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