For any point P inside a given triangle ABC, denote by x, y, and z the distances from P to the lines [BC], [AC], and [AB], respectively. Find the position of P for which the sum x^2 + y^2 + z^2 is a minimum.?

1 Answer
Feb 7, 2018

It's the Lemoine Point (Symmedian Point)

Explanation:

To construct this, draw the symmedians (the reflection of the medians by the bisections of the respective angles) and mark the intersection point.

Demonstration:

If AP is a symmedian then the distances to the sides AB and AC are proportional to the sides themselves.

So... The symmedian point would have the following property:

x/a=y/b=z/c

Knowing that we can start the proof.

We want some point that minimize the expression x^2+y^2+z^2. Taking a, b and c as the lengths of the sides of the triangle. Minimizing above expression is the same that minimizing the expression

E=(x^2+y^2+z^2)(a^2+b^2+c^2)
E=(ax+by+cz)^2 + (bx-ay)^2 + (cy-bz)^2 + (cx-az)^2

(ax+by+cz)=2 Delta is constant, once that Delta is the area of the triangle. So, we just need to minimize the part (bx-ay)^2 + (cy-bz)^2 + (cx-az)^2. If this is the symmedian point, we know that:

x/y=a/b, y/z=b/c" and " x/z=a/c

So...
(bx-ay)^2 + (cy-bz)^2 + (cx-az)^2 = 0

Which is the minimum value for the sum of three squares.

This would make E minimum and, once that (a^2 + b^2 + c^2) doesn't depends of x, y and z, this would make x^2+y^2+z^2 minimum.