Problem:
Let w1,w2,…,wn be complex numbers. A line L in the complex plane is called a mean line for the points w1,w2,…,wn if L contains points (complex numbers) z1,z2,…,zn such that
k=1∑n(zk−wk)=0
For the numbers w1=32+170i,w2=−7+64i,w3=−9+200i,w4=1+27i, and w5=−14+43i there is a unique mean line with y-intercept 3. Find the slope of this mean line.
Solution:
Let y=mx+b be a mean line for the complex numbers wk=uk+ivk, where uk and vk are real, and k=1,2,…,n. Assume that the complex numbers zk=xk+iyk, where xk and yk are real, are chosen on the line y=mx+b so that
k=1∑n(zk−wk)=0
Then
∑xk=∑uk,∑yk=∑vk, and yk=mxk+b(1≤k≤n)
where ∑ means summation as k ranges from 1 to n. Consequently,
∑vk=∑yk=∑(mxk+b)=m∑xk+nb=(∑uk)m+nb
Since in our case, n=5,b=3,∑uk=3, and ∑vk=504, it follows that 504=3m+15 and hence m=163.
Note. We have shown only that m=163 is necessary for y=mx+3 to be a mean line for the given set of points. The reader should find corresponding z1,z2,…,z5 to verify the sufficiency of this choice for m.
The problems on this page are the property of the MAA's American Mathematics Competitions