F1 : [0, b ]→ [0, ∞ ) and F2 :[, b a ]→ [0, ∞ ) , respectively (see Figure 3). To avoid intersecting U by the arcs L and R (except the points (0,0) and ( a,0)) it is necessary to assume that FxFx( )< ( ) for each x ∈ (0, b ], 1 (1) FxFx2 ( )< ( ) for each xba ∈ [ , ].
If U is the graph of a continuous function F a],0[: → ∞),0[ with F(0)= F () a = 0, then the U-arbelos is also called the F-arbelos or the F-belos ([7], [8]). Note that the classic arbelos and parbelos belong to this class. The case of U being the graph of a continuous function F: [0, a ]→ [0, ∞ ) has been deeply investigated in [7] and [8], where one can find more properties of F-arbelos and examples, including hyperbolic analogues of arbelos.
126 M. Różański, A. Samulewicz, M. Szweda, R. Wituła
Fig. 3.
Fig. 4. F-arbelos built of the lemniscate of Gerono (on the left) and F-arbelos built of non- convex function (on the right)
It turns out that the class of ( U,R,L)-arbelos bounded by three similar arcs U, L and R (which are not necessarily the graphs of real-valued functions) also satisfies Properties 2.1, 3.1, 2.2, 3.2 and the conditions corresponding to Properties 2.4 and 3.3.
5. Ab ovo et ad ovum: elliptic arbelos
Now consider elliptic arbelos or ellarbelos , the class of figures whose bounda- ries consist of three semi ellipses instead of three semicircles.
Fig. 5. Elliptic arbelos
Variations on the arbelos 127
We can assume that all considered ellipses are symmetric about the x-axis, however their major axes are not necessarily parallel. Then the arcs U, L, R form- ing the boundary of the ellarbelos are the graphs of the appropriate functions F, F1 and F2 (see Section 4). In particular, if all semi ellipses are pairwise similar then the ellarbelos is an F-arbelos and possesses all the properties mentioned in Section 4. In the definition of the ellarbelos, we do not assume that the semi ellipses are similar, but we require that both the L and R arcs meet the arc U in exactly one point. Consider ellipses 2 2 − 2 2 − 2 2 x + y = (x x1) + y = (x x2 ) + y = E : 2 2 ,1 E1 : 2 2 ,1 E2 : 2 2 ,1 (2) a b a1 b1 a2 b2 where aa,12 , a ,,, bb 12 b are positive numbers, x1, x 2 ∈ R and x1< x 2 .
Assume that E, E1 and E2 are pairwise tangent and the curves E1, E2 are con- tained in the closed region bounded by E. Then
a = a1 + a2 , x1 = −a2 , x2 = a1. (3)
Let F:[(−+ aaaa1212 ), +→ ][0,], bF 11212 :[ −− aaaa , −→ ][0,] b 1 and
Faaaa21212:[− , + ][0,] → b 2 be the parametrizations of the semi ellipses
U ={( yx ), ∈ E : y ≥ },0 L ={( yx ), ∈ E1 : y ≥ }0 and R={( xy , ) ∈ Ey2 : ≥ 0}, re- spectively. We intend to find conditions assuring that Ei , i ∈ },2,1{ has only one common point with E. Since the largest ellipse E is symmetric about the y-axis, we can restrict our consideration to E2 and x∈[a1, a1 + a2 ].
Define an auxiliary function ϕ ,0[: a2 ]→ R putting
2 2 ϕ t)( = F(( a1 + t)) − (F2 (a1 + t .))
