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Problem Of The Week # 303 - Mar 23, 2018

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Indicium Physicus
Staff member
Jan 26, 2012
Here is this week's POTW:


Four distinct lines $L_1, \, L_2,\,L_3,\,L_4$ are given in the plane, with $L_1$ and $L_2$ respectively parallel to $L_3$ and $L_4$. Find the locus of a point moving so that the sum of its perpendicular distances from the four lines is constant.


Remember to read the POTW submission guidelines to find out how to submit your answers!
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Indicium Physicus
Staff member
Jan 26, 2012
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:

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.

\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);
\begin{scope}[very thick,red]
\draw (A) -- (B);
\draw (C) -- (D);
\draw (E) -- (F);
\draw (G) -- (H);
\path (G) -- (C) node[midway] {$(1)$};
\path (H) -- (A) node[midway,left] {$(2)$};
\path (A) -- (B) node[midway,above left] {$(3)$};

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 $S_2$:
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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