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 + 2 BAC = 180o ,
In ADS,
ASD + 2b = 180o ,
so that from (2)
(*2 )
ASD + 2 ABD = 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
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
J. L. Coolidge, A Treatise On the Circle and the Sphere , AMS - Chelsea Publishing, 1971, p. 45
H. Dorrie, 100 Great Problems Of Elementary Mathematics , Dover Publications, NY, 1965, pp. 188-193
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 .
R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry) , Dover, 1960, p. 95
R. Honsberger, In Pólya's Footsteps , MAA, 1999, pp. 100-101
Copyright © 1996-2008 Alexander Bogomolny
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