Ellipse in Arbelos

Here is a problem (E 762) from the 1947 American Mathematical Monthly. It was proposed by J. R. Van Andel, Naval Air Experimental Station, Philadelphia, Pa. (The solution below is by Norman Anning, Ann Arbor, Michigan):

Let A₁ and A₂ be two circles with radii a₁ and a₂ and centers (a₁, 0) and (a₂, 0), respectively, with a₂ > a₁ > 0. Let C be any circle in the crescent shaped area M between A₁ and A₂, and tangent to both A₁ and A₂.

The locus of the center of C as it sweeps out M is an ellipse with semiaxes the arithmetic mean (a₁ + a₂)/2 and the geometric mean √(a₁a₂) of the radii a₁ and a₂.
If C_t is a circle of radius r_t and center P_t(x_t, y_t) where
- φ_t = a₁a₂ + t² (a₂ - a₁)²,
- r_t = a₁a₂(a₂ - a₁) / φ_t,
- x_t = a₁a₂(a₂ + a₁) / φ_t,
- y_t = 2t r_t.
then, for any real value of t, C_t lies in M and is tangent to A₁, A₂, and C_t-1.

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Solution

References

J. R. Van Andel,; Norman Anning, American Mathematical Monthly, Vol. 54, No. 9. (Nov., 1947), pp. 547-548.

Part (a) is elementary. A glance at a figure shows that the center of C is always in such a position that the sum of its distances from (a₁, 0) and (a₂ 0) is a₁ + a₂.

That the major semiaxis of the ellipse is (a₁ + a₂)/2 is indeed rather obvious. To determine the minor semiaxis, consider circle C with the center at the topmost point of the ellipse. The triangle of the centers of the three circles will then be isosceles, with the legs equal to (a₁ + a₂)/2 and the base a₂ - a₁. The altitude h of the triangle is found from the Pythagorean theorem:

h² = [(a₁ + a₂)/2]² - [(a₂ - a₁)/2]²,

so that indeed, h² = a₁a₂. (These two facts are easily extended to a more general shape.)

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For part (b), let (X, Y) be a generic point on C^t. Then

(1)	(X² + Y²)·φ_t - 2a₁a₂(a₁ + a₂)X - 4t a₁a₂(a₂ - a₁)Y + 4(a₁a₂)² = 0.

Apply to C_t the inversion

X = 4a₁a₂ x / (x² + y²),
Y = 4a₁a₂ y / (x² + y²).

Then (1) becomes

x² + y² - 2(a₁ + a₂)x - 4t(a₂ - a₁)y + 4φ_t = 0,

which may be rewritten as

(2)	(x - a₁ - a₂)² + (y - 2ta₂ + 2ta₁)² = (a₂ - a₁)².

With t as parameter, (2) is the family of equal circles which touch the parallel lines x = 2a₁ and x = 2a₂. In this family, for every t, the circle C_t is tangent to C_t-1 because the distance between their centers, 2(a₂ - a₁), is equal to the diameter of either.

Now invert again. Circle C₀ inverts into itself and (2) inverts into (1). The line x = 2a₂ inverts into the circle A₁, the inner boundary of the arbelos. Similarly, x = 2a₁ inverts into A₂, the outer boundary. Since it is well known that inversion turns circles into circles and preserves contacts, the proof of the stated theorem is complete.

The solution in the Monthly is followed by the following note:

One tracing the history of the problem would find it under arbelos. See R. Johnson's Modern Geometry, for instance. The neatest of the properties, y_t = 2tr_t, appears in book 4 of Pappus's Collection. See Ivor Thomas, Greek Mathematical Works, II (Loeb Classical Library, No. 362), p. 578.

The proposer pointed out that J. Steiner in Geometrische Betrachtungen (1826) discussed, in particular, the chains of circles corresponding to the sequences t = 0, 1, 2, ... and t = 1/2, 3/2, 5/2, ... Williams mentioned several additional properties of the figure which easily follow from the inversion. For instance, the line joining the points of contact of C_t with A₁ and A₂ passes through the fixed point (2a₁a₂/(a₁ + a₂), 0); the common internal tangent of C_t and C_t-1 passes through this same point; the four points consisting of the origin, the centers of A₁ and A₂, and the above point form a harmonic set. If the diameter of A₁ is taken as two-thirds that of A₂, then r₁ is one-seventh the diameter of A₂. Of this particular figure Victor Thébault has stated a very pretty property. Let the diameter of A₂ taken along the line of centers of A₁ and A₂ be OB, and let BM be the tangent from B to the circle A₁. Then the circle on BM as diameter is tangent to the circle C₁.

Ellipse

Conic Sections > Ellipse

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