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Difference between revisions of "2009 AIME II Problems/Problem 13"
(→‎Solution 2)
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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]],  
  
−
<math>\overline {AC_1}^2</math> = <math>8 - 8 cos \frac {\pi}{7}</math>,   
+
<math>\overline {AC_1}^2</math> = <math>8 - 8 \cos \frac {\pi}{7}</math>,   
−
<math>\overline {AC_2}^2</math> = <math>8 - 8 cos \frac {2\pi}{7}</math>,  
+
<math>\overline {AC_2}^2</math> = <math>8 - 8 \cos \frac {2\pi}{7}</math>,  
 
.
 
.
 
.
 
.
 
.
 
.
−
<math>\overline {AC_6}^2</math> = <math>8 - 8 cos \frac {6\pi}{7}</math>                                 
+
<math>\overline {AC_6}^2</math> = <math>8 - 8 \cos \frac {6\pi}{7}</math>                                 
  
−
So <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {2\pi}{7})\dots(1 - cos \frac{6\pi}{7})</math>. It can be rearranged to form
+
So <math>n</math> = <math>(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {2\pi}{7})\dots(1 - \cos \frac{6\pi}{7})</math>. It can be rearranged to form
  
−
<math>n</math> = <math>(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})</math>.  
+
<math>n</math> = <math>(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})</math>.  
  
−
<math>cos a</math> = - <math>cos (\pi - a)</math>, so we have
+
<math>\cos a</math> = - <math>\cos (\pi - a)</math>, so we have
  
−
<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>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>
  
−
= <math>(8^6)(sin^2 \frac {\pi}{7})(sin^2 \frac {2\pi}{7})(sin^2 \frac {3\pi}{7})</math>
+
= <math>(8^6)(\sin^2 \frac {\pi}{7})(\sin^2 \frac {2\pi}{7})(\sin^2 \frac {3\pi}{7})</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>
+
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>
  
 
=== Solution 3 ===
 
=== Solution 3 ===

Revision as of 10:09, 9 March 2014

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 the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\ldots, C_6$ are 6 of the 14th roots of unity. Let $\omega=\text{cis}\frac{360^{\circ}}{14}$; then $C_1,\ldots, C_6$ correspond to $\omega,\ldots, \omega^6$. Let $C_1',\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\omega^8\ldots, \omega^{13}$. Then the lengths of the segments are $|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|$. Noting that $B$ represents 1 in the complex plane, the desired product is $\begin{align*} BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6\\ &= |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

for $x=1$. However, the polynomial $(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$. Hence \[(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})=\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\cdots +x^2+1.\] Thus the product is $|x^{12}+\cdots +x^2+1|=7$ ($x=1$) when the radius is 1, and the product is $2^{12}7=28672$. Thus the answer is $\boxed {672}$.

Solution 2

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 3

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 $\frac {\sqrt {7}}{8}$, which is what we needed to find.

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

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