From Wolfram MathWorld, I know there is a Lemoine point of triangle, also called symmedian point, the sum of squared distances of this point to all the three sides is algebraically minimum.

Lemoine point

How to prove such a property?


Let $(d_a,d_b,d_c)$ be the distances of your point from the sides. You have to minimize the quantity: $$ d_a^2+d_b^2+d_c^2 $$ under the constraint: $$ a d_a + b d_b + c d_c = 2\Delta,$$ hence Lagrange multipliers gives that $(d_a,d_b,d_c)=\lambda(a,b,c)$, so your stationary point is the isogonal conjugate of the point having trilinear coordinates $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$, that is the centroid.


Let $(d_a,d_b,d_c)$ be the distances of your point from the sides. Area identity:

$$a d_a + b d_b + c d_c = 2\Delta $$

By Cauchy-Schwarz inequality $(a d_a + b d_b + c d_c )^2 \leq (a^2+b^2+c^2)(d_a^2+d_b^2+d_c^2)$ and therefore

$$ d_a^2+d_b^2+d_c^2 \geq \dfrac{4\Delta ^2}{a^2+b^2+c^2} $$

Equality conditions: $$\dfrac{d_a}{a}=\dfrac{d_b}{b}=\dfrac{d_c}{c}$$

This equality well-known property of Lemoine point and it shows that in Lemoine point, $d_a^2+d_b^2+d_c^2$ will be minimum.


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