V OM - I - Problem 7

In the plane, a line $ p $ and points $ A $ and $ B $ are given. Find a point $ M $ on the line $ p $ such that the sum of the squares $ AM^2 + BM^2 $ is minimized.
The geometric locus of points $ X $ in the plane for which $ AX^2 + BX^2 $ has a given value $ 2k^2 $ is a certain circle whose center lies at the midpoint $ S $ of segment $ AB $; the radius of this circle is larger the larger $ k $ is.
The smallest circle centered at $ S $ that intersects a line $ p $ is the circle tangent to line $ p $; the point of tangency is the projection $ M $ of point $ S $ onto line $ p $. Point $ M $ is the sought point; the problem always has a solution and only one.
Note. The theorem about the geometric locus, which we referred to in the above solution, can be proven as follows. Let $ AB = c $ and let $ X $ be any point in the plane (Fig. 22). Then

When $ X $ lies outside the line $ AB $, this equality is a known formula for the square of the median of a triangle (see problem 2). When $ X $ lies on the line $ AB $, verifying the equality is easy and is left to the reader.
From this equality, it follows: 1°) if $ AX^2 + BX^2 $ has a given value $ 2k^2 $, then point $ X $ lies on a circle centered at $ S $ with radius $ \sqrt{k^2 - \frac{1}{4} c^2} $; thus, $ k \geq \frac{1}{2} c $, and when $ k = \frac{1}{2} c $, the circle reduces to point $ S $; 2°) if point $ X $ lies on such a circle, i.e., if $ SX = \sqrt{k^2 - \frac{1}{4} c^2} $, then $ AX^2 + BX^2 = 2k^2 $, Q.E.D.