Problem:
The figure below depicts a regular 7-gon inscribed in a unit circle.
What is the sum of the 4 th powers of the lengths of all 21 of its edges and diagonals?
Answer Choices:
A. 49
B. 98
C. 147
D. 168
E. 196
Solution:
Recall that if v and w are arbitrary complex numbers, then the following identities hold:
β£wβ£2=wwΛ,v+wβ=vΛ+wΛ, and vw=vΛwΛ,
where wΛ denotes the complex conjugate of w. Hence if z is a complex number, then
β£zβ1β£2=(zβ1)(zβ1)β=(zβ1)(zΛβ1)=zzΛβzβzΛ+1=β£zβ£2βzβzΛ+1
From now on, assume that β£zβ£=1. For such z, the above gives β£zβ1β£2=2βzβzΛ. Hence
β£zβ1β£4=(2βzβzΛ)2.
Expanding and collecting like terms yield
β£zβ1β£4=z2β4z+6β4zΛ+zΛ2.
Now set
z=cis(72Οβ)=cos72Οβ+isin72Οβ=e72Οiβ
Thus 1,z,z2,z3,z4,z5,z6 are the vertices of a regular 7 -gon inscribed in the unit circle in the complex plane, zΛ=zβ1, and z7=1. The sum of the 4th powers of the lengths of the edges and diagonals with one endpoint at 1 is therefore
S=k=1β6ββ£β£β£βzkβ1β£β£β£β4.
If k=0 were allowed in this sum, then the contribution would be β£β£β£βz0β1β£β£β£β4=β£1β1β£4=0. Thus
S=k=0β6ββ£β£β£βzkβ1β£β£β£β4=k=0β6β(z2kβ4zk+6β4zβk+zβ2k)
=T(2)β4T(1)+6T(0)β4T(β1)+T(β2)
where
T(a)=k=0β6βzak
Suppose that a is an integer, and that 7 does not divide a. Then zaξ =1, so by the formula for the sum of a geometric progression,
T(a)=1βza1βz7aβ=1βza1β1β=0
Thus S=6T(0)=6β
7. The other vertices make the same contribution, which gives a total of 6β
72. However, each edge and each diagonal has two endpoints, so all contributions have been counted twice. Hence the requested total is 3β
72=(C)147β.
The problems on this page are the property of the MAA's American Mathematics Competitions
