An ε -characterization of a vertex formed by two non-overlapping geodesic arcs on surfaces with constant Gaussian curvature
aa r X i v : . [ m a t h . G M ] A p r AN ǫ CHARACTERIZATION OF A VERTEX FORMED BY TWONON-OVERLAPPING GEODESIC ARCS ON SURFACES WITHCONSTANT GAUSSIAN CURVATURE
ANASTASIOS N. ZACHOS
Abstract.
We determine a positive real number (weight) which correspondsto the intersection point (vertex) of two non-overlapping geodesic arcs, whichdepends on the two weights which correspond to two points of these geodesicarcs, respectively, and an infinitesimal number ǫ. As a limiting case for ǫ → , the triad of the corresponding weights yields a degenerate weighted Fermat-Torricelli tree which coincides with these two geodesic arcs. By applying thisprocess for a geodesic triangle on a circular cone, we derive an ǫ characteriza-tion of conical points in R . Introduction
Let △ A A A be a geodesic triangle on a surface S with constant Gaussiancurvature in R . We denote by A a point on S, by ( a ij ) g the length of the geodesicarc A i A j by ∠ A i A j A k the angle formed by the geodesic arcs A i A j and A j A k , atthe vertex A j , by ~U ij is the unit tangent vector of the geodesic arc A i A j at A i , for i, j, k = 0 , , , B i a positive real number (weight), which corresponds toeach vertex A i , for i = 1 , , . We state the weighted Fermat-Torricelli problem for a △ A A A on S. Problem 1.
Find a point A (weighted Fermat-Torricelli point) on S, such that f ( A ) = X i =1 B i ( a i ) g → min . (1.1)The solution of the weighted Fermat-Torricelli problem is called a branching so-lution (weighted Fermat-Torricelli tree) and consists of the three geodesic branches { A A , A A , A A } which meet at the weighted Fermat-Torricelli point A . The following two propositions give a characterization of the weighted Fermat-Torricelli point A on a smooth surface, which have been proved in [2],[4]). Proposition 1 (Floating Case) . [2] , [4] ) The following (I), (II), (III) conditionsare equivalent:(I) All the following inequalities are satisfied simultaneously: (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) > B , (1.2) Mathematics Subject Classification.
Key words and phrases. weighted Fermat-Torricelli problem, weighted Fermat-Torricelli point,surfaces with constant Gaussian curvature, sphere, hyperboloid, circular cylinder, circular cone. (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) > B , (1.3) (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) > B , (1.4) (II) The point A is an interior point of the triangle △ A A A and does not belongto the geodesic arcs A i A j for i, j = 1 , , (III) B ~U + B ~U + B ~U = ~ . Proposition 2 (Absorbed Case) . [2] , [4] ) The following (I), (II) conditions areequivalent.(I) One of the following inequalities is satisfied: (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) ≤ B , (1.5) or (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) ≤ B , (1.6) or (cid:13)(cid:13)(cid:13) B ~U + B ~U (cid:13)(cid:13)(cid:13) ≤ B . (1.7) (II) The point A is attained at A or A or A , respectively. The inverse weighted Fermat-Torricelli problem on S states that: Problem 2. [2] , [4] Given a point A which belongs to the interior of △ A A A on M , does there exist a unique set of positive weights B i , such that B + B + B = c = const, for which A minimizes f ( A ) = B ( a ) g + B ( a ) g + B ( a ) g A positive answer w.r. to the inverse weighted Fermat-Torricelli problem on a C complete surface is given by the following proposition ([2],[4]): Proposition 3. [2] , [4] The weight B i are uniquely determined by the formula: B i = C sin ∠ A i A A j sin ∠ A j A A k + sin ∠ A i A A k sin ∠ A j A A k , (1.8) for i, j, k = 1 , , and i = j = k. If B , B , B satisfy the inequalities of the floating case, we derive a weightedFermat-Torricelli tree { A A , A A , A A } . The location of the weighted Fermat-Torricelli (floating) tree for geodesic trian-gles on the K plane (sphere S K , hyperboloid H ) is given in [5], [6] and an analyticalsolution of the weighted Fermat-Torricelli problem for an equilateral geodesic trian-gle with equal lengths π is given in [8], for the weighted floating case. Concerning,the solution of the weighted Fermat-Torricelli problem for geodesic triangles on flatsurfaces of revolution (Circular cylinder, circular cone) we refer to [7].If B , B , B satisfy one of the inequalities of the absorbed case (1.5) or (1.6)or (1.7), we obtain a degenerate weighted Fermat-Torricelli tree { A A , A A } , { A A , A A } and { A A , A A } , respectively.For instance, if (1.5) is valid, the minimum value of B is determined by: N ǫ CHARACTERIZATION OF A VERTEX ON SURFACES 3 B = B + B + 2 B B cos ∠ A A A or B = f ( B , B ) . Thus, we consider the following problem:
Problem 3.
