Difference between revisions of "2009 AIME II Problems/Problem 13"

Latest revision as of 21:32, 3 October 2024

Problem

Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1$ , $C_2$ , $\dots$ , $C_6$ . All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000$ .

Solution

Solution 1

Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ 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 $C_1,\ldots, C_6$ are 6 of the 14th roots of unity. Let $\omega=\text{cis}\frac{360^{\circ}}{14}$ ; then $C_1,\ldots, C_6$ correspond to $\omega,\ldots, \omega^6$ . Let $C_1',\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\omega^8\ldots, \omega^{13}$ . Then the lengths of the segments are $|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|$ . Noting that $B$ represents 1 in the complex plane, the desired product is $\begin{align*} BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6'\\ &= |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| \end{align*}$

for $x=1$ . However, the polynomial $(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$ . 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 $|x^{12}+\cdots +x^2+1|=7$ when the radius is 1, and the product is $2^{12}\cdot 7=28672$ . Thus the answer is $\boxed {672}$ .

Solution 2

Let $O$ be the midpoint of $A$ and $B$ . Assume $C_1$ is closer to $A$ instead of $B$ . $\angle AOC_1$ = $\frac {\pi}{7}$ . Using the Law of Cosines,

$\overline {AC_1}^2$ = $8 - 8 \cos \frac {\pi}{7}$ , $\overline {AC_2}^2$ = $8 - 8 \cos \frac {2\pi}{7}$ , . . . $\overline {AC_6}^2$ = $8 - 8 \cos \frac {6\pi}{7}$

So $n$ = $(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {2\pi}{7})\dots(1 - \cos \frac{6\pi}{7})$ . It can be rearranged to form

$n$ = $(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {6\pi}{7})\dots(1 - \cos \frac {3\pi}{7})(1 - \cos \frac {4\pi}{7})$ .

Since $\cos a = - \cos (\pi - a)$ , we have

$n$ = $(8^6)(1 - \cos \frac {\pi}{7})(1 + \cos \frac {\pi}{7}) \dots (1 - \cos \frac {3\pi}{7})(1 + \cos \frac {3\pi}{7})$

= $(8^6)(1 - \cos^2 \frac {\pi}{7})(1 - \cos^2 \frac {2\pi}{7})(1 - \cos^2 \frac {3\pi}{7})$

= $(8^6)(\sin^2 \frac {\pi}{7})(\sin^2 \frac {2\pi}{7})(\sin^2 \frac {3\pi}{7})$

It can be shown that $\sin \frac {\pi}{7} \sin \frac {2\pi}{7} \sin \frac {3\pi}{7}$ = $\frac {\sqrt {7}}{8}$ , so $n$ = $8^6(\frac {\sqrt {7}}{8})^2$ = $7(8^4)$ = $28672$ , so the answer is $\boxed {672}$

Solution 3

Note that for each $k$ the triangle $ABC_k$ is a right triangle. Hence the product $AC_k \cdot BC_k$ is twice the area of the triangle $ABC_k$ . Knowing that $AB=4$ , the area of $ABC_k$ can also be expressed as $2c_k$ , where $c_k$ is the length of the altitude from $C_k$ onto $AB$ . Hence we have $AC_k \cdot BC_k = 4c_k$ .

By the definition of $C_k$ we obviously have $c_k = 2\sin\frac{k\pi}7$ .

From these two observations we get that the product we should compute is equal to $8^6 \cdot \prod_{k=1}^6 \sin \frac{k\pi}7$ , 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 $\prod_{k=1}^6 \sin \frac{k\pi}7 = \prod_{k=1}^3 (\sin \frac{k\pi}7)^2$ .

Let $\omega_k = \cos \frac{2k\pi}7 + i\sin \frac{2k\pi}7$ . The numbers $1,\omega_1,\omega_2,\dots,\omega_6$ are the $7$ -th complex roots of unity. In other words, these are the roots of the polynomial $x^7-1$ . Then the numbers $\omega_1,\omega_2,\dots,\omega_6$ are the roots of the polynomial $\frac{x^7-1}{x-1} = x^6+x^5+\cdots+x+1$ .

We just proved the identity $\prod_{k=1}^6 (x - \omega_k) = x^6+x^5+\cdots+x+1$ . Substitute $x=1$ . The right hand side is obviously equal to $7$ . Let's now examine the left hand side. We have:

$\begin{align*} (1-\omega_k)(1-\omega_{7-k})=|1-\omega_k|^2 & = \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 \\ & = 2-2\cos \frac{2k\pi}7 \\ & = 2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right) \\ & = 4\left( \sin \frac{k\pi}7 \right)^2 \end{align*}$

Therefore the size of the left hand side in our equation is $\prod_{k=1}^3 4 (\sin \frac{k\pi}7)^2 = 2^6 \prod_{k=1}^3 (\sin \frac{k\pi}7)^2$ . As the right hand side is $7$ , we get that $\prod_{k=1}^3 (\sin \frac{k\pi}7)^2 = \frac{7}{2^6}$ .

Solution 4 (Product of Sines)

Lemma 1: A chord $ab$ of a circle with center $O$ and radius $r$ has length $2r\sin\left(\dfrac{\angle AOB}{2}\right)$ .

