Congratulations to castor28 and Opalg for their correct solutions to this week's POTW, which was Problem 25 in the MAA Challenges. This was a tricky problem because of the special case of all four lines parallel. Also, extra kudos to both submitters for terrific TikZ illustrations. Here is castor28's solution:
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We will write $d(P,L_i)$ for the perpendicular distance between the point $P$ and the line $L_i$, and $D =\sum{d(P,L_i)}$ for the common sum of distances from a point in the locus to the given lines.
We start by looking at a simpler case, with only two intersecting lines. In a suitable coordinate system, we have the lines $L_1:y=0$ and $L_2: y = x\tan\alpha$, where $\alpha$ is the angle between the lines.
If $P = (x,y)$ is a point in the first "quadrant" (above $L_1$ and to the right of $L_2$), we have $d(P,L_1) = y$. Since $L_2$ is perpendicular to the unit vector $(\sin\alpha,-\cos\alpha)$, we have $d(P,L_2) = x\sin\alpha-y\cos\alpha$. This gives the equation $x\sin\alpha + y(1-\cos\alpha) = D$, which describes a straight line segment; that segment can be constructed easily from its intersections with the given lines.
A similar construction can be used in the other quadrants, and the complete locus is a parallelogram; the given lines are the diagonals of that parallelogram.
Consider now two parallel lines $L_i$ and $L_j$ separated by a distance $h$. If a point $P$ is between the two lines, we have $d(P,L_i)+d(P,L_j) = h$. On the other hand, if $P$ is outside, on the side of $L_i$, we have $d(P,L_i)+d(P,L_j) = h + 2d(P,L_i)$. Note that, in either case, the sum is at least equal to $h$.
Let us write $h_1$ for the distance between $L_1$ and $L_3$, and $h_2$ for the distance between $L_2$ and $L_4$. Because of the previous remark, if $D<h_1+h_2$, the locus will be empty. If $D=h_1+h_2$, the locus will be the whole parallelogram (interior and boundary) defined by the four lines.
Assume now that $D>h_1+h_2$, and write $D-h_1-h_2=2\Delta>0$, and let us look at the figure below.
\begin{tikzpicture}
\path (-2,0) (0,0) coordinate (O)
(1.732,3) coordinate (Q)
++(-1.155,0) coordinate (A) node[above left]{$A$}
++(6.309,0) coordinate (D) node[above right] {$D$}
++(0.5,0) coordinate (d1);
\path (-1.155,0) coordinate (H) node[above left] {$H$}
++(6.309,0) coordinate (E) node[below right] {$E$}
++(0.5,0) coordinate (e1);
\path (-0.577,-1) coordinate (G) node[below right] {$G$}
++(4,0) coordinate (F) node[below right] {$F$};
\path (Q) ++(0.577,1) coordinate (B) node[above left]{$B$}
++(0.289,0.5) coordinate (b1)
(B) ++(4,0) coordinate (C) node[above left]{$C$}
++(0.289,0.5) coordinate (c1);
\draw (H) ++(-0.5,0) -- (e1) node[midway,above]{$L_1$};
\draw (A) ++(-0.5,0) -- (d1) node[midway,below]{$L_3$};
\draw (G) ++(-0.289,-0.5) -- (b1) node[midway,right]{$L_2$};
\draw (F) ++(-0.289,-0.5) -- (c1) node[midway,left]{$L_4$};
\begin{scope}[very thick,blue]
\draw (H) -- (A);
\draw (B) -- (C);
\draw (D) -- (E);
\draw (F) -- (G);
\end{scope}
\begin{scope}[very thick,red]
\draw (A) -- (B);
\draw (C) -- (D);
\draw (E) -- (F);
\draw (G) -- (H);
\end{scope}
\path (G) -- (C) node[midway] {$(1)$};
\path (H) -- (A) node[midway,left] {$(2)$};
\path (A) -- (B) node[midway,above left] {$(3)$};
\end{tikzpicture}
The four lines divide the plane into nine regions. For a point $P$ in the central region $(1)$, we have $\sum{d(P,L_i)} = h_1 + h_2 < D$; there are no points of the locus in that region on on its boundary.
For a point $P$ in the left region $(2)$, we have:
$$\sum{d(P,L_i)} = h_1 + h_2 + 2d(P, L_2)$$
which shows that $d(P,L_2)=\Delta$. The locus of such points is a straight line segment parallel to $L_2$ ($AH$ in the figure). The other three blue segments can be constructed using the same argument.
For a point in the upper left region $(3)$, we have:
$$\sum{d(P,L_i)} = h_1 + h_2 + 2\big(d(P,L_2) + d(P,L_3)\big)$$
which shows that $d(P,L_2)+d(P,L_3)=\Delta$. We have seen that the locus of such points is a straight line segment. Since we know that the points $A$ and $B$ are on the segment, we may construct the segment $AB$, and the other three red segments can be constructed in the same way.
The conclusion is that, if $D>h_1 + h_2$, the locus is the octagon $ABCDEFGH$.
If the four lines are parallel, the whole figure is invariant under a translation in the common direction; the same must be true for the locus, which will therefore consist of a set of parallel lines (or bands). Take an $x$-axis perpendicular to the lines, and assume that $L_1,L_2,L_3,L_4$ intersect the axis at $a,b,c,d$ (in that order). Write $h_1=d-a$ for the distance between the outer lines $L_1$ and $L_4$, and $h_2 = c-b$ for the distance between the inner lines $L_2$ and $L_3$.
We split the sum into two parts: $\sum{d(P,L_i)}=S_1 + S_2$, where $S_1 = d(P,L_1)+d(P,L_4)$ and $S_2=d(P,L_2) + d(P,L_3)$. The argument used above for two parallel lines shows that $S_1$ is constant, minimum, and equal to $h_1$ between $L_1$ and $L_4$ and increases (with a slope of $\pm2$) when going away from that interval in either direction. The same argument applies to $S_2$. Adding the two functions together, we get the following table for the slopes of $S_1$ and $S_2$:
$$\begin{array}{c|c|c|c|c|c}
x&(-\infty,a)&(a,b)&(b,c)&(c,d)&(d,+\infty)\\
\hline
S'_1&-2&0&0&0&+2\\
S'_2&-2&-2&0&+2&+2\\
S'_1+S'_2&-4&-2&0&+2&+4
\end{array}$$
This shows that the graph of $S_1+S_2$ is concave upward and has a minimum region (with constant value $h_1+h_2$) between $L_2$ and $L_3$.
If $D<h_1+h_2$, the locus is empty. If $D=h_1+h_2$, the locus consists in the band between $L_2$ and $L_3$. If $D>h_1+h_2$, the locus consists of two parallel lines, one on the left of $L_2$ and the other on the right of $L_3$.
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