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 A_{1} and A_{2} be two circles with radii a_{1} and a_{2} and centers (a_{1}, 0) and (a_{2}, 0), respectively, with a_{2} > a_{1} > 0. Let C be any circle in the crescent shaped area M between A_{1} and A_{2}, and tangent to both A_{1} and A_{2}.
The locus of the center of C as it sweeps out M is an ellipse with semiaxes the arithmetic mean (a_{1} + a_{2})/2 and the geometric mean √(a_{1}a_{2}) of the radii a_{1} and a_{2}.
If C_{t} is a circle of radius r_{t} and center P_{t}(x_{t}, y_{t}) where
 φ_{t} = a_{1}a_{2} + t² (a_{2}  a_{1})²,
 r_{t} = a_{1}a_{2}(a_{2}  a_{1}) / φ_{t},
 x_{t} = a_{1}a_{2}(a_{2} + a_{1}) / φ_{t},
 y_{t} = 2t r_{t}.
then, for any real value of t, C_{t} lies in M and is tangent to A_{1}, A_{2}, and C_{t1}.



Solution
References
 J. R. Van Andel,; Norman Anning, American Mathematical Monthly, Vol. 54, No. 9. (Nov., 1947), pp. 547548.
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Copyright © 19962018 Alexander Bogomolny
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 (a_{1}, 0) and (a_{2} 0) is a_{1} + a_{2}.
That the major semiaxis of the ellipse is (a_{1} + a_{2})/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 (a_{1} + a_{2})/2 and the base a_{2}  a_{1}. The altitude h of the triangle is found from the Pythagorean theorem:

h² = [(a_{1} + a_{2})/2]²  [(a_{2}  a_{1})/2]²,

so that indeed, h² = a_{1}a_{2}. (These two facts are easily extended to a more general shape.)


For part (b), let (X, Y) be a generic point on C^{t}. Then
(1) 
(X² + Y²)·φ_{t}  2a_{1}a_{2}(a_{1} + a_{2})X  4t a_{1}a_{2}(a_{2}  a_{1})Y + 4(a_{1}a_{2})² = 0.

Apply to C_{t} the inversion

X = 4a_{1}a_{2} x / (x² + y²),
Y = 4a_{1}a_{2} y / (x² + y²).

Then (1) becomes

x² + y²  2(a_{1} + a_{2})x  4t(a_{2}  a_{1})y + 4φ_{t} = 0,

which may be rewritten as
(2) 
(x  a_{1}  a_{2})² + (y  2ta_{2} + 2ta_{1})² = (a_{2}  a_{1})².

With t as parameter, (2) is the family of equal circles which touch the parallel lines x = 2a_{1} and x = 2a_{2}. In this family, for every t, the circle C_{t} is tangent to C_{t1} because the distance between their centers, 2(a_{2}  a_{1}), is equal to the diameter of either.
Now invert again. Circle C_{0} inverts into itself and (2) inverts into (1). The line x = 2a_{2} inverts into the circle A_{1}, the inner boundary of the arbelos. Similarly, x = 2a_{1} inverts into A_{2}, 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 A_{1} and A_{2} passes through the fixed point (2a_{1}a_{2}/(a_{1} + a_{2}), 0); the common internal tangent of C_{t} and C_{t1} passes through this same point; the four points consisting of the origin, the
centers of A_{1} and A_{2}, and the above point form a harmonic set. If the diameter of A_{1} is taken as twothirds that of A_{2}, then r_{1} is oneseventh the diameter of A_{2}. Of this particular figure Victor Thébault has stated a very pretty property. Let the diameter of A_{2} taken along the line of centers of A_{1} and A_{2} be OB, and let BM be the tangent from B to the circle A_{1}. Then the circle on BM as diameter is tangent to the circle C_{1}.
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Copyright © 19962018 Alexander Bogomolny