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Easy Construction of Bicentric Quadrilateral: What Is This About?
A Mathematical Droodle


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Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

Easy Construction of Bicentric Quadrilateral

A quadrilateral is bicentric if it's both inscriptable and circumscriptable. (Inscriptable means admitting an incircle. Circumscriptable means cyclic, i.e., admitting a circumcircle.) Bicentric quadrilaterals might seem exotic, but the applet shows how such quadrilaterals can be constructed easily. (Another construction appears elsewhere.)


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Let ABCD be a cyclic quadrilateral with vertices on a given circle w. Assume ABCD is also orthodiagonal, i.e., AC BD. Then quadrilateral PQRS formed by the tangents to w at the vertices of ABCD is bicentric. (It's inscriptible by construction. The assertion therefore is about its being circumscriptible - cyclic.) The converse is also true: if PQRS is cyclic, then the diagonal of ABCD are orthogonal. Furthermore, if PQRS is cyclic, then the point of intersection E of the diagonals of ABCD, and the centers I and O of the two circles are collinear.

The latter assertion can be rephrased. We know that, the diagonals of a cyclic quadrilateral (ABCD in this case) and those of the quadrilateral formed by the points of tangency of ABCD and its incircle, are concurrent. Therefore we can say that, for a bicentric quadrilateral, the intersecition of the diagonals M, the incenter I and the circumcenter O are collinear.

Proof

First of all note the following angle identities

(1)QCB = BCQ = BAC = BDC = a.
(2)ADS = DAS = ABD = ACD = b.
(3)AEB = CED = g.

We have to show that the diagonals of ABCD are orthogonal, i.e., g = 90o, iff

  PQR + PSR = 180o,

or, which is the same, iff

(4) BQC + ASD = 180o

Now, in BCQ,

  BQC + 2a = 180o,

so that from (1)

(*1) BQC + 2BAC = 180o,

In ADS,

  ASD + 2b = 180o,

so that from (2)

(*2) ASD + 2ABD = 180o,

Thus, in ABE,

 
g= 180o - a - b
 = 180o - (180o - BQC)/2 - (180o - ASD)/2
 = (BQC + ASD)/2
 = (PQR + PSR)/2,

which proves the first part of the assertion. The fact that I, O, and E are collinear has been proven in [Dubrovsky] and [Honsberger]. An absolutely delicious proof appears as a consequence of another construction of bicentric quadrilaterals.

References

  1. G. Bennett, Bi-centric Quadrilaterals and the Pedal n-gon, in The Lighter Side of Mathematics, R.K.Guy and R.E.Woodrow, eds, MAA, 1994, p. 97
  2. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971, p. 45
  3. H. Dorrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY, 1965, pp. 188-193
  4. V. N. Dubrovsky, Solution to problem M1154, Kvant, n 8, 1989, pp 34-35 (in Russian), pdf is available at http://kvant.mccme.ru/1989/08/p34.htm.
  5. R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960, p. 95
  6. R. Honsberger, In Pólya's Footsteps, MAA, 1999, pp. 100-101

Copyright © 1996-2008 Alexander Bogomolny

28737578Page copy protected against web site content infringement by Copyscape


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