Difference between revisions of "2022 AMC 12B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | The figure below depicts a regular 7-gon inscribed in a unit circle. | + | The figure below depicts a regular <math>7</math>-gon inscribed in a unit circle. |
+ | <asy> | ||
+ | import geometry; | ||
+ | unitsize(3cm); | ||
+ | draw(circle((0,0),1),linewidth(1.5)); | ||
+ | for (int i = 0; i < 7; ++i) { | ||
+ | for (int j = 0; j < i; ++j) { | ||
+ | draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); | ||
+ | } | ||
+ | } | ||
+ | for(int i = 0; i < 7; ++i) { | ||
+ | dot(dir(i * 360/7),5+black); | ||
+ | } | ||
+ | </asy> | ||
+ | What is the sum of the <math>4</math>th powers of the lengths of all <math>21</math> of its edges and diagonals? | ||
− | + | <math>\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196</math> | |
− | ==Solution (Complex | + | ==Solution 1 (Complex Numbers)== |
− | There are 7 segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>. | + | There are <math>7</math> segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>. |
− | Therefore, the sum of the | + | Therefore, the sum of the <math>4</math>th powers of these lengths is |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | + | 7 \cdot 2^4 \sin^4 \frac{\pi}{7} | |
+ 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} | + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} | ||
− | + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} | + | + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4} |
− | & = \frac{7 \cdot 2^4}{(2i)^4} | ||
\left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 | \left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 | ||
+ \frac{7 \cdot 2^4}{(2i)^4} | + \frac{7 \cdot 2^4}{(2i)^4} | ||
Line 52: | Line 65: | ||
& \quad + 7 \cdot 6 \cdot 3 \\ | & \quad + 7 \cdot 6 \cdot 3 \\ | ||
& = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ | & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ | ||
− | & = \boxed{\textbf{(C) 147 | + | & = \boxed{\textbf{(C) }147}, |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 66: | Line 79: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution ( | + | ==Solution 2 (Trigonometry)== |
− | There are 7 segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>. | + | There are <math>7</math> segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>. |
− | Therefore, the sum of the | + | Therefore, the sum of the <math>4</math>th powers of these lengths is |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Line 103: | Line 115: | ||
& = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) | & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) | ||
- 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ | - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ | ||
− | & = \boxed{\textbf{(C) 147 | + | & = \boxed{\textbf{(C) }147}, |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 113: | Line 125: | ||
\] | \] | ||
</cmath> | </cmath> | ||
+ | |||
+ | For explanation see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement]. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 3 (Complex Numbers and Trigonometry)== | ||
+ | |||
+ | As explained in Solutions 1 and 2, what we are trying to find is <math>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}</math>. Using trig we get | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ | ||
+ | = & \sin^2 \frac{\pi}{7} \left(1 - \cos^2 \frac{\pi}{7} \right) + \sin^2 \frac{2\pi}{7} \left(1 - \cos^2 \frac{2\pi}{7} \right) + \sin^2 \frac{3\pi}{7} \left(1 - \cos^2 \frac{3\pi}{7} \right) \\ | ||
+ | = & \sin^2 \frac{\pi}{7} - \left(\frac{1}{2} \sin \frac{2\pi}{7}\right)^2 + \sin^2 \frac{2\pi}{7} - \left(\frac{1}{2} \sin \frac{4\pi}{7}\right)^2 + \sin^2 \frac{3\pi}{7} - \left(\frac{1}{2} \sin \frac{6\pi}{7}\right)^2\\ | ||
+ | = & \sin^2 \frac{\pi}{7} - \frac{1}{4} \sin^2 \frac{2\pi}{7} + \sin^2 \frac{2\pi}{7} - \frac{1}{4} \sin^2 \frac{4\pi}{7} + \sin^2 \frac{3\pi}{7} - \frac{1}{4} \sin^2 \frac{6\pi}{7} \\ | ||
+ | = & \frac{3}{4} \left(\sin^2 \frac{\pi}{7} + \sin^2 \frac{2\pi}{7} + \sin^2 \frac{3\pi}{7}\right) \\ | ||
+ | = & \frac{3}{4} \cdot \frac{1}{2} \left(1 - \cos \frac{2\pi}{7} + 1 - \cos \frac{4\pi}{7} + 1 - \cos \frac{6\pi}{7} \right)\\ | ||
+ | = & \frac{3}{4} \cdot \frac{1}{2} \left(3 - \left(-\frac{1}{2}\right)\right) \\ | ||
