Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Computing the product of sines)
(→‎Computing the product of sines)
Line 68: Line 68:
 
</cmath>
 
</cmath>
  
−
Therefore the size of the left hand side in our equation is <math>\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</math>. As the right hand side is <math>7</math>, we get that <math>\prod_{k=1}^6 \sin \frac{k\pi}7 = \frac{7}{2^6}</math>. However, since sin <math>x</math> = sin <math>\pi - x</math>, then  
+
Therefore the size of the left hand side in our equation is <math>\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</math>. As the right hand side is <math>7</math>, we get that <math>\prod_{k=1}^6 \sin \frac{k\pi}7 = \frac{7}{2^6}</math>. However, since <math>sin x</math> = <math>sin \pi - x</math>, then  
 
<math>\prod_{k=1}^3 \sin \frac{k\pi}7 </math> would be the square root of <math>\frac {7}{2^6}</math>, or <math>\frac {\sqrt {7}}{8}</math>.
 
<math>\prod_{k=1}^3 \sin \frac{k\pi}7 </math> would be the square root of <math>\frac {7}{2^6}</math>, or <math>\frac {\sqrt {7}}{8}</math>.
  

Revision as of 21:49, 20 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

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

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

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

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 \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 } \\ & = \sqrt{ 2-2\cos \frac{2k\pi}7 } \\ & = \sqrt{ 2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right) } \\ & = \sqrt{ 4\left( \sin \frac{k\pi}7 \right)^2 } \\ & = 2 \sin \frac{k\pi}7 \end{align*}

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 $\frac {\sqrt {7}}{8}$.

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_13&oldid=31412"