Difference between revisions of "Mock AIME 1 2005-2006/Problem 1"
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== Solution == | == Solution == | ||
− | Number the points <math>p_1</math>, <math>p_2</math>, <math>\dots</math>, <math> | + | 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>\angle p_nOp_{n+1}</math> = <math>\frac {\pi}{1003}</math>, and we need to find the maximum <math>n</math> such that <math>\angle p_1Op_{n+1} \le 60</math> degrees. This can be done in <math>\frac {\pi}{3}</math> divided by <math>\frac {\pi}{1003}</math> = <math>\frac {1003}{3}</math> = <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>, <math>\dots</math>, <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 <math>\boxed{668}</math>. |
Latest revision as of 21:32, 17 April 2009
Problem 1
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 ,
,
,
. Assume the center is
and the given point is
. Then
=
, and we need to find the maximum
such that
degrees. This can be done in
divided by
=
=
, so
+
=
. We can choose
,
,
,
, so
points. But we need to multiply by
to count the number of points on the other side of
, so the answer is
.