Butterfly Theorem

 

Given a chord of a circle, draw any other two chords and passing through its midpoint. Call the points where and meet and . Then is also the midpoint of . There are a number of proofs of this theorem, including those by W. G. Horner, Johnson (1929, p. 78), and Coxeter (1987, pp. 78 and 144). The latter concise proof employs projective geometry.

The following proof is given by Coxeter and Greitzer (1967, p. 46). In the figure at right, drop perpendiculars and from and to , and and from and to . Write , , and , and then note that by similar triangles

(1)

(2)

(3)

so

			(4)
			(5)

so . Q.E.D.

SEE ALSO: Butterfly Catastrophe, Butterfly Curve, Butterfly Effect, Butterfly Function, Butterfly Graph, Butterfly Lemma, Butterfly Polyiamond, Chord, Circle, Cyclic Quadrilateral, Midpoint, Quadrilateral. [Pages Linking Here]

REFERENCES:

Bogomolny, A. "The Butterfly Theorem." http://www.cut-the-knot.org/pythagoras/Butterfly.shtml.

Bogomolny, A. "A Better Butterfly Theorem." http://www.cut-the-knot.org/pythagoras/BetterButterfly.shtml.

Bogomolny, A. "Two Butterflies Theorem." http://www.cut-the-knot.org/pythagoras/BetterButterfly.shtml.

Coxeter, H. S. M. Projective Geometry, 2nd ed. New York: Springer-Verlag, pp. 78 and 144, 1987.

Coxeter, H. S. M. and Greitzer, S. L. "The Butterfly." §2.8 in Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 45-46, 1967.

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, p. 78, 1929.

ریاضیات زیباست

 

Bertrand's Problem

 

What is the probability that a chord drawn at random on a circle of radius (i.e., circle line picking) has length (or sometimes greater than or equal to the side length of an inscribed equilateral triangle; Solomon 1978, p. 2)? The answer depends on the interpretation of "two points drawn at random," or more specifically on the "natural" measure for the problem.

In the most commonly considered measure, the angles and are picked at random on the circumference of the circle. Without loss of generality, this can be formulated as the probability that the chord length of a single point at random angle measured from the intersection of the positive x-axis along the unit circle. Since the length as a function of (circle line picking) is given by

(1)

solving for gives , so the fraction of the top unit semicircle having chord length greater than 1 is

(2)

However, if a point is instead placed at random on a radius of the circle and a chord drawn perpendicular to it, then

(3)

The latter interpretation is more satisfactory in the sense that the result remains the same for a rotated circle, a slightly smaller circle inscribed in the first, or for a circle of the same size but with its center slightly offset. Jaynes (1983) shows that the interpretation of "random" as a continuous uniform distribution over the radius is the only one possessing all these three invariances

Holditch's Theorem

 

Let a chord of constant length be slid around a smooth, closed, convex curve , and choose a point on the chord which divides it into segments of lengths and . This point will trace out a new closed curve , as illustrated above. Provided certain conditions are met, the area between and is given by , as first shown by Holditch in 1858.

The Holditch curve for a circle of radius is another circle which, from the theorem, has radius

Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus. Any two circles can be made concentric by inversion by picking the inversion center as one of the limiting points.

Given two concentric circles with radii and , what is the probability that a chord chosen at random from the outer circle will cut across the inner circle? Depending on how the "random" chord is chosen, 1/2, 1/3, or 1/4 could all be correct answers.

1. Picking any two points on the outer circle and connecting them gives 1/3.

2. Picking any random point on a diagonal and then picking the chord that perpendicularly bisects it gives 1/2.

3. Picking any point on the large circle, drawing a line to the center, and then drawing the perpendicularly bisected chord gives 1/4.

So some care is obviously needed in specifying what is meant by "random" in this problem.

Given an arbitrary chord to the larger of two concentric circles centered on , the distance between inner and outer intersections is equal on both sides . To prove this, take the perpendicular to passing through and crossing at . By symmetry, it must be true that and are equal. Similarly, and must be equal. Therefore, equals . Incidentally, this is also true for homeoids, but the proof is nontrivial.

The three-chord lemma states that in the figure above,