So we want to minimize the function $f(A) = \sum_{i=1}^n AH_i$ over the plane. Clearly the minimum can only occur in a bounded area of the plane, and $f$ is continuous, so the minimum is reached (by Weierstrass theorem), therefore there do exist ideal point(s). Unicity is guaranteed by the fact that, if $A$, $B$, $H$ are points in the plane, and $M$ is the midpoint of $AB$, then $MH \leq \dfrac {AH + BH} {2}$ (by the triangle inequality, after building the parallelogram, or the known property of the median - vectorial solution also immediate), with equality if and only if $H,A,B$ collinear (and $H$ not on the segment $AB$). Assume therefore there are two ideal points $A,B$, with equal minimum sum $d$ of distances. Then $\sum_{i=1}^n MH_i \leq \sum_{i=1}^n \dfrac {AH_i + BH_i} {2} = \sum_{i=1}^n \dfrac {AH_i} {2} + \sum_{i=1}^n \dfrac {BH_i} {2} = d$, while equality cannot occur, since it would require collinearity of all $H_i$ with $A$ and $B$ - which is ruled out by the conditions of the problem.

Hello, Wouldn't this point be the Fermat point of the polygon?