How can we determine the values of B , B , such that f ( B , B ) gives the minimum value of B that corresponds to the vertex A on a surface withconstant Gaussian curvature S ? In this paper, we determine the value of B by introducing an infinitesimal realnumber ǫ, ( ǫ characterization of A ) such that: ∠ A A A = k ǫ k , ∠ A A A = ∠ A A A + 2 ǫ, ∠ A A A = k k ǫ k for a rational number k, by applying the solutionof the inverse weighted Fermat-Torricelli problem on S. By setting A to be thevertex of a (right) circular cone, we give an ǫ characterization of conical points in R . An ǫ characterization of the vertices of a triangle in R Let A be an interior point of △ A A A in R . We denote by ( a ij ) the length of the linear segment A i A j . We set ∠ A A A = ǫ, ∠ A A A = ∠ A A A + 2 ǫ ∠ A A A = kǫ. Theorem 1.
The weight B i = B i ( ǫ ) are uniquely determined by the formula: B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (2.1) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (2.2) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (2.3) where ∠ A A A = ∠ A A A + 2 ǫ (2.4) ∠ A A A = ∠ A A A ( ǫ ) = arccot( − ( a ) + ( a ) cos( ∠ A A A + 2 ǫ )( a ) sin( ∠ A A A + 2 ǫ ) ) , (2.5) ∠ A A A = 2 π − ∠ A A A − ǫ − ∠ A A A ( ǫ ) . (2.6) Proof.
From △ A A A , we obtain: ∠ A A A − ǫ + ∠ A A A + 2 ǫ + ∠ A A A − kǫ = π, which yields k = 1 . By applying the law of sines in △ A A A , △ A A A , we derive: ANASTASIOS N. ZACHOS ( a ) sin ǫ = ( a ) sin ∠ A A A = ( a ) − sin( ∠ A A A + 2 ǫ + ∠ A A A ) . (2.7)From (2.7), we obtain (2.5).By replacing (2.5), (2.4), (2.6) in (1.8), we obtain B i = B i ( ǫ ) , for i = 1 , , . (cid:3) Corollary 1.
For ǫ → , we derive a degenerate weighted Fermat-Torricelli tree { A A , A A } . An ǫ characterization of the vertices of a geodesic triangle onthe K -plane Let △ A A A be a geodesic triangle on the K plane. The K plane is a sphere S K of radius R = √ K and a hyperbolic plane H K with constant Gaussian curvature − K for K < R . We set κ = (cid:26) √ K if K > i √− K if K < K plane is given in [1].We set ∠ A A A = k ǫ k , ∠ A A A = ∠ A A A + 2 ǫ, ∠ A A A = k ǫ k . We set ǫ = (cid:26) k ǫ k if K > −k ǫ k if K < Theorem 2.
The weight B i = B i ( ǫ ) are uniquely determined by the formula: B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (3.1) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (3.2) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (3.3) where ∠ A A A = ∠ A A A + 2 ǫ (3.4) ∠ A A A = arccot( − sin( κ ( a ) g ) + 2 sin( κ ( a ) g ) cos( k ǫ k ) cos( ∠ A A A + 2 ǫ )2 sin κ ( a ) g ) cos( k ǫ k ) sin( ∠ A A A + 2 ǫ ) ) , (3.5) ∠ A A A = 2 π − ∠ A A A − ǫ − ∠ A A A ( ǫ ) . (3.6) Proof.
