Easy Construction of Bicentric Quadrilateral:
What Is This About?
A Mathematical Droodle
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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 = α. |
(2) | ∠ADS = ∠DAS = ∠ABD = ∠ACD = β. |
(3) | ∠AEB = ∠CED = γ.
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We have to show that the diagonals of ABCD are orthogonal, i.e., γ = 90°, iff
or, which is the same, iff
Now, in ΔBCQ,
so that from (1)
| (*1) |
∠BQC + 2∠BAC = 180°,
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In ΔADS,
so that from (2)
| (*2) |
∠ASD + 2∠ABD = 180°,
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Thus, in ΔABE,
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| γ | = 180° - α - β
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| | = 180° - (180° - ∠BQC)/2 - (180° - ∠ASD)/2
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| | = (∠BQC + ∠ASD)/2
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| | = (∠PQR + ∠PSR)/2,
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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
Bicentric Quadrilateral
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Copyright © 1996-2012 Alexander Bogomolny
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