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> | ||
Line 70: | Line 70: | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | (1-\omega_k)(1-\ | + | (1-\omega_k)(1-\omega_{7-k})=|1-\omega_k|^2 |
& = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 | & = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 | ||
\ | \ | ||
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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>. | 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 21:32, 3 October 2024
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 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.