A Circle With Two Centers
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At points B and C of an angle BAC erect perpendiculars and let them intersect in D. Since AB and AC are not parallel, the perpendiculars are not parallel either. Thus, unless AB and AC form a 0° or a 180° angle, point D is well defined.
Draw a circle circumscribing ΔBCD. Besides B, the circle intersects AB in another point, say, E. It intersects AC in F different from C. Now, since ∠DBE is right, DE is a diameter of the circle. The midpoint of DE is the center of the circle. A similar argument applies to DF. As a result, we arrive at the conclusion that the circle at hand has two centers. How come?
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Copyright © 1996-2012 Alexander Bogomolny
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The problem is with the drawing. Since both angles ABD and ACD are right, they both subtend a diameter of the circle, which is AD. Thus in all cases the points E and F coincide with A. (The applet cheats by drawing BD and CD not quite perpendicular to AB and AC.)
References
- V. M. Bradis et al, Lapses in Mathematical Reasoning, Dover, 1999, pp. 137-138
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|Activities| |Contact| |Front page| |Contents| |Geometry| |Fallacies| |Store|
Copyright © 1996-2012 Alexander Bogomolny
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