Question

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.?

Answer 1

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.