Clearly E2 is contained in the closed region bounded by E if and only if ϕ t)( ≥ 0 for all t ∈ ,0[ a2 ]. Notice that
(a + t) 2 t 2 t = b 2 − 1 − b 2 − ϕ )( 1 2 2 1 2 (4) (a1 + a2 ) a2 is a quadratic polynomial and
2 2 2 2 (2 a + t) 2 2t b b 2 ba t −= b 1 + b = 2 − t − 1 ϕ )(' 2 2 2 2 2 2 2 . (5) (a1 + a2 ) a2 a2 (a1 + a2 ) (a1 + a2 )
128 M. Różański, A. Samulewicz, M. Szweda, R. Wituła
Note that ϕ(a2 )= 0 . Therefore, if ϕ ≥ 0on ,0[ a2 ] then ϕ must be decreasing on an interval (a2 − ε ,a2 ] for some ε > 0 . Hence ϕ (' a2 ) ≤ .0 The latter is equiv- alent to
a1 b≥ b 2 1 + . (6) a2
Now assume that (6) holds. If the coefficient of t in the equality (5) is non- positive, i.e. if b b 2 ≤ , (7) a2 a 1+ a 2 then ϕ '< 0 on [0,a2 ] and ϕ is a strictly decreasing function with ϕ(a2 )= 0. Thus
ϕ ≥ 0 for all t∈[0, a 2 ]. On the other hand, if b b 2 > (8) a2 a 1+ a 2 then (5), (6) and (8) imply that ϕ' is an increasing linear function with ϕ (' a2 ) ≤ ,0 so ϕ t)(' ≤ 0 for all t ∈ ,0[ a2 ]. This means that ϕ is decreasing and, since
ϕ(a2 ) = ,0 all its values are non-negative.
Recapitulating, ϕ ≥ 0 on ,0[ a2 ] if and only if (6) holds. Lemma 5.1. Three ellipses 22+ 22 − 22 x+= y() xay2 += () xay1 += E:22 1, E1 : 22 1, E 2 : 22 1 (aab12+ ) ab 11 ab 22 are pairwise tangent if and only if
a2 a1 b ≥ maxb1 1 + , b2 1 + . (9) a a 1 2
Henceforth we assume that the semi ellipses forming the ellarbelos satisfy the condition (9).
a2 a1 Example 5.2. The borderline case of (9) is b = b1 1+ = b2 1 + . The ellar- a1 a2 belos satisfying this condition, namely with a1 + a2 = ,9 b = ,3 a1 = ,5 b1 = ,5
Variations on the arbelos 129
a2 = ,4 b2 = ,2 is depicted in Figure 6. Enlarging the value of b1 or b2 would re- sult in an output of E1 or E2 beyond the region bounded by E.
Fig. 6. Ellarbelos with optimally selected parameters
Property 5.3. The area of ellarbelos equals
1 1 π (( a1 + a2 )b − π ba 11 − π ba 22 ) = π (( a1 + a2 )(b − b1 − b2 ) + π ba 21 + π ba 12 ). 2 2 The above equality is shown in Figure 7.
Fig. 7. Illustrations for the above properties for ellarbelos, on the left side, when b > b1+b2, on the right side, when b < b1+b2
In particular, the area of the classic arbelos is equal to the area of the ellipse whose semiaxes correspond to the radii of the smaller semicircles (see Fig. 8).
130 M. Różański, A. Samulewicz, M. Szweda, R. Wituła
Fig. 8. Illustration for the above properties for arbelos
Remark 5.4. If all semi ellipses bounding an ellarbelos are similar, then there is a shear mapping that transforms it into an arbelos. Thus for every ellarbelos of this kind, there exist two congruent ellipses that are pre-images of the twin circles of Archimedes associated with the arbelos. Question 5.5. Assume that all semi ellipses enclosing an ellarbelos are similar. For which eccentricities do all ellarbelos have twin circles corresponding to the twin circles of Archimedes? Does the answer depend on whether the foci of all el- lipses are colinear ( a > b ) or not ( a < b )?
6. Space: above and beyond
By a simple 3D-elliptic arbelos , or briefly 3D-ellarbelos , we mean the upper half of the solid obtained by rotating an ellarbelos around the x-axis (see Fig. 9). In other words, 3D-ellarbelos is enclosed by the plane z = 0 and three pairwise tan- gent ellipsoids of revolution. We assume additionally that a ≥ b, a1 ≥ b1 and
a2 ≥ b2 , i.e. all ellipsoids are prolate spheroids or spheres. The 3D-ellarbelos is regular provided that the semiellipsoids enclosing it are similar.
