Quadrilaterals Formed by Perpendicular Bisectors
What Is This About?
A Mathematical Droodle
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 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.
What if applet does not run? |
The Problem
The perpendicular bisectors of the sides of a quadrilateral ABCD form a quadrilateral A_{1}B_{1}C_{1}D_{1}, and the perpendicular bisectors of the sides of A_{1}B_{1}C_{1}D_{1} form a quadrilateral A_{2}B_{2}C_{2}D_{2}. Show that A_{2}B_{2}C_{2}D_{2} is similar to ABCD and find the ratio of similitude.
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, A_{1}B_{1}C_{1}D_{1} would degenerate into a point.
For a proof, note that A_{1} is the intersection of the perpendicular bisectors of AB and AD. It is therefore the circumcenter of ΔABD. Similarly, C_{1} is the circumcenter of ΔBCD. It follows that A_{1}C_{1} is the perpendicular bisector of the diagonal BD. In other words, the diagonal A_{1}C_{1} of quadrilateral A_{1}B_{1}C_{1}D_{1} 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: A_{1}B_{1}C_{1}D_{1} and A_{2}B_{2}C_{2}D_{2}: the diagonal B_{2}D_{2} of the former is the perpendicular bisectors of the diagonal A_{1}C_{1} of the latter. The important thing is that this implies BD||B_{2}D_{2}.
Now, the corresponding sides of the quadrilaterals ABCD and A_{2}B_{2}C_{2}D_{2} are similarly parallel. For example, A_{2}D_{2} is the perpendicular bisector of A_{1}D_{1} which, in turn, is the perpendicular bisector of AD. We conclude that, e.g., triangles A_{2}B_{2}D_{2} and ABD are similar so that their angles at A and A_{2} 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 A_{2}B_{2}C_{2}D_{2}. 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) | If ABCD is not cyclic, then the first iterate A_{1}B_{1}C_{1}D_{1} is not degenerate. |
(2) | A_{1}B_{1}C_{1}D_{1} is never cyclic. |
(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 A_{1}, B_{1}, C_{1}, D_{1} coincide, then all four points coincide and the quadrilateral ABCD is cyclic.
Proof
The vertices of A_{1}B_{1}C_{1}D_{1} 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 A_{1}, B_{1}, C_{1}, and D_{1} are distinct, then no three of them can be collinear, and so A_{1}B_{1}C_{1}D_{1} does not degenerate into a point.
Proof
Suppose that A_{1}, B_{1}, and C_{1} are collinear. Since the sides of A_{1}B_{1}C_{1}D_{1} are perpendicular bisectors of ABCD, this means that two consecutive bisectors are the same line; in this case line A_{1}B_{1} coincides with C_{1}D_{1}. 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
References
- D. Bennett, Dynamic Geometry Renews Interest in an Old Problem, in Geometry Turned On, MAA Notes 41, 1997, pp. 25-28
- B. Grünbaum, Quadrangles, Pentagons, and Computers, Geombinatorics 3 (1993), pp. 4-9
- J. King, Quadrilaterals Formed by Perpendicular Bisectors, in Geometry Turned On, MAA Notes 41, 1997, pp. 29-32
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
63597742 |