By applying the law of sines in △ A A A , we get:sin( κ ( a ) g )sin k ǫ k = sin( κ ( a ) g )sin ∠ A A A (3.7)or N ǫ CHARACTERIZATION OF A VERTEX ON SURFACES 5 sin( κ ( a ) g )sin | ǫ k
12 cos | ǫ k = sin( κ ( a ) g )sin ∠ A A A . (3.8)By applying the law of sines in △ A A A , we get:sin( κ ( a ) g )sin k ǫ k = − sin( κ ( a ) g )sin( ∠ A A A + 2 ǫ + ∠ A A A ) (3.9)By replacing (3.9) in (3.8), we derive: − sin( κ ( a ) g )sin( ∠ A A A + 2 ǫ + ∠ A A A ) 12 cos | ǫ k = sin( κ ( a ) g )sin ∠ A A A . (3.10)By solving (3.10) w.r. to ∠ A A A , we obtain (3.5). (cid:3) An ǫ characterization of the vertices of a geodesic triangle onflat surfaces of revolution In this section, we shall give an ǫ characterization of the vertices of geodesictriangles on flat surfaces of revolution (Circular Cylinder S and Circular Cone C )which are flat Euclidean Surfaces with zero Gaussian curvature.4.1. An ǫ characterization of the vertices of a geodesic triangle on a cir-cular cylinder. The parametric form of a (right) circular cylinder S of unit radiusand axis (axis of revolution) the z-axis: ~r ( u, v ) = (cos v, sin v, u ) . The geodesics of the circular cylinder are the straight lines on the circular cylinderparallel to the z -axis, the circles obtained by intersecting the circular cylinder withplanes parallel to the xy -plane and circular helixes of the parametric form ~r ( t ) =(cos t, sin t, bt + c ) . Let △ A A A be a geodesic triangle on S which is composed of three circularhelixes.We set A = (1 , , , A = (cos ϕ , sin ϕ , z ) A = (cos ϕ , sin ϕ , z ) and ~r ij = (cos t, sin t, b ij t ) the circular helix on S from A i to A j for i, j = 1 , , , i = j and ϕ , ϕ ∈ (0 , π ) . The coefficient b ij is called the step of the helix from A i to A j . The step of thehelices b from A to A and b from A to A are given by: b = z ϕ and b = z ϕ . A cylindrical law of cosines for geodesic triangles on S composed of three circularhelixes is given in [7]. Proposition 4. [7]
The following formula holds for △ A A A on S: (1 + b )( ϕ − ϕ ) = (1 + b ) ϕ + (1 + b ) ϕ − q (1 + b )(1 + b ) ϕ ϕ cos α . (4.1)We denote by ( a ij ) S the length of the geodesic arc from A i to A j . ANASTASIOS N. ZACHOS
Theorem 3.
The weight B i = B i ( ǫ ) are uniquely determined by the formula: B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.2) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.3) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.4) where ∠ A A A = ∠ A A A + 2 ǫ (4.5) ∠ A A A = ∠ A A A ( ǫ ) = arccot( − p b ϕ + p b ϕ cos( ∠ A A A + 2 ǫ )( p b ϕ sin( ∠ A A A + 2 ǫ ) ) , (4.6) ∠ A A A = 2 π − ∠ A A A − ǫ − ∠ A A A ( ǫ ) . (4.7) Proof.