+ | = & \frac{21}{16}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Like in the second solution, we also use the fact that <math>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</math>, which admittedly might need some explanation. | ||
+ | |||
+ | For explanation see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement]. | ||
+ | |||
+ | Notice that | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\ | ||
+ | & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | In the brackets we have the sum of the roots of the polynomial <math>x^7 - 1 = 0</math>. These sum to <math>0</math> by [[Vieta’s formulas]], and the desired identity follows. See [[Roots of unity]] if you have not seen this technique. | ||
+ | |||
+ | Going back to the question: <cmath>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{\textbf{(C) }147}.</cmath> | ||
+ | ~obscene_kangaroo | ||
+ | |||
+ | == Solution 4 (Trigonometry) == | ||
+ | This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers: | ||
+ | |||
+ | <cmath>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</cmath> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\ | ||
+ | S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7} | ||
+ | + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ | ||
+ | &= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right) | ||
+ | + \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right) | ||
+ | + \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right) | ||
+ | + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ | ||
+ | &= \frac{1}{2}(3 + S) | ||
+ | + \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right) | ||
+ | + \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right) | ||
+ | + \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\ | ||
+ | &= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\ | ||
+ | &= \frac{1}{2}(3 + S) + 2S. \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | We end up with <cmath>2S^2 - 5S - 3 = 0.</cmath> | ||
+ | Using the quadratic formula, we find the solutions for <math>S</math> to be <math>-\frac{1}{2}</math> and <math>3</math>. Because <math>3</math> is impossible, <math>S = -\frac{1}{2}</math>. | ||
+ | With this result, following similar to steps to Solutions 2 and 3 will get <math>\boxed{\textbf{(C) }147}</math> | ||
+ | |||
+ | ~lordf | ||
+ | |||
+ | == Solution 5 (Law of Cosines) == | ||
+ | |||
+ | Let <math>x</math>, <math>y</math>, and <math>z</math> be the lengths of the chords with arcs <math>\frac{2\pi}{7}</math>, <math>\frac{4\pi}{7}</math> and <math>\frac{6\pi}{7}</math> respectively. | ||
+ | |||
+ | Then by the law of cosines we get: | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ | ||
+ | y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ | ||
+ | z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). | ||
+ | \end{align*}</cmath> | ||
+ | The answer is then just <math>7(x^4+y^4+z^4)</math> (since there's <math>7</math> of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by <math>7</math>. | ||
+ | <cmath>\begin{align*} | ||
+ | & \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\ | ||
+ | &= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ | ||
+ | &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ | ||
+ | &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\ | ||
+ | &= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\ | ||
+ | &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\ | ||
+ | &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\ | ||
+ | &= \boxed{\textbf{(C) }147}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | (Use the identity that <math> \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}</math>.) | ||
+ | |||
+ | For explanation of this identity see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement]. | ||
+ | |||
+ | - SAHANWIJETUNGA | ||
+ | |||
+ | == Solution 6 (Ruler Measure) == | ||
+ | |||
+ | Hope you had a ruler handy! This problem can be done with a ruler and basic estimation. | ||
+ | |||
+ | First, measuring the radius of the circle obtains <math>2.9</math> cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by <math>2.9</math>. | ||
+ | |||
+ | Measuring the sides of the circle gets <math>2.5</math> cm. The shorter diagonals are <math>4.5</math> cm, and the longest diagonals measure <math>5.6</math> cm. Thus, we'd like to estimate <cmath>7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.</cmath> | ||
