William H. Richardson
A problem in an old textbook of mine2 reads ``Show that any quadrangle may be projected into a parallelogram.'' The solution of this problem usually went as follows: Let $PQRS$ be the given quadrangle with exterior diagonal points $T$ and $U$. Let $O$ be any point not in the plane of $PQRS$ and join the points $P, Q, R, S, T, U$ with $O$. See Figure 1. Now if the set of lines $OP, OQ, OR, OS$ are cut by a plane $\alpha$ parallel to the plane $OTU$ we get points $P'$ on $OP$, $Q'$ on $OQ$, $R'$ on $OR$ and $S'$ on $OS$, which when joined form a parallelogram. The fact that $P'Q'R'S'$ is a parallelogram is established as follows: $T$ is on the line of intersection of the planes of $OUT$ and $PQRS$, therefore its image $T'$ on $\alpha$ must be an ideal point since $\alpha$ and the plane $OUT$ are parallel. Now $T'$ would be the intersection of $P'Q'$ and $R'S'$, so it follows that $P'Q'$ and $R'S'$ are parallel. A similar argument establishes that $P'S'$ and $Q'R'$ are parallel.
Figure 1: $O$ not in the plane of $PQRS$

What intrigued me about this problem was the fact that although we were dealing with a three dimensional solution to the problem, the two dimensional drawing in Figure 1 made the problem appear solvable in the plane. This led to the investigation of the case in which the point $O$ is in the plane of $PQRS$. More precisely, given a quadrangle $PQRS$ and a point $O$ in its plane, construct a line $X_1X_2$ such that there is a parallelogram $P'Q'R'S'$ which is copolar with $PQRS$ from the point $O$ and coaxial from the line $X_1X_2$. See Figure 2.
Alternate View of $O$ not in the plane of $PQRS$

Figure 2: $O$ in the plane of $PQRS$

If $O$ is in the plane of $PQRS$, the construction is as follows. Join $O$ with the vertices $P, Q, R, S$ and the exterior diagonal points $T$ and $U$. On the line $OP$ pick a point $P'$ and construct lines $P'X_1$ and $P'X_2$ parallel to $OT$ and $OU$, respectively. $P'X_1$ meets $OQ$ in a point $Q'$ and $P'X_2$ meets $OS$ in a point $S'$. Join the points $X_1$ and $X_2$. The line $X_1X_2$ meets $ST$ at $X_3$ and $QU$ at $X_4$. Join the points $X_3$ and $S'$ which meet $OR$ at $R'$. Join $Q'$ and $X_4$. This line intersects $OR$ at $R'$ and the figure $P'Q'R'S'$ is the required parallelogram.

Proof. Triangles $OP'Q'$ and $UX_2X_4$ are coaxial from the line $PQ$ since $OP'$ meets $UX_2$ in $P$, $OQ'$ meets $UX_4$ in $Q$ and $P'Q'$ meets $X_2X_4$ in $X_1$. By Desargues' Theorem, these triangles are copolar. By construction, $P'X_2$ is parallel to $OU$. Therefore, the pole is an ideal point, and thus $Q'X_4$ is parallel to $P'X_2$. Since triangles $OP'S'$ and $TX_1X_3$ are coaxial from line $PS$, we see, by a similar argument, that $S'X_3$ is parallel to $P'X_1$. All that remains to prove is that $Q'X_4, S'X_3$ and $OR$ are concurrent at $R'$.

Suppose $S'X_3$ meets $OR$ at $R'$. Triangles $OR'S'$ and $UX_4X_2$ are coaxial from the line $RS$, and therefore they must be copolar. Line $S'X_2$ was constructed parallel to $OU$; therefore $R'X_4$ is parallel to $OU$. On the other hand, it has already been established that $Q'X_4$ is parallel to $OU$. Since one and only one line can be drawn parallel to a given line through a given point, $R'$ is on line $Q'X_4$. Therefore, lines $OR, Q'X_4, S'X_3$ are concurrent on $R'$.

It is worth noting that if $O$ is chosen to be on the circle with diameter $TU$, the parallelogram $P'Q'R'S'$ will be a rectangle.

Figure 3: $O$ on circle with diameter $TU$

Moreover, it appears that there exist points on the circle for which the rectangle becomes a square. Below, in Figure 4, we see rectangles that are produced by choices of $O$ close to $T$ and close to $U$. In the drawing on the left $P'Q'<P'S'$; whereas, on the right $P'Q'>P'S'$. As $O$ moves from the position on the left to the position on the right, the dimensions of the rectangle change continuously from $P'Q'<P'S'$ to $P'Q'>P'S'$. In Figure 5 we see $O$ near a point that produces a square.

Figure 4: $O$ chosen in different positions on circle with diameter $TU$

Figure 5: $O$ chosen close to a position that makes $P'Q'R'S'$ a square

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This document was generated using the LaTeX2HTML translator Version 2002-2-1 (1.70)

Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The translation was initiated by Bill Richardson on 2007-04-16

Bill Richardson 2007-04-16