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

(Created page with "==Problem== The figure below depicts a regular 7-gon inscribed in a unit circle. What is the sum of the 4th powers of the lengths of all 21 of its edges and diagonals? ==So...")
 
Line 65: Line 65:
 
& = -1 .
 
& = -1 .
 
\end{align*}
 
\end{align*}
 +
</cmath>
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Solution (Trig approach)==
 +
 +
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>.
 +
 +
Therefore, the sum of the 4th powers of these lengths is
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
where the second from the last equality follows from the property that
 +
<cmath>
 +
\[
 +
\cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7}
 +
= - \frac{1}{2} .
 +
\]
 
</cmath>
 
</cmath>
  

Revision as of 14:38, 17 November 2022

Problem

The figure below depicts a regular 7-gon inscribed in a unit circle.

What is the sum of the 4th powers of the lengths of all 21 of its edges and diagonals?

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 (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)

Video Solution

https://youtu.be/yRbweIYtLU8

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

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