+ | |||
+ | We know <math>\left(\frac{2.5}{2.9}\right)^4</math> is slightly less than <math>1.</math> Let's approximate it as 1 for now. Thus, <math>7\left(\frac{2.5}{2.9}\right)^4 \approx 7.</math> | ||
+ | |||
+ | Next, <math>\left(\frac{4.5}{2.9}\right)^4</math> is slightly more than <math>\left(\frac{4.5}{3}\right)^4.</math> We know <math>\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},</math> slightly more than <math>5,</math> so we can approximate <math>\left(\frac{4.5}{2.9}\right)^4</math> as <math>5.5.</math> Thus, <math>7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.</math> | ||
+ | |||
+ | Finally, <math>\left(\frac{5.6}{2.9}\right)^4</math> is slightly less than <math>\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.</math> We say it's around <math>15,</math> so then <math>7\left(\frac{5.6}{2.9}\right)^4 \approx 105.</math> | ||
+ | |||
+ | Adding what we have, we get <math>105 + 38.5 + 1 = 144.5</math> as our estimate. We see <math>\boxed{\textbf{(C) }147}</math> is very close to our estimate, so we have successfully finished the problem. | ||
+ | |||
+ | ~sirswagger21 | ||
+ | |||
+ | == Solution 7 (Pythagorean Theorem and Trig) == | ||
+ | First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle <math>\frac{2\pi}{7},</math> which are <cmath>\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)</cmath> for integer <math>n.</math> Then, we notice there are three types of diagonals: the ones with chords of arcs <math>\dfrac{2\pi}{7}, \dfrac{4\pi}{7},</math> and <math>\dfrac{6\pi}{7}.</math> We notive there are here are <math>7</math> of each type of diagonal. Then, we use the [[pythagorean theorem]] to find the distance from <math>\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)</math> to <math>(1, 0)</math>: | ||
+ | <cmath>\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4 </cmath> | ||
+ | <cmath>=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4 </cmath> | ||
+ | <cmath>=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.</cmath> | ||
+ | By the cosine double angle identity, <math>\cos{2\theta}=2\cos^2\theta-1.</math> This means that <math>2\cos^2\theta=\cos{2\theta}+1.</math> Substituting this in, | ||
+ | <cmath>4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.</cmath> Summing this up for <math>n=1,2,3,</math> | ||
+ | <cmath>2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6</cmath> <cmath>=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18</cmath> | ||
+ | <cmath>=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18</cmath> | ||
+ | <cmath>=2(-0.5)-8(-0.5)+18=21.</cmath> | ||
+ | (These equalities are based on <math>\cos\theta=\cos(2\pi-\theta)</math> and <math>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.</math>) (For explanation of this see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement].) | ||
+ | Finally, because there are <math>7</math> of each type of diagonal, the answer is <math>7\cdot 21=\boxed{147}.</math> | ||
+ | |||
+ | ~[[BS2012]] | ||
+ | |||
+ | == Solution 8 (Roots of Unity) == | ||
+ | |||
+ | Place the figure on the complex plane and let <math>w = e^{\frac{2\pi}{7}}</math> (a <math>7</math>th root of unity). The vertices of the <math>7</math>-gon are <math>w^0,w^1,w^2,\dots,w^6</math>. We wish to find | ||
+ | <cmath>\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.</cmath> | ||
+ | The second expression is more convenient to work with. The factor of <math>\tfrac{1}{2}</math> is because it double-counts each edge (and includes when <math>i=j</math>, but the length is <math>0</math> so it doesn't matter). | ||
+ | |||
+ | Recall the identity <math>|z|^2 = z\overline{z}</math>. Since <math>|w| = 1</math>, <math>\overline{w} = w^{-1}</math>, and | ||
+ | <cmath>\begin{align*} | ||
+ | \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\ | ||
+ | &= (2-w^{i-j}-w^{j-i})^2 \\ | ||
+ | &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\ | ||
+ | &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | But notice that, for a fixed value of <math>i</math>, over all values of <math>j</math> from <math>0</math> to <math>6</math>, by properties of modular arithmetic with <math>w^7 = w^0</math> (or expanding), | ||
