Difference between revisions of "2009 AIME II Problems/Problem 13"
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− | 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 | + | 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 x<math> = </math>sin \pi - x<math>, then |
− | <math>\prod_{k=1}^3 \sin \frac{k\pi}7 < | + | </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}$, which is what we needed to find. |
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=12|num-a=14}} | {{AIME box|year=2009|n=II|num-b=12|num-a=14}} |
Revision as of 21:53, 20 April 2009
Contents
[hide]Problem
Let and be the endpoints of a semicircular arc of radius . The arc is divided into seven congruent arcs by six equally spaced points , , , . All chords of the form or are drawn. Let be the product of the lengths of these twelve chords. Find the remainder when is divided by .
Solution
Solution 1
Let be the midpoint of and . Assume is closer to instead of . = . Using the Law of Cosines,
= , = , . . . =
So = . It can be rearranged to form
= .
= - , so we have
=
=
=
It can be shown that = , so = = = , so the answer is
Solution 2
Note that for each the triangle is a right triangle. Hence the product is twice the area of the triangle . Knowing that , the area of can also be expressed as , where is the length of the altitude from onto . Hence we have .
By the definition of we obviously have .
From these two observations we get that the product we should compute is equal to , which is the same identity as in Solution 1.
Computing the product of sines
In this section we show one way how to evaluate the product .
Let . The numbers are the -th complex roots of unity. In other words, these are the roots of the polynomial . Then the numbers are the roots of the polynomial .
We just proved the identity . Substitute . The right hand side is obviously equal to . Let's now examine the left hand side. We have:
Therefore the size of the left hand side in our equation is . As the right hand side is , we get that . However, since sin xsin \pi - x\prod_{k=1}^3 \sin \frac{k\pi}7 \frac {7}{2^6}\frac {\sqrt {7}}{8}$, 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 |