Unrolling the cylinder S in terms of the vertex A , we derive an isometricmapping from S to R , which yields:( a ij ) S = ( a ij ) , f or i, j = 1 , , . (4.8)From (4.8), we have: ( a ) S = ( a ) (4.9)and ( a ) S = ( a ) (4.10)The Euclidean distances ( a ) and ( a ) are given by:( a ) = q z + ϕ (4.11)or ( a ) = q b + 1 ϕ (4.12)and ( a ) = q z + ϕ (4.13)or ( a ) = q b + 1 ϕ (4.14)Thus, the length of the circular helix with parametric form ~r from A to A and with parametric form ~r from A to A is given by:( a ) s = q b ϕ . (4.15)and ( a ) s = q b ϕ . (4.16) N ǫ CHARACTERIZATION OF A VERTEX ON SURFACES 7
By replacing (4.15) and (4.16) in 2.5, we obtain (4.6) and by applying Theorem 1,we derive (4.2), (4.3) and (4.4). The weights B , B , B depend on the angle ∠ A A A , ǫ and the step of the helices b and b . (cid:3) An ǫ characterization of the vertices of a geodesic triangle on a cir-cular cone. We consider the parametric form of a (right) circular cone C with aunit base radius. ~r ( u, v ) = (cid:0)(cid:0) − uH (cid:1) cos v, (cid:0) − uH (cid:1) sin v, u (cid:1) , < u ≤ H, < v < π. The geodesic equations on C are given in [3, Exercise 5.2.14, Subsection 5.6.2pp. 222, pp. 247-248].Let △ A A A be a geodesic triangle on C. We denote by P the center of the unit bases circle of S ′ , H the distance AP, by A ip the intersection of the line defined by the linear segment AA i with the unitbases circle c ( P,
1) for i = 0 , , , , by ϕ the angle ∠ A P A p by ϕ the angle ∠ A P A p and by ϕ the angle ∠ A P A p . By unrolling the circular cone C w.r. to A A, (cut along A A ) we derive anisometric mapping from C to R . Thus, we get: ( a ij ) g = ( a ij ) (4.17)By setting A = (0 , , we obtain: ϕ = 2 π √ H , (4.18) ∠ A AA = ϕ √ H , (4.19) ∠ A AA = ϕ √ H (4.20)and ∠ A AA = ϕ √ H . (4.21)where A i A = p H . (4.22) Theorem 4.
The weight B i = B i ( ǫ ) are uniquely determined by the formula: B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.23) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.24) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.25) where ∠ A A A = ∠ A A A + 2 ǫ (4.26) ANASTASIOS N. ZACHOS ∠ A A A = ∠ A A A ( ǫ ) = arccot( − ( a ) c + ( a ) c cos( ∠ A A A + 2 ǫ )( a ) c sin( ∠ A A A + 2 ǫ ) ) , (4.27) ∠ A A A = 2 π − ∠ A A A − ǫ − ∠ A A A ( ǫ ) . (4.28) and ( a ) c = r (1 + H ) + ( A A ) − p H ( A A ) cos( ϕ √ H ) , (4.29)( a ) c = r (1 + H ) + ( A A ) − p H ( A A ) cos( ϕ √ H ) . (4.30) Proof.
From the law of cosines in △ A AA , we get:( a ) c = p ( A A ) + ( A A ) − A A )( A A ) cos( ∠ A AA ) . (4.31)By replacing (4.22) and (4.19) in (4.31), we obtain (4.29).From the law of cosines in △ A AA , we get:( a ) c = p ( A A ) + ( A A ) − A A ( A A ) cos( ∠ A AA ) . (4.32)By replacing (4.22) and (4.20) in (4.32), we obtain (4.30). Unrolling C along A A yields an isometric mapping from C to R and by applying theorem 1, weobtain (4.33), (4.34) and (4.35). The weights B , B , B depend on ǫ, ϕ , ϕ and H. (cid:3) By setting A ≡ A, we obtain an ǫ characterization of the conical vertex A in R . Proposition 5.
The weight B i = B i ( ǫ ) are uniquely determined by the formula: B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.33) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.34) B = C sin ∠ A A A sin ∠ A A A + sin ∠ A A A sin ∠ A A A , (4.35) where ∠ A A A = ∠ A A A + 2 ǫ (4.36) ∠ A A A = ∠ A A A ( ǫ ) = arccot( − AA + AA cos( ∠ A A A + 2 ǫ )( AA sin( ∠ A A A + 2 ǫ ) ) , (4.37) ∠ A A A = 2 π − ∠ A A A − ǫ − ∠ A A A ( ǫ ) . (4.38) Remark 1.
The deviation of k B − B ( ǫ ) k where B = p B ( ǫ ) + B ( ǫ ) + 2 B ( ǫ ) B ( ǫ ) cos ∠ A A A gives an error estimate which depends on ǫ. N ǫ CHARACTERIZATION OF A VERTEX ON SURFACES 9
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