 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Random Choice (Posted on 2015-10-02) P and Q are roots of Z2015 = 1 chosen, at random, from the 2015 complex (including one real) roots of 1. The same root may or may not be chosen twice.

Determine the probability that |P+Q| ≥ √(2+√3)

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution Comment 1 of 1
Candidates for P and Q lie on the unit circle centered at the origin on the Argand plane, spaced equidistantly around the circle with one such root being at (1,0), i.e., 1 + 0i.

The sum of two such numbers is the diagonal of a parallelogram as a vector sum on the plane, with the origin to each of the points taken as two of the sides. The diagonal is the one including the origin, and the length of the diagonal is the absolute value of the total.

WLOG, one of the points can be taken as (1,0). If the other vector makes angle theta with that of (1,0), the length of the diagonal can be found via the law of cosines:

L^2 = 1 + 1 - 2*cos(180° - theta)
= 2 + 2*cos(theta)

To find the most this can be to solve the puzzle, set

2 + sqrt(3) <= 2 + 2 * cos(theta)

cos(theta) >= sqrt(3)/2

theta <= 30°

This can be either above or below the vector for (1,0) for a total of 60° or 1/6 of the circle.

2015/6 = 335.8333...

We know it must be an odd number of points as it's centered on one of them: (1,0). So 335 of the roots satisfy the condition and the probability is 335/2015 = 67/403 ~=  .1662531017369727.

 Posted by Charlie on 2015-10-02 10:25:48 Please log in:

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