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


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Explanation

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Copyright © 1996-2017 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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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The Problem

The perpendicular bisectors of the sides of a quadrilateral ABCD form a quadrilateral A1B1C1D1, and the perpendicular bisectors of the sides of A1B1C1D1 form a quadrilateral A2B2C2D2. Show that A2B2C2D2 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, 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 ΔABD. Similarly, C1 is the circumcenter of ΔBCD. 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) If ABCD is not cyclic, then the first iterate A1B1C1D1 is not degenerate.
(2) A1B1C1D1 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 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

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  • Parallelogram and Similar Triangles
  • Two Triples of Similar Triangles
  • Three Similar Triangles
  • Similar Triangles on Sides and Diagonals of a Quadrilateral
  • Point on Bisector in Right Angle
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