Perpendicular bisector construction of a quadrilateral

In geometry, the perpendicular bisector construction of a quadrilateral is a construction which produces a new quadrilateral from a given quadrilateral using the perpendicular bisectors to the sides of the former quadrilateral. This construction arises naturally in an attempt to find a replacement for the circumcenter of a quadrilateral in the case that is non-cyclic.

Definition of the construction

Suppose that the vertices of the quadrilateral $Q$ are given by $Q_{1},Q_{2},Q_{3},Q_{4}$. Let $b_{1},b_{2},b_{3},b_{4}$ be the perpendicular bisectors of sides $Q_{1}Q_{2},Q_{2}Q_{3},Q_{3}Q_{4},Q_{4}Q_{1}$ respectively. Then their intersections $Q_{i}^{(2)}=b_{i+2}b_{i+3}$, with subscripts considered modulo 4, form the consequent quadrilateral $Q^{(2)}$. The construction is then iterated on $Q^{(2)}$ to produce $Q^{(3)}$ and so on.

First iteration of the perpendicular bisector construction

An equivalent construction can be obtained by letting the vertices of $Q^{(i+1)}$ be the circumcenters of the 4 triangles formed by selecting combinations of 3 vertices of $Q^{(i)}$.

Properties

1. If $Q^{(1)}$ is not cyclic, then $Q^{(2)}$ is not degenerate.

2. Quadrilateral $Q^{(2)}$ is never cyclic. Combining #1 and #2, $Q^{(3)}$ is always nondegenrate.

3. Quadrilaterals $Q^{(1)}$ and $Q^{(3)}$ are homothetic, and in particular, similar. Quadrilaterals $Q^{(2)}$ and $Q^{(4)}$ are also homothetic.

3. The perpendicular bisector construction can be reversed via isogonal conjugation. That is, given $Q^{(i+1)}$, it is possible to construct $Q^{(i)}$.

4. Let $\alpha ,\beta ,\gamma ,\delta$ be the angles of $Q^{(1)}$. For every $i$, the ratio of areas of $Q^{(i)}$ and $Q^{(i+1)}$ is given by

$(1/4)(\cot(\alpha )+\cot(\gamma ))(\cot(\beta )+\cot(\delta )).$

5. If $Q^{(1)}$ is convex then the sequence of quadrilaterals $Q^{(1)},Q^{(2)},\ldots$ converges to the isoptic point of $Q^{(1)}$, which is also the isoptic point for every $Q^{(i)}$. Similarly, if $Q^{(1)}$ is concave, then the sequence $Q^{(1)},Q^{(0)},Q^{(-1)},\ldots$ obtained by reversing the construction converges to the Isoptic Point of the $Q^{(i)}$'s.

6. If $Q^{(1)}$ is tangential then $Q^{(2)}$ is also tangential.

References

• J. Langr, Problem E1050, Amer. Math. Monthly, 60 (1953) 551.
• V. V. Prasolov, Plane Geometry Problems, vol. 1 (in Russian), 1991; Problem 6.31.
• V. V. Prasolov, Problems in Plane and Solid Geometry, vol. 1 (translated by D. Leites), available at http://students.imsa.edu/~tliu/math/planegeo.eps.
• D. Bennett, Dynamic geometry renews interest in an old problem, in Geometry Turned On, (ed. J. King), MAA Notes 41, 1997, pp. 25–28.
• J. King, Quadrilaterals formed by perpendicular bisectors, in Geometry Turned On, (ed. J. King), MAA Notes 41, 1997, pp. 29–32.
• G. C. Shephard, The perpendicular bisector construction, Geom. Dedicata, 56 (1995) 75–84.
• A. Bogomolny, Quadrilaterals formed by perpendicular bisectors, Interactive Mathematics Miscellany and Puzzles, http://www.cut-the-knot.org/Curriculum/Geometry/PerpBisectQuadri.shtml.
• B. Grünbaum, On quadrangles derived from quadrangles—Part 3, Geombinatorics 7(1998), 88–94.
• O. Radko and E. Tsukerman, The Perpendicular Bisector Construction, the Isoptic Point and the Simson Line of a Quadrilateral, Forum Geometricorum 12: 161–189 (2012).

External links

• Perpendicular-Bisectors of Circumscribed Quadrilateral Theorem at Dynamic Geometry Sketches, interactive dynamic geometry sketches.