Difference between revisions of "2022 AMC 12B Problems/Problem 24"
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The figure below depicts a regular 7-gon inscribed in a unit circle. | 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? | ||
− | + | <math>\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196</math> | |
==Solution (Complex numbers approach)== | ==Solution (Complex numbers approach)== |
Revision as of 21:23, 18 November 2022
Contents
[hide]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 , 7 segments whose lengths are , 7 segments whose lengths are .
Therefore, the sum of the 4th 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 (Trig approach)
There are 7 segments whose lengths are , 7 segments whose lengths are , 7 segments whose lengths are .
Therefore, the sum of the 4th powers of these lengths is where the second from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.