2009 AIME II Problems/Problem 13

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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

Solution 1

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

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

$cos a$ = - $cos (\pi - a)$ , 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 $\boxed {672}$

Solution 2

Note that for each $k$ the triangle $ABC_k$ is a right triangle. Hence the product $AC_k \cdot BC_k$ is twice the area of the triangle $ABC_k$ . Knowing that $AB=4$ , the area of $ABC_k$ can also be expressed as $2c_k$ , where $c_k$ is the length of the altitude from $C_k$ onto $AB$ . Hence we have $AC_k \cdot BC_k = 4c_k$ .

By the definition of $C_k$ we obviously have $c_k = 2\sin\frac{k\pi}7$ .

From these two observations we get that the product we should compute is equal to $8^6 \cdot \prod_{k=1}^6 \sin \frac{k\pi}7$ , which is the same identity as in Solution 1.

Computing the product of sines

In this section we show one way how to evaluate the product $\prod_{k=1}^6 \sin \frac{k\pi}7$ .

Let $\omega_k = \cos \frac{2k\pi}7 + i\sin \frac{2k\pi}7$ . The numbers $1,\omega_1,\omega_2,\dots,\omega_6$ are the $7$ -th complex roots of unity. In other words, these are the roots of the polynomial $x^7-1$ . Then the numbers $\omega_1,\omega_2,\dots,\omega_6$ are the roots of the polynomial $\frac{x^7-1}{x-1} = x^6+x^5+\cdots+x+1$ .

We just proved the identity $\prod_{k=1}^6 (x - \omega_k) = x^6+x^5+\cdots+x+1$ . Substitute $x=1$ . The right hand side is obviously equal to $7$ . Let's now examine the left hand side. We have:

$\begin{align*}|1-\omega_k| & = \left| 1-\cos \frac{2k\pi}7 - i\sin \frac{2k\pi}7 \right| \\& = \sqrt{ \left( 1-\cos \...$

Therefore the size of the left hand side in our equation is $\prod_{k=1}^6 |1-\omega_k| = \prod_{k=1}^6 2 \sin \frac{k\pi}7 = 2^6 \prod_{k=1}^6 \sin \frac{k\pi}7$ . As the right hand side is $7$ , we get that $\prod_{k=1}^6 \sin \frac{k\pi}7 = \frac{7}{2^6}$ . However, since sin $x$ = sin $\pi - x$ , then $\prod_{k=1}^3 \sin \frac{k\pi}7$ would be the square root of $\frac {7}{2^6}$ , or $\frac {\sqrt {7}}{8}$ , which is what we needed to find.

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

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