a full day after this puzzle was posted...

Edit: No! I found an earlier one here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1184086625) - made by yourself, which is probably not that much of a coincidence. So yes, he is (or at least was) incorrigible...

Title: Re: hard
Post by Barukh on Aug 23^rd, 2007, 11:48am

Nicely done, pex! :D