A Newly Born Pair of Siblings to Archimedes' Twins:
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A Mathematical Droodle
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Copyright © 1996-2012 Alexander Bogomolny
Arbelos is a geometric shape bounded by three semicircles. Arbelos was researched yet by Archimedes in his Book of Lemmas. In particular, Archimedes discovered two equal circles inscribed into arbelos that, since then, have been known as Archimedes' twins. In 1974, L. Bankoff found a sibling of the same size. The family has been expanded by C. W. Dodge and coauthors in 1999 and additional quadruplets have been delivered by F. Power in 2005.
Quang Tuan Bui just (November 12, 2011) came up with another pair.
An arbelos is comprised of three semicircles (O1), (O2), O, with points of tangency A, C, B. Let r1, r2 the radii of (O1), (O2) respectively. Perpendicular bisector of diameter AC intersects semicircle (O) at D1. D1A and D1C intersect semicircle (O1) at E1, F1, respectively. Denote by (K1) the circle with E1F1 as a diameter. Construct similarly D2, E2, F2, (K2). Then
- Two circles (K1), (K2) are congruent with Archimedes' twin circles i.e their radius: r = r1·r2/(r1 + r2).
- Their centers K1, K2 are equidistant from the center O of the big circle: K1O = K2O.
Proof
To prove the first assertion, observe that (because of the symmetry in O1D1) E1F1||AC. Draw CE1 and BD1. Both are perpendicular to AD1 and are, therefore, also parallel. We have several proportions:
E1F1/AC = D1E1/AD1 = BC/AB = 2·r2/(2·r1 + 2·r2) = r2/(r1 + r2).
But AC = 2r1, so that
E1F1 = 2r1r2/(r1 + r2),
half of which is exactly the radius of Archimedes' twins. The expression is symmetric in r1 and r2, meaning that a similar derivation for the length of E2F2 will give the same result.
For the second assertion, apply the Pythagorean theorem to, say, ΔOO1K1. Indeed, we can find the legs K1O1and O1O in terms of r1 and r2.
K1O1/D1O1 = AE1/AD1 = AC/AB = 2·r1/(2·(r1 + r2)) = r1/(r1 + r2)
D1O1² = AO1·BO1 = r1·(r1 + 2·r2)
K1O1² = r1·(r1 + 2·r2)·( r1/(r1 + r2))².
Now, O1O2 = r1 + r2; it follows that O1O = r2. The Pythagorean theorem - K1O² = K1O1² + O1O² - then yields
K1O² = (r1 + r2)² - r1·r2·(2·r1 + r2)·(r1 + 2·r2)/(r1 + r2)².
This expression is again symmetric in r1 and r2, so that K2O² evaluates to the same quantity: K1O = K2O.
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Copyright © 1996-2012 Alexander Bogomolny
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