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Geometry Difficulty 5.5 AIME, harder Prove it Ireland

Given nn points a1,a2,,ana_1, a_2, \dots, a_n in the complex plane, determine a point the sum of whose squared distances from the points and the real axis is a minimum.

Solution

Let zz stand for an arbitrary complex number and denote by Im(z)\operatorname{Im}(z) its imaginary part. Then we want to minimize the expression
f(z)=Im(z)2+k=1nzak2. f(z) = |\operatorname{Im}(z)|^2 + \sum_{k=1}^{n} |z - a_k|^2.
If mm denotes the centroid of the given points, so that m=1nk=1nakm = \frac{1}{n} \sum_{k=1}^{n} a_k, then,
using w2=wwˉ|w|^2 = w \bar{w}, we easily obtain
k=1nzak2=nzm2+k=1nmak2. \sum_{k=1}^{n} |z - a_k|^2 = n|z - m|^2 + \sum_{k=1}^{n} |m - a_k|^2.
Hence,
f(z)=Im(z)2+nzm2+k=1nmak2. f(z) = |\operatorname{Im}(z)|^2 + n|z - m|^2 + \sum_{k=1}^{n} |m - a_k|^2.
But, if m=a+ibm = a + ib, z=x+iyz = x + iy, where a,b,x,ya, b, x, y are real numbers, then
Im(z)2+nzm2=y2+n[(xa)2+(yb)2]=n(xa)2+y2+n(yb)2,|\operatorname{Im}(z)|^2 + n|z - m|^2 = y^2 + n[(x - a)^2 + (y - b)^2] = n(x - a)^2 + y^2 + n(y - b)^2,
which takes its minimum when x=a,y=nb/(n+1)x = a, y = nb/(n + 1). Hence

\min f = \frac{n^2 b^2}{(n+1)^2} + \frac{n b^2}{(n+1)^2} + \sum_{k=1}^{n} |m - a_k|^2 = \frac{n(\operatorname{Im}(m))^2}{n+1} + \sum_{k=1}^{n} |m - a_k|^2.

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