XXXIX OM - II - Problem 3

Inside an acute triangle $ ABC $, consider a point $ P $ and its projections $ L, M, N $ onto the sides $ BC, CA, AB $, respectively. Determine the point $ P $ for which the sum $ |BL|^2 + |CM|^2 + |AN|^2 $ is minimized.
Let $ P $ be any point inside the triangle $ ABC $. Considering the right triangles $ ANP $, $ BLP $, $ CMP $ (Figure 7), we observe that






Similarly, from the triangles $ AMP $, $ BNP $, $ CLP $ we get






Thus, by equating the right sides of the obtained equalities:






The left side of equation (1) is precisely the quantity we are considering in the problem; let's denote it by $ f(P) $. We need to find the point $ P $ for which $ f(P) $ reaches its minimum.
The sum of the left and right sides of equation (1) is $ 2f(P) $. Therefore,






Now notice that






The value of this expression is the smallest when the difference $ |BL| - |CL| $ is zero (since $ |BC| $ is constant) - that is, when $ L $ is the midpoint of segment $ BC $. Similarly, the other two sums in parentheses on the right side of (2) have the smallest value when points $ M $ and $ N $ are the midpoints of sides $ CA $ and $ AB $.
These conditions can be satisfied simultaneously: the projections of point $ P $ onto the sides of the triangle coincide with the midpoints of these sides if and only if $ P $ lies at the intersection of the perpendicular bisectors of the sides. Hence the answer: the quantity $ f(P) $ reaches its minimum when $ P $ is the circumcenter of triangle $ ABC $ (it lies in the interior of triangle $ ABC $, since the triangle is acute).