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Difference between revisions of "Mock AIME 1 2005-2006/Problem 1"

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== Solution ==
 
== Solution ==
  
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Number the points <math>p_1</math>, <math>p_2</math>, <math>\dots</math>, <math>p_2006</math>. Assume the center is <math>O</math> and the given point is <math>p_1</math>. Then <math>\anglep_nOp_(n+1)</math> = <math>\frac {\pi}{1003}</math>, and we need to find the maximum <math>n</math> such that <math>\anglep_1Op_n+1 \le 60</math> degrees (<math>n+1</math> is given so that there are <math>n</math> repetitions of <math>\frac {pi}{1003}</math>). This can be done in <math>\frac {\frac {\pi}{3}}{\frac {\pi}{1003}</math> = <math>\frac {1003}{3} = </math>334.333\dots<math>, so </math>n<math> + </math>1<math> = </math>335<math>. We can choose </math>p_2<math>, </math>p_3<math>, \dots, </math>p_335<math>, so </math>334<math> points. But we need to multiply by </math>2<math> to count the number of points on the other side of </math>p_1$, so the answer is \boxed{668}.
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Number the points <math>p_1</math>, <math>p_2</math>, <math>\dots</math>, <math>p_2006</math>. Assume the center is <math>O</math> and the given point is <math>p_1</math>. Then <math>\anglep_nOp_(n+1)</math> = <math>\frac {\pi}{1003}</math>, and we need to find the maximum <math>n</math> such that <math>\anglep_1Op_n+1 \le 60</math> degrees (<math>n+1</math> is given so that there are <math>n</math> repetitions of <math>\frac {\pi}{1003}</math>). This can be done in <math>\frac {\pi}{3}</math> divided by <math>\frac {\pi}{1003} = </math>\frac {1003}{3} = <math>334.333\dots</math>, so <math>n</math> + <math>1</math> = <math>335</math>. We can choose <math>p_2</math>, <math>p_3</math>, \dots, <math>p_335</math>, so <math>334</math> points. But we need to multiply by <math>2</math> to count the number of points on the other side of <math>p_1</math>, so the answer is \boxed{668}.

Revision as of 20:26, 17 April 2009

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 $\anglep_nOp_(n+1)$ (Error compiling LaTeX. Unknown error_msg) = $\frac {\pi}{1003}$, and we need to find the maximum $n$ such that $\anglep_1Op_n+1 \le 60$ (Error compiling LaTeX. Unknown error_msg) degrees ($n+1$ is given so that there are $n$ repetitions of $\frac {\pi}{1003}$). 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}.

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