+ | each of the <math>w^\bullet</math>-terms takes on every value in <math>w^0,w^1,w^2,\dots,w^6</math> exactly once. Since <math>\sum_{j=0}^6 w^j = 0</math>, | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\ | ||
+ | &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\ | ||
+ | &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\ | ||
+ | &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\ | ||
+ | &= \frac{1}{2}\cdot7\cdot7\cdot6 \\ | ||
+ | &= \boxed{\textbf{(C) }147}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Fun fact: This generalizes to any regular <math>n</math>-gon with <math>n \ge 3</math> to obtain | ||
+ | <cmath>\frac{1}{2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}6 = 3n^2,</cmath> | ||
+ | though for even <math>n</math>, <math>w^{2i-2j}</math> and <math>w^{2j-2i}</math> take on every value in <math>w^0,w^2,w^4,\dots,w^{n-2}</math> exactly twice. These are the <math>\tfrac{n}{2}</math>th roots of unity, so they still sum to <math>0</math>. | ||
+ | |||
+ | This also explains why <math>n < 3</math> doesn't follow the pattern. For <math>n = 2</math> (one edge of length <math>2</math>), since <math>w^{2i-2j} = (w^2)^{i-j} = 1^{i-j} = 1</math> and similarly for <math>w^{2j-2i}</math>, the constant term in the summation becomes <math>8</math> instead of <math>6</math>. For <math>n = 1</math> (no edges), the constant term becomes <math>0</math>. | ||
+ | |||
+ | -maxamc | ||
+ | |||
+ | ~edited by [[User:emerald_block|emerald_block]] | ||
+ | |||
+ | == Solution 9 (Inductive Reasoning) == | ||
+ | |||
+ | This is how I solve this problem: | ||
+ | |||
+ | It's easy to solve for <math>3</math>-gon, <math>4</math>-gon, and <math>6</math>-gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon) | ||
+ | |||
+ | For <math>3</math>-gon, the sum is equal to <math>3</math> times the <math>4</math>th power of an edge. Thus, <cmath>S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.</cmath> | ||
+ | |||
+ | For <math>4</math>-gon, the sum is equal to <math>4</math> times the <math>4</math>th power of an edge, and <math>2</math> times the <math>4</math>th power of the diagonal. Thus, <cmath>S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.</cmath> | ||
+ | |||
+ | For <math>6</math>-gon, the sum is equal to <math>6</math> times the <math>4</math>th power of an edge, <math>6</math> times the <math>4</math>th power of the short diagonal, and <math>3</math> times the <math>4</math>th power of the long diagonal. Thus, <cmath>S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.</cmath> | ||
+ | |||
+ | Then, I quickly noticed that <math>27=3\,\cdot\,3^2</math>, <math>48=3\,\cdot\,4^2</math>, and <math>108=3\,\cdot\,6^2</math>. So reasonably, it will work out this formula, <math>S_n=3n^2</math>. (This step is purely out of guessing, maybe have a look at Solution 8 for more info...) | ||
+ | |||
+ | By inductive reasoning, we got <math>S_7=3\,\cdot\,7^2=\boxed{\textbf{(C) }147}</math>. | ||
+ | |||
+ | - Prof. Joker | ||
+ | |||
+ | ==Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)== | ||
+ | |||
+ | The sum of all roots of unity is <math>0</math>, therefore, the sum of the real parts of the roots of unity is <math>0</math>. | ||
+ | |||
+ | Consider the roots of unity of <math>x^n-1=0</math> where <math>n</math> is odd on the unit circle, the cosine (real parts) of all the roots under the x-axis is the same as the cosine of all the roots above the x-axis because of symmetry. Hence, <math>2</math> times the sum of the cosines of all the roots above the x-axis plus <math>1</math> is <math>0</math>. | ||
+ | |||
+ | In the context of this problem, <math>2(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}) + 1 = 0</math>. Hence, <math>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 17:06, 15 November 2023
Contents
- 1 Problem
- 2 Solution 1 (Complex Numbers)
- 3 Solution 2 (Trigonometry)
- 4 Solution 3 (Complex Numbers and Trigonometry)
- 5 Solution 4 (Trigonometry)
- 6 Solution 5 (Law of Cosines)
- 7 Solution 6 (Ruler Measure)
- 8 Solution 7 (Pythagorean Theorem and Trig)
- 9 Solution 8 (Roots of Unity)
- 10 Solution 9 (Inductive Reasoning)
- 11 Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)
- 12 Video Solution
- 13 See Also
Problem
The figure below depicts a regular -gon inscribed in a unit circle.
