Quadrilaterals Formed by Perpendicular Bisectors: What Is This About?
A Mathematical Droodle

Explanation

Copyright © 1996-2008 Alexander Bogomolny

The applet is supposed to suggest a problem with a curious history. The problem was posed by Josef Langr as problem E1050 in MAA's American Mathematical Monthly (v. 60) in 1953. No solution appeared for about 40 years. B. Grünbaum wrote about the problem in 1993 as an example of an unproven problem whose correctness could not be doubted. In 1994 at a joint math meeting D. Schattschneider used the problem to demonstrate the utility of the dynamic geometry software. She also proved several particular cases of the problem, but the general problem remained yet unsolved. It looks like, by that time, the problem made it into the mathematical folklore. It reached Dan Bennett by the word of mouth and its simplicity had piqued his interest. He published a solution in 1997 to a major part of the problem under an additional assumption that was promptly removed by J. King who (independently) also supplied a proof based on the same ideas.

Inexplicably, the solution is beautifully simple.

The Problem

D. Bennett's Solution

All three quadrilaterals involved are assumed to be not degenerate. In particular, ABCD is assumed not to be cyclic as, otherwise, A₁B₁C₁D₁ would degenerate into a point.

For a proof, note that A₁ is the intersection of the perpendicular bisectors of AB and AD. It is therefore the circumcenter of ABD. Similarly, C₁ is the circumcenter of BCD. It follows that A₁C₁ is the perpendicular bisector of the diagonal BD. In other words, the diagonal A₁C₁ of quadrilateral A₁B₁C₁D₁ is the perpendicular bisector of the diagonal BD of the quadrilateral ABCD. This is the main point of the argument which can be applied to the "next" pair of quadrilaterals: A₁B₁C₁D₁ and A₂B₂C₂D₂: the diagonal B₂D₂ of the former is the perpendicular bisectors of the diagonal A₁C₁ of the latter. The important thing is that this implies BD||B₂D₂.

Now, the corresponding sides of the quadrilaterals ABCD and A₂B₂C₂D₂ are similarly parallel. For example, A₂D₂ is the perpendicular bisector of A₁D₁ which, in turn, is the perpendicular bisector of AD. We conclude that, e.g., triangles A₂B₂D₂ and ABD are similar so that their angles at A and A₂ are equal. The argument is generic and applies to other diagonals and triangles, so that all four angles of ABCD are equal to the corresponding (i.e. formed by parallel sides) angles of A₂B₂C₂D₂. The two quadrilaterals are thus similar. Since their sides are pairwise parallel, they are also homothetic.

J. King's Improvements

J. King has established two important results:

(1) and (2) show that Bennett's argument holds under a natural restriction that ABCD is neither degenerate nor cyclic.

(1) follows from two lemmas:

Lemma 1

If two or more of the points A₁, B₁, C₁, D₁ coincide, then all four points coincide and the quadrilateral ABCD is cyclic.

Proof

The vertices of A₁B₁C₁D₁ serve as circumcenters of triangles formed by the vertices of ABCD. If any two coincide, the corresponding circles also coincide, so that all four given points A, B, C, and D are concyclic.

Lemma 2

If the points A₁, B₁, C₁, and D₁ are distinct, then no three of them can be collinear, and so A₁B₁C₁D₁ does not degenerate into a point.

Proof

Suppose that A₁, B₁, and C₁ are collinear. Since the sides of A₁B₁C₁D₁ are perpendicular bisectors of ABCD, this means that two consecutive bisectors are the same line; in this case line A₁B₁ coincides with C₁D₁. But this is impossible since it would mean that the lines AB and AD must also be the same, which contradicts the fact that ABCD is not degenerate.

(2) requires a somewhat deeper analysis of plane transformations associated with the quadrilateral ABCD. For now I just refer to J. King's original paper.

Remark

Both B. Grünbaum and D. Schattschneider suggested that further interesting results may hold for n-gons with n > 4. I wrote an applet that might help investigate those cases.

References

1. D. Bennett, Dynamic Geometry Renews Interest in an Old Problem, in Geometry Turned On, MAA Notes 41, 1997, pp. 25-28
2. B. Grünbaum, Quadrangles, Pentagons, and Computers, Geombinatorics 3 (1993), pp. 4-9
3. J. King, Quadrilaterals Formed by Perpendicular Bisectors, in Geometry Turned On, MAA Notes 41, 1997, pp. 29-32

Copyright © 1996-2008 Alexander Bogomolny