A translation of Zalgaller's "The shortest space curve of unit width" (1994)
aa r X i v : . [ m a t h . HO ] O c t A translation of Zalgaller’s “The shortest space curve of unit width”(1994)
Steven Finch
October 7, 2019
Abstract.
This is an English translation of V. A. Zalgaller’s article“On a problem of the shortest space curve of unit width” that appeared in
Matematicheskaya Fizika, Analiz, Geometriya v. 1 (1994) n. 3–4, 454–461.We refer interested readers to Ghomi (2018) for up-to-date discussion; thecurve L of length 3 . ... in Zalgaller (1994) still appears to be shortest,whereas the closed curve L is provably not of unit width. I am thankful toNatalya Pluzhnikov for her dedicated work and to the B. Verkin Institute forLow Temperature Physics and Engineering of the National Academy of Sciencesof Ukraine for permission to post this translation on the arXiv.
1. The width of a space curve is the width of its convex hull. The class of curvesin R of bounded width 1 is nonempty and compact; therefore, it contains at leastone curve of smallest length. We pose the question of finding such a curve and statea conjecture on its shape. For planar curves this question was solved long time ago(see [1] or [2]).2. Consider a unit cube (Figure 1). The three-link polygonal line L = ABCD is an example of a curve of finite length and width 1. (The convex hull of L is aregular tetrahedron of width 1.) The length is | L | = 3 √ ≈ . C -smooth curves. Let L be such a curve, and Φ itsconvex hull. Each support plane to Φ has contact with L . The points on the surfaceof the solid body Φ may be conic, ridged, or smooth. Only points of L itself or ofsegments joining pairs of its corner points may be ridged. To verify that Φ has widthat least 1, it is sufficient to check that for every support plane P to Φ that passesthrough a ridged point of the surface Φ there is a point x of L at a distance at least1 from P .For the polygonal line L this verification is trivial.4. We determine a curve of width 1 that is shorter than L , using first the followingapproach. We fix three real numbers that satisfy the following six conditions: a >
12 ; b < a ; c <
1; (1-3)1 translation of Zalgaller’s “The shortest space curve of unit width” (1994) c + t ≥ , where t = ( a − b ) / √ a −
1; (4)4 a (cid:0) c + t (cid:1) > ( a + b ) + c + t ; (5)4 (cid:0) a t + b c + c t (cid:1) > ( a + b ) + c + t . (6)Then, in Cartesian coordinates ( x, y, z ), we choose four points: A ( a, , , B ( − b, − t, c ) , C ( b, t, c ) , D ( − a, , . (7)Projected to the plane z = 0, these points form a parallelogram with vertices A ( a, , , B ( − b, − t, , C ( b, t, , D ( − a, , . A suitable choice of t ensures that the sides AC and DB of this parallelogram areat distance 1 / AB and DC are at distance d = (cid:12)(cid:12)(cid:12) −−→ B A × −−→ B D (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −−→ B A (cid:12)(cid:12)(cid:12) = 2 at p ( a + b ) + t < translation of Zalgaller’s “The shortest space curve of unit width” (1994) d , d of points A and B fromthe straight line DC : d = (cid:12)(cid:12)(cid:12) −−→ DA × −−→ DC (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −−→ DC (cid:12)(cid:12)(cid:12) = 2 a √ c + t p ( a + b ) + c + t ,d = (cid:12)(cid:12)(cid:12) −−→ DB × −−→ DC (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −−→ DC (cid:12)(cid:12)(cid:12) = 2 √ a t + b c + c t p ( a + b ) + c + t . According to conditions (5) and (6), d > d > Z , Z , Z in space, of radii 1. Their axesare the straight lines AB , AD , DC , where the points A , B , C , D are those in (7).We construct the curve L ( a, b, c ) from three C -smooth parts g AB , g BC , g CD . Thepart g AB will be the shortest among the contours that join the points A and B andencompass the cylinder Z from the outside. This shortest contour ”spans” Z , andhence it consists of a segment AP , an arc ] P Q of a helix on Z , and the segment Q B .Similarly, the part g BC = BP + ] P Q + Q C is the shortest contour among thosethat join B and C and enclose the cylinder Z . Here BP , Q C are straight linesegments and ] P Q is an arc of a helix on Z . Finally, the part g CD = CP + ] P Q + Q D is the mirror image of g AB with respect to the coordinate axis z . Figure 2 showsan approximate shape of the curve L ( a, b, c ).We calculate the length of the curves