2009 AIME II Problems/Problem 13

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 $\boxed {672}$

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2009 AIME II Problems/Problem 13

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