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

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It can be shown that <math>sin \frac {\pi}{7} sin \frac {2\pi}{7} sin \frac {3\pi}{7}</math> = <math>\frac {\sqrt {7}}{8}</math>, so <math>n</math> = <math>8^6(\frac {\sqrt {7}}{8})^2</math> = <math>7(8^4)</math> = <math>28672</math>, so the answer is <math>\boxed {672}</math>
 
It can be shown that <math>sin \frac {\pi}{7} sin \frac {2\pi}{7} sin \frac {3\pi}{7}</math> = <math>\frac {\sqrt {7}}{8}</math>, so <math>n</math> = <math>8^6(\frac {\sqrt {7}}{8})^2</math> = <math>7(8^4)</math> = <math>28672</math>, so the answer is <math>\boxed {672}</math>
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== See Also ==
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{{AIME box|year=2009|n=II|num-b=11|num-a=13}}

Revision as of 21:40, 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,

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

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 $\boxed {672}$

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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