Cut The Knot!An interactive column using Java applets
by Alex Bogomolny
The mathematical object that goes by the name of complete quadrilateral is neither complete nor quadrilateral, at least not in the sense in which the word "quadrilateral" appears in, say, Brahmagupta's theorem about cyclic quadrilaterals. Simply put, it's not a 4-sided polygon. (The latter is sometimes called a simple quadrilateral.) Seldom mentioned in elementary courses, it plays an important role in projective geometry, where it is used for a ruler only construction of harmonic conjugates and as a corner stone of the projective coordinate system [Möbius, p. 91, Kline, p. 127-128, Courant, p. 179].
A complete quadrilateral is a configuration of four straight lines in general position and six points at which the lines intersect. This is the (projective) configuration that Hilbert and Cohn-Vossen used to denote as (6243) [Hilbert, p. 96] meaning a system of 6 points and 4 lines, 2 lines through each point, 3 points per line. A triangle in this notations appears as (3232) and has the property that the straight lines joining the points of the configuration are necessarily the lines of the configuration. This is obviously not the case with the configuration (6243). There are three lines joining points of the configuration that do not count among the four configuration lines. (These are known as the diagonals of the complete quadrilateral.) This is why the configuration may be called incomplete. Adjoining the three lines to the configuration does not save the situation because such an operation adds new points of intersection. Adding them leaves room for new lines and so on. In fact, all but trivial configurations are incomplete.
A typical quadrilateral (4-sided polygon) is represented by the symbol (4242) and is therefore a typical quadrangle (4-angled polygon). However, the complete quadrangle is the configuration dual to (6243), i.e., (4362), which is the configuration of 4 points and 6 straight lines, 2 points on a line, 3 lines through a point.
In plane geometry, there are quite a few curious theorems associated with the complete quadrilateral [Wells, p. 34-35], some of which are illustrated by the applet below.
The practice of omitting one of the given lines is reminiscent of Clifford's chain and Frank Morley's research. The complete quadrilateral is nothing but a 4-line in Morley's terminology. In a 1903 paper he showed that
(In the applet, the four lines are each defined by two draggable points. Dragging one of the points rotates the line around the other. The line may be also translated by dragging it anywhere away from the points. The four triangles are also shown in translated positions to avoid cluttering the diagram. Try moving -- not dragging -- the cursor over one of the translated triangles.)
Finally, the applet helps verify a problem that appeared in the latest issue of Monthly (Dec 2002, v 109, N 10, p. 921):
Now, the condition that none of the triangles involved is equilateral seems artificial. Indeed, in the remaining portion of the column, I shall outline a proof of a simplified statement:
Which is a nice addition to the "Complete Quadrilateral" collection. First observe the effect of translation of one of the lines on the four triangles. When one of the lines is translated, one of the triangles remains fixed, while the other three undergo a homothetic transformation. But a homothety maps a line (Euler's in particular) onto a parallel line.
Thus to start with, translate one of the lines of the configuration until it passes through one of the configuration points it was not incident with before. The configuration (see the applet below) is no longer the complete quadrilateral, but consists of two "Siamese" triangles that share the baseline and one of the sides. (Check the "Show names" box.) We can see in fact three such pairs of twins --(ABC, ACD) being one pair -- or a set of triplets: ABC, ACD, ABD.
The original problem is equivalent to showing that if the Euler line of ABC is parallel to AD, then the Euler line of ACD is parallel to AB.
A key observation here is that wherever the Euler lines appear to be parallel to the corresponding triangle side lines, they cross on the common baseline. To ascertain whether this is indeed the case, note that on the Euler line there is a slew of remarkable points [Kimberling, p. 128]. Of course any two determine the line uniquely. As in the problem statement, I shall consider the three most important: the centroid, the circumcenter and the orthocenter of a triangle.
The next applet illustrates the following
Given ABD and a point C on BD. Through the centroid (circumcenter, orthocenter) of ACD draw a line parallel to AB. Similarly, draw a line parallel to AD through the centroid (circumcenter, orthocenter) of ABC. Then the two lines meet on BD.
Assuming the proposition true, it follows that if a line parallel to AB is drawn through, the centroid (circumcenter, orthocenter) of ACD to its intersection with BD and the latter point is joint to the centroid (circumcenter, orthocenter) of ABC, then the resulting line is parallel to AD.
Therefore, if the line parallel to AB happens to be the Euler line of ACD, then the three lines drawn from its intersection with BD to the centroid, circumcenter, and orthocenter of ABC are all parallel to AD and hence coalesce into the Euler line of that triangle. This proves our simplified version of the problem.
Proof of Proposition
I'll use the dynamic approach that worked for us so well with another Monthly problem. Let's fix ABD, but allow point C glide over BD. With each position of C, we associate two lines: one parallel to AB, the other to AD passing through the denominationally corresponding points of triangles ACD and ABC. For the centroid, the circumcenter and the orthocenter it is quite clear that if the lines intersect on BD for one position of C, then the same is true for all other positions as well. (What is lost or gained on the left on the left from the common point of intersection, is gained or lost on its right.)
Thus the problem is reduced to finding a position for C, for which the claim is obvious. For the centroid we may take C to be the midpoint of BD. The lines in question then will be the midlines of ABD drawn from C. For the circumcenter and the orthocenter C could be taken to be the foot of the altitude from A. Then both ABC and ACD are right. Their circumcenters lie on AB and AD, respectively, such that the lines in question are again the midlines of ABD. As regard the lines through the orthocenter, the situation is even simpler, since in this case, the orthocenters of both triangles ABC and ACD lie at C.
Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles to which he brought more than 10 years of college instruction and, at least as much, programming experience. In September 2002 the site has welcomed its 5,000,000th visitor. Alex holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at firstname.lastname@example.org
Copyright © 1996-2002 Alexander Bogomolny