Difference between revisions of "2009 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
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]], | ||
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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 === | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | By the definition of <math>C_k</math> we obviously have <math>c_k = 2\sin\frac{k\pi}7</math>. | ||
+ | |||
+ | From these two observations we get that the product we should compute is equal to <math> 8^6 \cdot \prod_{k=1}^6 \sin \frac{k\pi}7 </math>, 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 <math>\prod_{k=1}^6 \sin \frac{k\pi}7 </math>. | ||
+ | |||
+ | Let <math>\omega_k = \cos \frac{2k\pi}7 + i\sin \frac{2k\pi}7</math>. The numbers <math>1,\omega_1,\omega_2,\dots,\omega_6</math> are the <math>7</math>-th complex roots of unity. In other words, these are the roots of the polynomial <math>x^7-1</math>. Then the numbers <math>\omega_1,\omega_2,\dots,\omega_6</math> are the roots of the polynomial <math>\frac{x^7-1}{x-1} = x^6+x^5+\cdots+x+1</math>. | ||
+ | |||
+ | We just proved the identity <math>\prod_{k=1}^6 (x - \omega_k) = x^6+x^5+\cdots+x+1</math>. | ||
+ | Substitute <math>x=1</math>. The right hand side is obviously equal to <math>7</math>. Let's now examine the left hand side. | ||
+ | We have: | ||
+ | |||
+ | <cmath> | ||
+ | \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*} | ||
+ | </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>. | ||
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=12|num-a=14}} | {{AIME box|year=2009|n=II|num-b=12|num-a=14}} |
Revision as of 20:43, 18 April 2009
Contents
[hide]Problem
Let and be the endpoints of a semicircular arc of radius . The arc is divided into seven congruent arcs by six equally spaced points , , , . All chords of the form or are drawn. Let be the product of the lengths of these twelve chords. Find the remainder when is divided by .
Solution
Solution 1
Let be the midpoint of and . Assume is closer to instead of . = . Using the Law of Cosines,
= , = , . . . =
So = . It can be rearranged to form
= .
= - , so we have
=
=
=
It can be shown that = , so = = = , so the answer is
Solution 2
Note that for each the triangle is a right triangle. Hence the product is twice the area of the triangle . Knowing that , the area of can also be expressed as , where is the length of the altitude from onto . Hence we have .
By the definition of we obviously have .
From these two observations we get that the product we should compute is equal to , 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 .
Let . The numbers are the -th complex roots of unity. In other words, these are the roots of the polynomial . Then the numbers are the roots of the polynomial .
We just proved the identity . Substitute . The right hand side is obviously equal to . Let's now examine the left hand side. We have:
Therefore the size of the left hand side in our equation is . As the right hand side is , we get that .
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 |