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"

m (→‎Solution 1)
(→‎Solution)
Line 8: Line 8:
  
 
=== Solution 1 ===
 
=== Solution 1 ===
 +
 +
Let the radius be 1 instead. All lengths will be halved so we will multiply by <math>2^{12}</math> 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 <math>C_1,\ldots, C_6</math> are 6 of the 14th roots of unity. Let <math>\omega=\text{cis}\frac{360^{\circ}}{14}</math>; then <math>C_1,\ldots, C_6</math> correspond to <math>\omega,\ldots, \omega^6</math>. Let <math>C_1',\ldots, C_6'</math> be their reflections across the diameter. These points correspond to <math>\omega^8\ldots, \omega^{13}</math>. Then the lengths of the segments are <math>|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|</math>. The product is
 +
\[
 +
|(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})|
 +
\]
 +
for <math>x=1</math>.
 +
However, the polynomial <math>(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})</math> has as its zeros all 14th roots of unity except for <math>-1</math> and <math>1</math>. 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 <math>|x^{12}+\cdots +x^2+1|=7</math> (<math>x=1</math>) when the radius is 1, and the product is <math>2^{12}7=28672</math>. Thus the answer is <math>\boxed {672}</math>.
 +
 +
=== Solution 2 ===
  
 
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]],  
Line 32: Line 45:
 
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 2 ===
+
=== Solution 3 ===
  
 
Note that for each <math>k</math> the triangle <math>ABC_k</math> is a right triangle. Hence the product <math>AC_k \cdot BC_k</math> is twice the area of the triangle <math>ABC_k</math>. Knowing that <math>AB=4</math>, the area of <math>ABC_k</math> can also be expressed as <math>2c_k</math>, where <math>c_k</math> is the length of the altitude from <math>C_k</math> onto <math>AB</math>. Hence we have <math>AC_k \cdot BC_k = 4c_k</math>.
 
Note that for each <math>k</math> the triangle <math>ABC_k</math> is a right triangle. Hence the product <math>AC_k \cdot BC_k</math> is twice the area of the triangle <math>ABC_k</math>. Knowing that <math>AB=4</math>, the area of <math>ABC_k</math> can also be expressed as <math>2c_k</math>, where <math>c_k</math> is the length of the altitude from <math>C_k</math> onto <math>AB</math>. Hence we have <math>AC_k \cdot BC_k = 4c_k</math>.

Revision as of 20:29, 11 February 2011

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 $|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|$. The product is \[ |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| \] 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_13&oldid=36864"