Mock AIME 1 2005-2006/Problem 1

Revision as of 21:28, 17 April 2009 by Aimesolver (talk | contribs) (Solution)

Problem 1

$2006$ points are evenly spaced on a circle. Given one point, find the maximum number of points that are less than one radius distance away from that point.

Solution

Number the points $p_1$, $p_2$, $\dots$, $p_2006$. Assume the center is $O$ and the given point is $p_1$. Then $\angle p_nOp_(n+1)$ = $\frac {\pi}{1003}$, and we need to find the maximum $n$ such that $\angle p_1Op_n+1 \le 60$ degrees. This can be done in $\frac {\pi}{3}$ divided by $\frac {\pi}{1003} =$\frac {1003}{3}$=$334.333\dots$, so$n$+$1$=$335$. We can choose$p_2$,$p_3$, \dots,$p_335$, so$334$points. But we need to multiply by$2$to count the number of points on the other side of$p_1$, so the answer is \boxed{668}.