Squares on Sides of a Quadrilateral
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Copyright © 1996-2018 Alexander BogomolnyDiscussion
Erect similarly oriented squares on the sides of a quadrilateral $ABCD.$ Denote their centers successively $O_{1},$ $O_{2},$ $O_{3},$ and $O_{4}.$ Then
(1)
$O_{1}O_{3} = O_{2}O_{4}$
and
(2)
$O_{1}O_{3} \perp O_{2}O_{4}.$
Proof
Let $M$ be the midpoint of $AC.$ By the Finsler-Hadwiger theorem,
$MO_{1} = MO_{2}$ and $MO_{1} \perp MO_{2}$
and also
$MO_{3} = MO_{4}$ and $MO_{3} \perp MO_{4}.$
Therefore, triangles $MO_{1}O_{3}$ and $MO_{2}O_{4}$ are equal and one is a rotation of the other through $90^{\circ}.$ Hence, (1) and (2) hold.
Remark
If any two adjacent vertices of a quadrilateral coalesce into a point, the statement just proven becomes a property of the Bride's Chair configuration.
The theorem we just proved is attributed to Van Aubel (Von Aubel in [Gardner, p. 176-178]) could also be found in [de Villiers, Yaglom, Finney] among others.
The argument falls through in case one of $O_{1}O_{3}$ or $O_{2}O_{4}$ is of zero length. This case is considered elsewhere. There it is shown that the two distances are either both 0 or not. The former case occurs iff the diagonals of the quadrilateral are equal and perpendicular.
The problem has a simple consequence.
There is an additional exploration of this configuration.
As a consequence, the quadrilateral with vertices at the midpoints of the sides of quadrilateral $O_{1}O_{2}O_{3}O_{4}$ is a square. (This was problem 11328, Am Math Monthly 114 December 2007.)
There is an interesting degenerate case.
References
- E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #320b
- M. de Villiers, The Role of Proof in Investigative, Computer-based Geometry: Some Personal Reflections, in Geometry Turned On, MAA Notes 41, 1997, pp. 15-24
- R. L. Finney, Dynamic Proofs of Euclidean Theorems, Math Magazine, 43, pp. 177-185.
- M. Gardner, Mathematical Circus, Vintage, 1981
- I. M. Yaglom, Geometric Transformations I, MAA, 1962
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