Suppose $A_1,...,A_n$ are $n$ points on the line $y=0$ on 2D plane. How can I find the point $P$ on the line $y=c$ (for some given $c$) so that the product of distances from $P$ to $A_1,...,A_n$ (i.e., $|PA_1|\times|PA_2|\times...\times|PA_n|$) is minimized?
Any solution to finding the approximate location of $P$ is appreciated as well.
This is a partial answer.
Suppose any point $A_i$ has coordinates $(x_i,0)$ and $P$ has coordinates $(x,c)$. Then we want to minimize the function $$ f(x)= \prod_{k=1}^n \sqrt{(x-x_k)^2+c^2}$$ Squaring and taking the log on both sides,
$$\log\left[(f^2)’\right] = \sum_{k=1}^n \log\left((x-x_k)^2+c^2\right) \\ \implies (f^2)’ = f^2 \cdot \sum_{k=1}^n \frac{2(x-x_k)}{(x-x_k)^2+c^2} \overset{\text{set}}= 0 \\ \sum_{k=1}^n \frac{x-x_k}{(x-x_k)^2+c^2} = 0 \hspace{2 cm} \because f^2\ne 0 \\ x\sum_{k=1}^n \frac{1}{(x-x_k)^2+c^2} - \sum_{k=1}^n \frac{x_k}{(x-x_k)^2+c^2} = 0$$ This will turn out to be an $n+1$ degree polynomial. You can solve for $x$ using numerical methods.
Let $A_k = (x_k , y_k)$ and $P = (x, y)$. The distance between $A_k$ and $P$ is $$d_k = \sqrt{ (x - x_k)^2 + (y - y_k)^2 }$$ and the product of all $k$ distances is $$d = \prod_{k=1}^{n} d_k = \sqrt{ \prod_{k=1}^{n} (x - x_k)^2 + (y - y_k)^2 }$$ and therefore the product of the distances is minimised when the products of the squares of the distances are minimized, $$d^2 = \prod_{k=1}^{n} (x - x_k)^2 + (y - y_k)^2$$ One set of solutions occurs when there exists $k$ for which $x_k = x$ and $y_k = y$, as then term $k$ is zero, and the entire product is zero.
Let's assume there are no $k$ for which $y_k = y$, and therefore the product is nonzero (positive) for all $x$.
The product of the distances and distances squared both reach an extremum when the derivative of the product of the squared distances (with respect to $x$) is zero: $$\frac{d}{d x} \prod_{k=1}^n (x - x_k)^2 + (y - y_k)^2 = 2 \left( \prod_{k=1}^{n} (x - x_k)^2 + (y - y_k)^2 \right)\left(\sum_{k=1}^{n} \frac{x - x_k}{(x - x_k)^2 + (y - y_k)^2}\right) = 0$$ Because we already decided the first parenthesized part (the product of the distances) is nonzero, this can only be true if the second parenthesized part is zero. Therefore, the (interesting) extrema occur when $$S = \sum_{k=1}^{n} s_k = \sum_{k=1}^{n} \frac{x - x_k}{(x - x_k)^2 + (y - y_k)^2} = 0 \tag{1}\label{1}$$
Because the product increases without limit when $x$ is smaller than any $x_k$, or greater than any $x_k$, we know that the minimum occurs for $$\min(x_1,\dots,x_n) \le x \le \max(x_1,\dots, x_n)$$
Examining each term $s_k$ we find that $$\begin{cases} x \lt x_k - \lvert y - y_k \rvert, & \frac{-1}{2\lvert y - y_k\rvert} \lt s_k \lt 0, & s_k \text{ negative and decreasing} \\ x = x_k - \lvert y - y_k \rvert, & s_k = \frac{-1}{2\lvert y - y_k \rvert} & s_k \text{ minimum} \\ x_k - \lvert y - y_k \rvert \lt x \lt x_k, & \frac{-1}{2\lvert y - y_k\rvert} \lt s_k \lt 0, & s_k \text{ negative and increasing} \\ x = x_k, & s_k = 0 & ~ \\ x_k \lt x \lt x_k + \lvert y - y_k \rvert, & 0 \lt s_k \lt \frac{1}{2\lvert y - y_k \rvert}, & s_k \text{ positive and increasing} \\ x = x_k + \lvert y - y_k \rvert, & s_k = \frac{1}{2\lvert y - y_k \rvert} & s_k \text{ maximum} \\ x_k + \lvert y - y_k \rvert \lt x, & 0 \lt s_k \lt \frac{1}{2\lvert y - y_k\rvert}, & s_k \text{ positive and decreasing} \\ \end{cases}$$ Let's say we create an ordered set $\chi_i$, which contains all unique $x_k$ and $x_k \pm y_k - c$ in increasing order: $\chi_i \lt \chi_{i+1}$. Let's say there are $N$ of these ($N \le 3 n$).
Within each interval $\chi_i \lt x \lt \chi_{i+1}$, each term $s_k$ belongs to one of the four classes above ($s_k = 0$ or minimum or maximum is impossible, since $x \ne x_k$ and $x \ne x_k \pm \lvert y_k - y \rvert$). $$\begin{aligned} S &= S_{--} + S_{-+} + S_{++} + S_{+-} \\ s_k \text{ in } & \begin{cases} S_{--}, ~\text{ if }~ \chi_{i+1} \le x_k - \lvert y - y_k \rvert \\ S_{-+}, ~\text{ if }~ x_k - \lvert y - y_k \rvert \le \chi_{i}, ~ \chi_{i+1} \le x_k \\ S_{++}, ~\text{ if }~ x_k \le \chi_{i}, \chi_{i+1} \le x_k + \lvert y - y_k \rvert \\ S_{+-}, ~\text{ if }~ x_k + \lvert y - y_k \rvert \le \chi_{i} \\ \end{cases} \\ \end{aligned}$$ Because of how we defined $\chi_i$, it is not possible for $\chi_{i} \lt x_k \lt \chi_{i+1}$ or $\chi_{i} \lt x_k \pm \lvert y - y_k \rvert \lt \chi_{i+1}$.
Now, if $$- S_{--}(\chi_{i}) - S_{-+}(\chi_{i+1}) \gt S_{++}(\chi_{i}) + S_{+-}(\chi_{i+1}) \tag{2a}\label{2a}$$ or $$S_{+-}(\chi_{i}) + S_{++}(\chi_{i+1}) \gt -S_{-+}(\chi_{i}) - S_{--}(\chi_{i+1}) \tag{2b}\label{2b}$$ where $S_{--}(x)$ means evaluating the sum at $x$, then $S$ cannot have a root in $\chi_{i} \lt x \lt \chi_{i+1}$.
This means that starting at each interval $\chi_{i}, \chi_{i+1}$, and subdividing (halving) each interval that does not fulfill $\eqref{2a}$ or $\eqref{2b}$, will converge to an arbitrarily short interval that does contain $S = 0$.
Each interval can contain more than one root, but this should find them all, since it considers the full range of values the sub-sums have in each interval, and discards those that cannot sum to zero. Also, one needs to evaluate the product of the distances at each $\chi_i$, as the above procedure only considers the interval between.
