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Difference between revisions of "2009 AIME II Problems/Problem 13"

(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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== 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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== 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
  
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<math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 + cos \frac {\pi}{7}) </math>\dots<math> (1 - cos \frac {3\pi}{7})(1 + cos \frac {3\pi}{7})</math>
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<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 22:34, 17 April 2009

Problem

Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$. The arc is divided into seven congruent arcs by six equally spaced points $C_1$, $C_2$, $\dots$, $C_6$. All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000$.


Solution

Let $O$ be the midpoint of $A$ and $B$. Assume $C_1$ is closer to $A$ instead of $B$. $\angle AOC_1$ = $\frac {\pi}{7}$. Using the Law of Cosines,

$\overline {AC_1}$ = $8 - 8 cos \frac {\pi}{7}$ $\overline {AC_2}$ = $8 - 8 cos \frac {2\pi}{7}$ . . . $\overline {AC_6}$ = $8 - 8 cos \frac {6\pi}{7}$

So $n$ = $(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {2\pi}{7})\dots(1 - cos \frac{6\pi}{7})$. It can be rearranged to form

$n$ = $(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {6\pi}{7})\dots(1 - cos \frac {3\pi}{7})(1 - cos \frac {2\pi}{4})$.

$cos a$ = - $cos (\pi - a)$, so we have

$n$ = $(8^6)(1 - cos \frac {\pi}{7})(1 + cos \frac {\pi}{7}) \dots (1 - cos \frac {3\pi}{7})(1 + cos \frac {3\pi}{7})$

= $(8^6)(1 - cos^2 \frac {\pi}{7})(1 - cos^2 \frac {2\pi}{7})(1 - cos^2 \frac {3\pi}{7})$

= $(8^6)(sin^2 \frac {\pi}{7})(sin^2 \frac {2\pi}{7})(sin^2 \frac {3\pi}{7})$

It can be shown that sin $\frac {\pi}{7}$ sin $\frac {2\pi}{7}$ sin $\frac {3\pi}{7}$ = $\frac {\sqrt {7}}{8}$, so $n$ = $8^6(\frac {\sqrt {7}}{8})^2$ = $7(8^4)$ = $28672$, so the answer is $\boxed {672}$

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