What is the sum of the
th powers of the lengths of all
of its edges and diagonals?
Solution 1 (Complex Numbers)
There are segments whose lengths are
,
segments whose lengths are
,
segments whose lengths are
.
Therefore, the sum of the th powers of these lengths is
where the fourth from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Trigonometry)
There are segments whose lengths are
,
segments whose lengths are
,
segments whose lengths are
.
Therefore, the sum of the th powers of these lengths is
where the second from the last equality follows from the property that
For explanation see supplement.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Complex Numbers and Trigonometry)
As explained in Solutions 1 and 2, what we are trying to find is . Using trig we get
Like in the second solution, we also use the fact that
, which admittedly might need some explanation.
For explanation see supplement.
Notice that
In the brackets we have the sum of the roots of the polynomial
. These sum to
by Vieta’s formulas, and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question:
~obscene_kangaroo
Solution 4 (Trigonometry)
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
We end up with
Using the quadratic formula, we find the solutions for
to be
and
. Because
is impossible,
.
With this result, following similar to steps to Solutions 2 and 3 will get
~lordf
Solution 5 (Law of Cosines)
Let ,
, and
be the lengths of the chords with arcs
,
and
respectively.
Then by the law of cosines we get:
The answer is then just
(since there's
of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by
.
(Use the identity that .)
For explanation of this identity see supplement.
- SAHANWIJETUNGA
Solution 6 (Ruler Measure)
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by
.
Measuring the sides of the circle gets cm. The shorter diagonals are
cm, and the longest diagonals measure
cm. Thus, we'd like to estimate
We know is slightly less than
Let's approximate it as 1 for now. Thus,
Next, is slightly more than
We know
slightly more than
so we can approximate
as
Thus,
Finally, is slightly less than
We say it's around
so then
Adding what we have, we get as our estimate. We see
is very close to our estimate, so we have successfully finished the problem.
~sirswagger21
Solution 7 (Pythagorean Theorem and Trig)
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle which are
for integer
Then, we notice there are three types of diagonals: the ones with chords of arcs
and
We notive there are here are
of each type of diagonal. Then, we use the pythagorean theorem to find the distance from
to
:
By the cosine double angle identity,
This means that
Substituting this in,
Summing this up for
(These equalities are based on
and
) (For explanation of this see supplement.)
Finally, because there are
of each type of diagonal, the answer is
Solution 8 (Roots of Unity)
Place the figure on the complex plane and let (a
th root of unity). The vertices of the
-gon are
. We wish to find
The second expression is more convenient to work with. The factor of
is because it double-counts each edge (and includes when
, but the length is
so it doesn't matter).
Recall the identity . Since
,
, and
But notice that, for a fixed value of , over all values of
from
to
, by properties of modular arithmetic with
(or expanding),
each of the
-terms takes on every value in
exactly once. Since
,
Fun fact: This generalizes to any regular -gon with
to obtain
though for even
,
and
take on every value in
exactly twice. These are the
th roots of unity, so they still sum to
.
This also explains why doesn't follow the pattern. For
(one edge of length
), since
and similarly for
, the constant term in the summation becomes
instead of
. For
(no edges), the constant term becomes
.
-maxamc
~edited by emerald_block
Solution 9 (Inductive Reasoning)
This is how I solve this problem:
It's easy to solve for -gon,
-gon, and
-gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For -gon, the sum is equal to
times the
th power of an edge. Thus,
For -gon, the sum is equal to
times the
th power of an edge, and
times the
th power of the diagonal. Thus,
For -gon, the sum is equal to
times the
th power of an edge,
times the
th power of the short diagonal, and
times the
th power of the long diagonal. Thus,
Then, I quickly noticed that ,
, and
. So reasonably, it will work out this formula,
. (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
By inductive reasoning, we got .
- Prof. Joker
Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)
The sum of all roots of unity is , therefore, the sum of the real parts of the roots of unity is
.
Consider the roots of unity of where
is odd on the unit circle, the cosine (real parts) of all the roots under the x-axis is the same as the cosine of all the roots above the x-axis because of symmetry. Hence,
times the sum of the cosines of all the roots above the x-axis plus
is
.
In the context of this problem, . Hence,
Video Solution
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.