Difference between revisions of "2009 AIME II Problems/Problem 13"
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=== 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]], | ||
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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 | + | === 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 21:29, 11 February 2011
Contents
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 the radius be 1 instead. All lengths will be halved so we will multiply by 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
are 6 of the 14th roots of unity. Let
; then
correspond to
. Let
be their reflections across the diameter. These points correspond to
. Then the lengths of the segments are
. The product is
\[
|(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})|
\]
for
.
However, the polynomial
has as its zeros all 14th roots of unity except for
and
. 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
(
) when the radius is 1, and the product is
. Thus the answer is
.
Solution 2
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 3
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
= sin
, then
would be the square root of
, or
, 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 |