Difference between revisions of "2009 AIME II Problems/Problem 13"
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\begin{align*} | \begin{align*} | ||
BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= | BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= | ||
− | BC_1\cdots BC_6 \cdot BC_1'\cdots BC_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})| | |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| | ||
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(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. | (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. | ||
</cmath> | </cmath> | ||
− | Thus the product is <math>|x^{12}+\cdots +x^2+1|=7</math> | + | Thus the product is <math>|x^{12}+\cdots +x^2+1|=7</math> when the radius is 1, and the product is <math>2^{12}\cdot 7=28672</math>. Thus the answer is <math>\boxed {672}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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<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 | + | Since <math>\cos a = - \cos (\pi - a)</math>, 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> | ||
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=== Computing the product of sines === | === 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>. | + | In this section we show one way how to evaluate the product <math>\prod_{k=1}^6 \sin \frac{k\pi}7 = \prod_{k=1}^3 (\sin \frac{k\pi}7)^2 </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>. | 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>. | ||
Line 70: | Line 70: | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | |1-\omega_k| | + | (1-\omega_k)(1-\omega_{k-7})=|1-\omega_k|^2 |
− | & = | + | & = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 |
− | \left | ||
\\ | \\ | ||
− | & = | + | & = 2-2\cos \frac{2k\pi}7 |
\\ | \\ | ||
− | & = \ | + | & = 2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right) |
\\ | \\ | ||
− | & = | + | & = 4\left( \sin \frac{k\pi}7 \right)^2 |
− | |||
− | |||
− | |||
− | |||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | Therefore the size of the left hand side in our equation is <math>\prod_{k=1}^ | + | Therefore the size of the left hand side in our equation is <math>\prod_{k=1}^3 4 (\sin \frac{k\pi}7)^2 = 2^6 \prod_{k=1}^3 (\sin \frac{k\pi}7)^2</math>. As the right hand side is <math>7</math>, we get that <math>\prod_{k=1}^3 (\sin \frac{k\pi}7)^2 = \frac{7}{2^6}</math>. |
− | < | + | |
+ | ===Solution 4 (Product of Sines)=== | ||
+ | |||
+ | <i><b>Lemma 1:</b> A chord <math>ab</math> of a circle with center <math>O</math> and radius <math>r</math> has length <math>2r\sin\left(\dfrac{\angle AOB}{2}\right)</math>.</i> | ||
+ | |||
+ | <i><b>Proof:</b> Denote <math>H</math> as the projection from <math>O</math> to line <math>AB</math>. Then, by definition, <math>HA=HB=r\sin\left(\dfrac{\angle AOB}{2}\right)</math>. Thus, <math>AB = 2r\sin\left(\dfrac{\angle AOB}{2}\right)</math>, which concludes the proof.</i> | ||
+ | |||
+ | <i><b>Lemma 2:</b> <math>\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n-1}}</math></i> | ||
+ | |||
+ | <i><b>Proof:</b> Let <math>w=\text{cis}\;\dfrac{\pi}{n}</math>. Thus, | ||
+ | <cmath>\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n} = \prod_{k=1}^{n-1} \dfrac{w^k-w^{-k}}{2i} = \dfrac{w^{\frac{n(n-1)}{2}}}{(2i)^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k}) = \dfrac{1}{2^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k})</cmath> | ||
+ | Since, <math>w^{-2k}</math> are just the <math>n</math>th roots of unity excluding <math>1</math>, by Vieta's, <math>\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n}=\dfrac{1}{2^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k}) = \dfrac{n}{2^{n-1}}</math>, thus completing the proof. | ||
+ | </i> | ||
+ | |||
+ | By Lemma 1, the length <math>AC_k=2r\sin\dfrac{k\pi}{14}</math> and similar lengths apply for <math>BC_k</math>. Now, the problem asks for <math>\left(\prod_{k=1}^6 \left(4\sin\dfrac{k\pi}{14}\right)\right)^2</math>. This can be rewritten, due to <math>\sin \theta = \sin (\pi-\theta)</math>, as <math>\prod_{k=1}^6 \left(4\sin\dfrac{k\pi}{14}\right) \cdot \prod_{k=8}^{13} \left(4\sin\dfrac{k\pi}{14}\right) = \dfrac{1}{\sin \dfrac{7\pi}{14}}\cdot \prod_{k=1}^{13} \left(4\sin\dfrac{k\pi}{14}\right) = \prod_{k=1}^{13} \left(4\sin\dfrac{k\pi}{14}\right).</math> By Lemma 2, this furtherly boils down to <math>4^{12}\cdot \dfrac{14}{2^{13}} = 7\cdot 2^{12} = \boxed{672} \; \text{(mod }1000\text{)}</math> | ||
+ | |||
+ | <b>~Solution by sml1809</b> | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/TrKxzgR7V8U?si=FFOBCJxjGrg9sWGC | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
== See Also == | == See Also == |
Latest revision as of 20:27, 17 May 2024
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
. Noting that
represents 1 in the complex plane, the desired product is
for .
However, the polynomial
has as its zeros all 14th roots of unity except for
and
. Hence
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
=
.
Since , 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 2.
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
.
Solution 4 (Product of Sines)
Lemma 1: A chord of a circle with center
and radius
has length
.
Proof: Denote as the projection from
to line
. Then, by definition,
. Thus,
, which concludes the proof.
Lemma 2:
Proof: Let . Thus,
Since,
are just the
th roots of unity excluding
, by Vieta's,
, thus completing the proof.
By Lemma 1, the length and similar lengths apply for
. Now, the problem asks for
. This can be rewritten, due to
, as
By Lemma 2, this furtherly boils down to
~Solution by sml1809
Video Solution
https://youtu.be/TrKxzgR7V8U?si=FFOBCJxjGrg9sWGC
~MathProblemSolvingSkills.com
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.