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Let there be given three points $x_1, x_2, x_3$ and a line $l$ on a plane. Does there exist an explicit method of finding such a point $p$ on $l$, that sum of distances of $p$ from $x_1, x_2, x_3$ is possibly minimal? Such result for two given points and a line is trivial and known as the Heron's theorem commonly used in optics to find a path of a ray of light.

An example of three points and a point $p$ on the line

All I know about the solution is that if $p$ minimizes the sum of distances, then $$\cos\alpha+\cos\beta+\cos\gamma = 0$$

However, that is not really helpful neither for construction nor finding the solution analytically. Some rare situations may be solved using Brianchon's theorem.

That is, if we add symmetrical reflections of $x_1, x_2, x_3$ on the opposite sides of $l$ and the six points are vertices of a hexagon circumscribing a conic section, then the intersection of the main diagonals minimizes the sum of distances. However it is far from general solution.

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  • $\begingroup$ Hi, welcome. Without loss of generality take the line $l$ to be the $x$-axis. Is your problem not a simple minimization problem of a quadratic loss function in one variable? $\endgroup$
    – Jim
    Commented Aug 9, 2019 at 11:32
  • $\begingroup$ Hi, let us denote the given points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and let $l$ lie on the $OX$ axis. Then what we wish to achieve is $$\frac{d}{dx} \sum_{i=1}^{3}\sqrt{(x-x_i)^2+(0-y_i)^2}=0$$. However this is an equation of degree 8, not a quadratic. $\endgroup$ Commented Aug 9, 2019 at 12:36
  • $\begingroup$ What happens when you use the chain rule? $\endgroup$
    – Jim
    Commented Aug 15, 2019 at 10:41
  • $\begingroup$ I'm not sure I understand your question, the formula above is an equation of one variable. After differentiation we obtain $$\sum_{i=1}^{3}\frac{x-x_i}{\sqrt{(x-x_i)^2+y_i^2}}=0$$ $\endgroup$ Commented Aug 17, 2019 at 11:39

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