Difference between revisions of "2009 AIME II Problems/Problem 13"
Aimesolver (talk | contribs) (New page: '''Problem''' Let <math>A</math> and <math>B</math> be the endpoints of a semicircular arc of radius <math>2</math>. The arc is divided into seven congruent arcs by six equally spaced poi...) |
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+ | == Problem == | ||
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Let <math>A</math> and <math>B</math> be the endpoints of a semicircular arc of radius <math>2</math>. The arc is divided into seven congruent arcs by six equally spaced points <math>C_1</math>, <math>C_2</math>, <math>\dots</math>, <math>C_6</math>. All chords of the form <math>\overline {AC_i}</math> or <math>\overline {BC_i}</math> are drawn. Let <math>n</math> be the product of the lengths of these twelve chords. Find the remainder when <math>n</math> is divided by <math>1000</math>. | Let <math>A</math> and <math>B</math> be the endpoints of a semicircular arc of radius <math>2</math>. The arc is divided into seven congruent arcs by six equally spaced points <math>C_1</math>, <math>C_2</math>, <math>\dots</math>, <math>C_6</math>. All chords of the form <math>\overline {AC_i}</math> or <math>\overline {BC_i}</math> are drawn. Let <math>n</math> be the product of the lengths of these twelve chords. Find the remainder when <math>n</math> is divided by <math>1000</math>. | ||
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+ | == Solution == | ||
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Let <math>O</math> be the midpoint of <math>A</math> and <math>B</math>. Assume <math>C_1</math> is closer to <math>A</math> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], | Let <math>O</math> be the midpoint of <math>A</math> and <math>B</math>. Assume <math>C_1</math> is closer to <math>A</math> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], | ||
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<math>cos a</math> = - <math>cos (\pi - a)</math>, so we have | <math>cos a</math> = - <math>cos (\pi - a)</math>, so we have | ||
− | <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 + cos \frac {\pi}{7}) | + | <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 + cos \frac {\pi}{7}) \dots (1 - cos \frac {3\pi}{7})(1 + cos \frac {3\pi}{7})</math> |
= <math>(8^6)(1 - cos^2 \frac {\pi}{7})(1 - cos^2 \frac {2\pi}{7})(1 - cos^2 \frac {3\pi}{7})</math> | = <math>(8^6)(1 - cos^2 \frac {\pi}{7})(1 - cos^2 \frac {2\pi}{7})(1 - cos^2 \frac {3\pi}{7})</math> |
Revision as of 21:34, 17 April 2009
Problem
Let and be the endpoints of a semicircular arc of radius . The arc is divided into seven congruent arcs by six equally spaced points , , , . All chords of the form or are drawn. Let be the product of the lengths of these twelve chords. Find the remainder when is divided by .
Solution
Let be the midpoint of and . Assume is closer to instead of . = . Using the Law of Cosines,
= = . . . =
So = . It can be rearranged to form
= .
= - , so we have
=
=
=
It can be shown that sin sin sin = , so = = = , so the answer is