and minimize it by suitable choice of thevalues of a , b , and c .6. To find the lengths of the parts under consideration, we consider a situationwhere on the sides of the circular cylinder Z of radius 1 there are points E and F whose positions with respect to Z in the projection to the plane of a cross-section(Figure 3) are determined by the magnitudes h = E ′ E ′′ = F ′ F ′′ , u = OE ′′ , v = OF ′′ , ( h < , h + u ≥ , h + v ≥ , and the projections of the points E and F to the axis of Z are at distance w fromthis axis. In this case, in the same notation (which is clear from Figure 3) we have ϕ = arctan hu , ϕ = arctan hv , ψ = arccos 1 √ h + u ,ψ = arccos 1 √ h + v , p = ] P ′ Q ′ = π − ϕ − ϕ − ψ − ψ , (9) translation of Zalgaller’s “The shortest space curve of unit width” (1994) translation of Zalgaller’s “The shortest space curve of unit width” (1994) p = E ′ P ′ = √ h + u − , p = Q ′ F ′ = √ h + v − . Unfolding the cylinder (Figure 4) over the planar curve E ′ P ′ Q ′ F ′ (Figure 3) wefind the length ℓ ( h, u, v, w ) of the space curve that joins the points E and F , encom-passing the cylinder Z from the outside: ℓ ( h, u, v, w ) = p ( p + p + p ) + w . (10)7. The length ℓ of the arc g BC is especially easy to find. In this case h = c , u = v = t , w = 2 b , whence ℓ = ℓ ( c, t, t, b ) . (11)8. To find the length ℓ of the part g AB , we draw a plane τ through the points D , C , C and consider the projection (shown in Figure 5) to the plane of the cross-sectionof the cylinder Z that passes through the point D . Here A ′ A ′′ = B ′ B ′′ = d (see (8)).Let ν denote the normal vector to the plane τ , ν = −→ BA × −−→ DC c = ( t, − ( a + b ) , . Then the vector r = −−→ DC × νc = ( a + b, t, − s ) , where s = (( a + b ) + t ) /c , is the direction vector of a straight line orthogonal to −−→ DC , in the plane τ . The magnitude of the projection of the vector −−→ DA in the direction r gives us the value u = DA ′′ (Figure 5): u = −−→ DA · r | r | = 2 a ( a + b ) p ( a + b ) + t + s . (12)Similarly, v = DB ′′ = −−→ BD · r | r | = 2 ( b + ab + t ) p ( a + b ) + t + s . (13)Finally, w = −→ BA · −−→ DC (cid:12)(cid:12)(cid:12) −−→ DC (cid:12)(cid:12)(cid:12) = ( a + b ) − c + t p ( a + b ) + c + t (14)and the length of the part g AB is ℓ = ℓ ( d, u , v , w ) . translation of Zalgaller’s “The shortest space curve of unit width” (1994) translation of Zalgaller’s “The shortest space curve of unit width” (1994)
79. The total length is | L ( a, b, c ) | = 2 ℓ + ℓ .Computer analysis shows that the length | L ( a, b, c ) | for admissible values of a , b , c attains a minimum at the point ( a , b , c ) where a ≈ . , b ≈ . , c ≈ . . The curve L = L ( a , b , c ) has length | L | ≈ . L with a shorter curve. Since the distance d between the parallel planes ABB and DCC (Figure 2), in view of (8), is less than1, we change the construction of the part g AB of the curves L ( a, b, c ) as follows: wewill encompass, from the outside, not the cylinder Z but the one that is left afterwe cut Z by the plane parallel to τ , at distance (1 + d ) / τ . Then Figure 5changes to Figure 6. The projection of the shorter arc g AB enclosing the truncatedcylinder to the plane of a cross-section of the cylinder (Figure 6) is the polygonal line A ′ M ′ N ′ B ′ spanned by the rectangle M ′′ M ′ N ′ N ′′ (shaded in Figure 6), where DM ′′ = DN ′′ = √ DN ′ − N ′ N ′′ = 12 √ − d − d . (15)11. Consider the curves L ( a, b, c ) for which the part g BC is constructed in thesame way as for L ( a, b, c ), and the part g AB is the shortest three-link polygonal Here c + t ≈ . BP , Q C on the part g BC are very short, but not zero. translation of Zalgaller’s “The shortest space curve of unit width” (1994) AM N B that encompasses from the outside not the cylinder Z , and not eventhe truncated cylinder, but only the parallelepiped with cross-section the rectangle M ′′ M ′ N ′ N ′′ (Figure 6).To find the length ℓ of this part g AB we obtain, consecutively, q = M ′ N ′ = √ − d − d , q = A ′ M ′ = s(cid:18) − d (cid:19) + (cid:16) u − q (cid:17) ,q = N ′ B ′ = s(cid:18) − d (cid:19) + (cid:16) v − q (cid:17) , ℓ = p ( q + q + q ) + w , where the values d , u , v , w are determined from (8), (12), (13), and (14).12. The total length is | L ( a, b, c ) | = 2 ℓ + ℓ .Computer analysis shows that the length | L ( a, b, c ) | for admissible a , b , c attainsa minimum at the point ( a , b , c ) where a ≈ . , b ≈ . , c ≈ . . Conjecture 1 . The curve L = L ( a , b , c ) is the shortest space curve ofwidth 1 in R .14. For the above-indicated values of a , b , c , we have t ≈ . d ≈ . g BC we have ϕ = ϕ ≈ . ψ = ψ ≈ . p = p ≈ . p ≈ . ℓ ≈ . BP = Q C = ℓ p / ( p + p + p ) ≈ . g AB , u ≈ . , v ≈ . , w ≈ . , q ≈ . ,q ≈ . , q ≈ . , ℓ ≈ . . The total length | L | = 2 ℓ + ℓ = 3 . ∠ DM ′ A and ∠ DN ′ B in Figure 6 are greater than π/
