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Difference between revisions of "2022 AMC 12B Problems/Problem 24"

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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>
 
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{(B)} \ 147}</math> is very close to our estimate, so we circle it and are happy that we successfully cheesed an AMC 12B problem 24.
+
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 circle it and are happy that we successfully cheesed an AMC 12B problem 24.
  
 
~sirswagger21
 
~sirswagger21

Revision as of 22:45, 20 November 2022

Problem

The figure below depicts a regular 7-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 4th powers of the lengths of all 21 of its edges and diagonals?

$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution (Complex numbers approach)

There are 7 segments whose lengths are $2 \sin \frac{\pi}{7}$, 7 segments whose lengths are $2 \sin \frac{2 \pi}{7}$, 7 segments whose lengths are $2 \sin \frac{3\pi}{7}$.

Therefore, the sum of the 4th powers of these lengths is \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{3 \pi}{7} \\ & = \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{2 \pi}{7}} - e^{i \frac{2 \pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{3 \pi}{7}} - e^{i \frac{4 \pi}{7}} \right)^4 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6 - 4 e^{- i \frac{2 \pi}{7}} + e^{- i \frac{4 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{8 \pi}{7}} - 4 e^{i \frac{4 \pi}{7}} + 6 - 4 e^{- i \frac{4 \pi}{7}} + e^{- i \frac{8 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{12 \pi}{7}} - 4 e^{i \frac{6 \pi}{7}} + 6 - 4 e^{- i \frac{6 \pi}{7}} + e^{- i \frac{12 \pi}{7}} \right) \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{i \frac{8 \pi}{7}} + e^{i \frac{12 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{8 \pi}{7}} + e^{-i \frac{12 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{i \frac{2 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ & = \boxed{\textbf{(C) 147}} , \end{align*} where the fourth from the last equality follows from the property that \begin{align*} e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} & = e^{-i \frac{6 \pi}{7}} \sum_{j=0}^6 e^{i \frac{2 \pi j}{7}} - 1 \\ & = 0 - 1 \\ & = -1 . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (Trig approach)

There are 7 segments whose lengths are $2 \sin \frac{\pi}{7}$, 7 segments whose lengths are $2 \sin \frac{2 \pi}{7}$, 7 segments whose lengths are $2 \sin \frac{3\pi}{7}$.

Therefore, the sum of the 4th powers of these lengths is \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{3 \pi}{7} \\ & = 7 \cdot 2^4 \left( \frac{1 - \cos \frac{2 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{4 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{6 \pi}{7}}{2} \right)^2 \\ & = 7 \cdot 2^2 \left( 1 - 2 \cos \frac{2 \pi}{7} + \cos^2 \frac{2 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{4 \pi}{7} + \cos^2 \frac{4 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{6 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \cos^2 \frac{2 \pi}{7} + \cos^2 \frac{4 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \frac{1 + \cos \frac{4 \pi}{7} }{2} + \frac{1 + \cos \frac{8 \pi}{7} }{2} + \frac{1 + \cos \frac{12 \pi}{7} }{2} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{12 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ & = \boxed{\textbf{(C) 147}} , \end{align*} where the second from the last equality follows from the property that \[ \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} = - \frac{1}{2} . \]

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3

As explained in the first two solutions, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$. Using trig we get \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*} Like in the second solution, we also use the fact that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$, which admittedly might need some explanation. Notice that \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*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$. These sum to $0$ 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: $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{\textbf{(C) 147}}$. ~obscene_kangaroo

Solution 4 (ruler cheese)

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 $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$.

Measuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \[7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.\]

We know $\left(\frac{2.5}{2.9}\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 7.$

Next, $\left(\frac{4.5}{2.9}\right)^4$ is slightly more than $\left(\frac{4.5}{3}\right)^4.$ We know $\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},$ slightly more than $5,$ so we can approximate $\left(\frac{4.5}{2.9}\right)^4$ as $5.5.$ Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 38.5.$

Finally, $\left(\frac{5.6}{2.9}\right)^4$ is slightly less than $\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\left(\frac{5.6}{2.9}\right)^4 \approx 105.$

Adding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\boxed{\textbf{(C)} \ 147}$ is very close to our estimate, so we circle it and are happy that we successfully cheesed an AMC 12B problem 24.

~sirswagger21

Video Solution

https://youtu.be/nO5p_xfXykI

~ ThePuzzlr

https://youtu.be/yRbweIYtLU8

~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

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