Proof: Denote $H$ as the projection from $O$ to line $AB$ . Then, by definition, $HA=HB=r\sin\left(\dfrac{\angle AOB}{2}\right)$ . Thus, $AB = 2r\sin\left(\dfrac{\angle AOB}{2}\right)$ , which concludes the proof.

Lemma 2: $\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n-1}}$

Proof: Let $w=\text{cis}\;\dfrac{\pi}{n}$ . Thus, $\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n} = \prod_{k=1}^{n-1} \dfrac{w^k-w^{-k}}{2i} = \dfrac{w^{\frac{n(n-1)}{2}}}{(2i)^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k}) = \dfrac{1}{2^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k})$ Since, $w^{-2k}$ are just the $n$ th roots of unity excluding $1$ , by Vieta's, $\prod_{k=1}^{n-1} \sin \dfrac{k\pi}{n}=\dfrac{1}{2^{n-1}}\prod_{k=1}^{n-1} (1-w^{-2k}) = \dfrac{n}{2^{n-1}}$ , thus completing the proof.

By Lemma 1, the length $AC_k=2r\sin\dfrac{k\pi}{14}$ and similar lengths apply for $BC_k$ . Now, the problem asks for $\left(\prod_{k=1}^6 \left(4\sin\dfrac{k\pi}{14}\right)\right)^2$ . This can be rewritten, due to $\sin \theta = \sin (\pi-\theta)$ , as $\prod_{k=1}^6 \left(4\sin\dfrac{k\pi}{14}\right) \cdot \prod_{k=8}^{13} \left(4\sin\dfrac{k\pi}{14}\right) = \dfrac{1}{\sin \dfrac{7\pi}{14}}\cdot \prod_{k=1}^{13} \left(4\sin\dfrac{k\pi}{14}\right) = \prod_{k=1}^{13} \left(4\sin\dfrac{k\pi}{14}\right).$ By Lemma 2, this furtherly boils down to $4^{12}\cdot \dfrac{14}{2^{13}} = 7\cdot 2^{12} = \boxed{672} \; \text{(mod }1000\text{)}$

~Solution by sml1809

Video Solution

https://youtu.be/TrKxzgR7V8U?si=FFOBCJxjGrg9sWGC

~MathProblemSolvingSkills.com

@@ Line 8: / Line 8: @@
 === 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>. Noting that <math>B</math> represents 1 in the complex plane, the desired product is
+<cmath>
+\begin{align*}
+BC_1\cdots BC_6 \cdot AC_1\cdots AC_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})|
+\end{align*}</cmath>
+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
+<cmath>
+(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>
+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 ===
 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]],
-<math>\overline {AC_1}^2</math> = <math>8 - 8 cos \frac {\pi}{7}</math>,
+<math>\overline {AC_1}^2</math> = <math>8 - 8 \cos \frac {\pi}{7}</math>,
-<math>\overline {AC_2}^2</math> = <math>8 - 8 cos \frac {2\pi}{7}</math>,
+<math>\overline {AC_2}^2</math> = <math>8 - 8 \cos \frac {2\pi}{7}</math>,
 .
 .
 .
-<math>\overline {AC_6}^2</math> = <math>8 - 8 cos \frac {6\pi}{7}</math>
+<math>\overline {AC_6}^2</math> = <math>8 - 8 \cos \frac {6\pi}{7}</math>
-So <math>n</math> = <math>(8^6)(1 - cos \frac {\pi}{7})(1 - cos \frac {2\pi}{7})\dots(1 - cos \frac{6\pi}{7})</math>. It can be rearranged to form
+So <math>n</math> = <math>(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {2\pi}{7})\dots(1 - \cos \frac{6\pi}{7})</math>. It can be rearranged to form
-<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 {2\pi}{4})</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</math> = - <math>cos (\pi - a)</math>, so we have
+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>
-= <math>(8^6)(1 - cos^2 \frac {\pi}{7})(1 - cos^2 \frac {2\pi}{7})(1 - cos^2 \frac {3\pi}{7})</math>
+= <math>(8^6)(1 - \cos^2 \frac {\pi}{7})(1 - \cos^2 \frac {2\pi}{7})(1 - \cos^2 \frac {3\pi}{7})</math>
-= <math>(8^6)(sin^2 \frac {\pi}{7})(sin^2 \frac {2\pi}{7})(sin^2 \frac {3\pi}{7})</math>
+= <math>(8^6)(\sin^2 \frac {\pi}{7})(\sin^2 \frac {2\pi}{7})(\sin^2 \frac {3\pi}{7})</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>
+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 ===
+=== 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>.
@@ Line 38: / Line 56: @@
 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.
+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 2.
 === 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>.
@@ Line 52: / Line 70: @@
 <cmath>
 \begin{align*}
-|1-\omega_k|
+(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 - 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 }
+& = 2-2\cos \frac{2k\pi}7
 \\
-& = \sqrt{  2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right) }
+& = 2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right)
 \\
-& = \sqrt{  4\left( \sin \frac{k\pi}7 \right)^2 }
+& = 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>. However, since sin <math>x</math> = sin <math>\pi - x</math>, then
+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>.
-<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}</math>, which is what we needed to find.
+===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 ==
 {{AIME box|year=2009|n=II|num-b=12|num-a=14}}
+{{MAA Notice}}

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