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.

1. First of all, we have the Theorem of Complete Quadrilateral: the midpoints of the three diagonals are collinear.

2. Next we consider the four triangles formed by the four lines (omitting one of them at a time.) The orthocenters of the triangles are collinear and the line (Ortholine in the applet) is perpendicular to the line (Midline in the applet) of the three mid-diagonals. (This is known as the Gauss-Bodenmiller theorem.)

3. Also, the ortholine serves as the common radical axis of the three circles constructed on the diagonals as diameters, such that whenever the circles intersect, all three of them intersect in two points on the ortholine.

4. The circumcircles of the four triangles meet in a point, the Miquel point of the complete quadrilateral.

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

1. the perpendiculars from the 9-point centers of the four triangles to the respective lines omitted from the 4-line in order to obtain the triangles, meet in a point. The common point lies on the ortholine.

(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):

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.

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

Proposition

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 joined 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 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.

Property #6 has a very nice

Assume the Euler line of ABC intersects side AB in M and side AC in N. Then the Euler line of ANM is parallel to BC.

References

1. J. Aubrey, Brief Lives, Penguin Books, 2000
2. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1996
3. J. Fauvel, J. Gray (eds), The History of Mathematics. A Reader, The Open University, 1987
4. J. Fauvel et al (eds), Möbius and His Band, Oxford University Press, 1993
5. D. Hilbert and S. Cohn-Vossen, Geometry and Imagination, Chelsea Publishing Co, NY 1990.
6. C. Kimberling, Triangle Centers and Central Triangles, Utilitas Mathematica Publishing Inc., 1998
7. M. Kline, Mathematical Thought from Ancient to Modern Times, Oxford University Press, 1972
8. F. Morley, Orthocentric Properties of the Plane n-Line, Trans Amer Math Soc, 4 (1903) 1-12.
9. D. Wells, Curious and Interesting Geometry, Penguin Books, 1991

Copyright © 1996-2008 Alexander Bogomolny