2. Therefore, the curve L , when it encloses the parallelepiped with cross-section M ′′ M ′ N ′ N ′′ , also encloses the cylinder Z truncated by M ′ N ′ . This allows us toverify that the width of the curve L is not less than 1.15. To give a better description of L , we find the coordinates of the points M and N . To this end, we expand the vectors −−→ AM and −−→ BN (see Subsection 8 and Figure 7)with respect to the basis ( r, −−→ DC, ν ): −−→ AM = − r | r | (cid:16) u − q (cid:17) − −−→ DC (cid:12)(cid:12)(cid:12) −−→ DC (cid:12)(cid:12)(cid:12) w q q + q + q + ν | ν | − d , translation of Zalgaller’s “The shortest space curve of unit width” (1994) −−→ BN = r | r | (cid:16) v − q (cid:17) − −−→ DC (cid:12)(cid:12)(cid:12) −−→ DC (cid:12)(cid:12)(cid:12) w q q + q + q + ν | ν | − d . From this, for L we obtain M ≈ (0 . , − . , . ,N ≈ (0 . , − . , . . The general form of the curve L is shown in Figure 8.16. A similar question can be posed for closed curves. In the plane, any curve ofconstant width 1 is the shortest closed curve of width 1. In space, consider a regulartetrahedron with edge √ L = ABCDA is an example of a closed curve of width 1. Mark the midpoints O , O , O , O ofthe sides of L . On the edge AC , which is not part of L , we mark the point A ′ atdistance 1 from the plane BCD , and the point C ′ at distance 1 from the plane ABD .Similarly, on the edge BD , which also is not part of L , we mark the point B ′ atdistance 1 from the plane ACD , and D ′ at distance 1 from the plane ABC .We compose L from four congruent C -smooth parts A ′ B ′ , B ′ C ′ , C ′ D ′ , D ′ A ′ .It suffices to describe the structure of A ′ B ′ . It is constructed as the shortest curve A ′ P O Q B ′ that joins the points A ′ and B ′ and encloses the circular cylinder Z ofradius 1 with axis CD . Thus, this part has the form A ′ B ′ = A ′ P + P O Q + Q B ′ , translation of Zalgaller’s “The shortest space curve of unit width” (1994) translation of Zalgaller’s “The shortest space curve of unit width” (1994) A ′ P and Q B ′ are straight line segments, and P O Q is an arc of the helixon Z . The parts B ′ C ′ = B ′ P O Q C ′ , C ′ D ′ = C ′ P O Q D ′ , D ′ A ′ = D ′ P O Q A ′ are constructed in a similar way.17. Conjecture 2 . The curve L has width 1 and is the shortest closed spacecurve of width 1 in R .The length | L | can be easily found by the rules of Subsection 6.The author is grateful to computer programmer V. G. Khachaturov for his helpin preparing this article. The work was supported by the RFFR grant 94-01-0104.Written by V. A. Zalgaller, St. PetersburgTranslated by Natalya Pluzhnikov Addendum
The denominator of d in (8) was mistakenly given by Zalgaller (1994) as (cid:12)(cid:12)(cid:12) −−→ B D (cid:12)(cid:12)(cid:12) ;I suspect that this error is merely typographical because the subsequent radical ex-pression for d was correct (except for a stray negative sign). The radical expressionfor d was missing a 2 in the numerator.In Subsection 8, a label (14) was absent with regard to the formula for w and acomma was missing between v and w in the definition of ℓ .In Subsection 14, the numerical values of q and q were wrongly interchanged.Ghomi [3, 4] recently disproved Zalgaller’s Conjecture 2. References [1] V. A. Zalgaller, How to get out of the woods? On a problem of Bellman (inRussian),
Matematicheskoe Prosveshchenie , v. 6 (1961) n. 6, 191–195.[2] A. Adhikari and J. Pitman, The shortest planar arc of width 1,
Amer. Math.Monthly
96 (1989) 309–327; MR0992078.[3] M. Ghomi, The length, width, and inradius of space curves,
Geom. Dedicata196(2018) 123–143; arXiv:1605.01144; MR3853631.[4] M. Ghomi, Open problems in geometry of curves and surfaces, in progress (2019),http://people.math.gatech.edu/˜ghomi/Papers/.Steven FinchMIT Sloan School of ManagementCambridge, MA, USA