Fig. 9. 3D-elliptic arbelos
The area of the prolate ellipsoid with the semi-principal axes a, b, b equals 2π ab 2πb2 + arcsin, ε (10) ε
Variations on the arbelos 131
a 2 − b 2 where ε = is the eccentricity of the ellipse. a
The difference S1 − S 2 between the upper surface and the lower one equals
π a + a b π ba S S πb 2 ( 1 2 ) ε πb 2 11 ε 1 − 2 = + arcsin − 1 + arcsin 1 ε ε1 π ba −πb 2 + 22 arcsinε − π (( a + a )b − π ba − π ba ), 2 ε 2 1 2 11 22 2 where ε1,ε 2 are the eccentricities of the ellipses E1, E2 , respectively. In particular, if all the eccentrities are equal then b b b = 1 = 2 , (11) a1 + a2 a1 a2 and ε SSπ bbb2 2 2 π arcsin aabbbabab 12−=() −−+ 12 −1 (( 12 + )( −−++ 121221 ) ) ε
ab1 ab 2 arcsin ε =2π ⋅+ π −+ 1() ab12 ab 21 a1+ a 2 a 1 + a 2 ε
arcsinε arcsinε arcsinε where = > .1 Note that →1as ε → .0 ε sin(arcsinε ) ε
If all the ellipses are circles then S1− S 2 = 2π aa 12 .
According to Property 5.3, π aa 21 is the area of the arbelos bounded by semicir- cles of radii a1 + a2 , a1, a2 . Property 6.1. If all semi ellipsoids enclosing 3D-ellarbelos are hemispheres, then the difference between the areas of the upper surface and the lower one equals double the area of the arbelos bounded by semicircles of the same radii as the hemispheres. Property 6.2. The regular 3D-ellarbelos has the same volume as the cylinder whose base is the revolved ellarbelos and the height equals the diameter of the ellarbelos (see Fig. 10). Proof: Denote by V and V0 , respectively, the volumes of the regular 3D-ellarbelos and the largest semiellipsoid. Then
3 3 2 V =V0 − λ V0 − 1( − λ) V0 = λ 1(3 − λ)V0 = λ 1(2 − )πλ ab , where λ = a1 / a = b1 b./ By (4) the area of the ellarbelos equals S E = λ 1( − )πλ ab, so indeed V = S E ⋅ 2a = S E ⋅ (2 a1 + a2 ).
132 M. Różański, A. Samulewicz, M. Szweda, R. Wituła
Fig. 10. Equal volumes for a 3D-arbelos composed of spheres and an arbelosoidal cylinder
One can claim that increasing the dimension should result in increasing the number of spheres. However, the case with four pairwise tangent hemispheres seems much more complicated than the ones considered previously. Even deter- mining the relationships between their radii is more complex.
Denote by R the radius of the outer sphere and by r1 r2 ,, r3 the radii of the inner spheres. Using Heron's formula to calculate the areas of the appropriate triangles, we obtain
(r1 + r2 + ) rrrr 3213 = R(R − r1 − r ) rr 212 + R(R − r1 − ) rrr 313 + R R − r − rrr ()2 323 (12) and, consequently,
R= rrr123/(2( rrrr 1231 ++−−− r 2 r 3 ) rr 12 rr 13 rr 23 ) (13) The formula (13) can be derived from Descartes' theorem as well ([9, chapter 4.1]):
2 2 2 2 2 (c1 + c2 + c3 + cR ) = 2(c1 + c2 + c3 + cR ), 1 1 where ck = , k = 3,2,1 and cR −= . rk R Question 6.3. Which properties of abelos (parbelos, F-arbelos, ellarbelos, 3D-ellarbelos) have 3D-arbelos with four spheres?
7. Conclusions
The present paper concerns the generalizations of the ancient notion of arbelos. The analogues known from other articles (parbelos, F-arbelos) have been briefly
Variations on the arbelos 133 described. We discuss in detail elliptic arbelos (ellarbelos). As far as we know, they can be useful in determining the static moments of arc rod constructions or in prob- lems of structural stability and durability of constructions. We also undertook an attempt of generalizing the notion of arbelos to the three-dimensional case, never- theless there is still a number of problems we are going to proceed to study. We hope that our results may be helpful to engineers, especially mechanical ones, and we look forward to being inspired by their work.
References
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