Hitting probabilities for random convex bodies and lattices of triangles
aa r X i v : . [ m a t h . P R ] D ec Hitting probabilities for random convex bodies andlattices of triangles
Uwe B¨asel
Abstract
In the first part of this paper, we obtain symmetric formulae for theprobabilities that a plane convex body hits exactly 1, 2, 3, 4, 5 or 6triangles of a lattice of congruent triangles in the plane. Furthermore,a very simple formula for the expectation of the number of hit trianglesis derived. In the second part, we calculate the hitting probabilitiesin the cases where the convex body is a rectangle, an ellipse and ahalf disc. Already known results for a line segment (needle) follow asspecial cases of the rectangle and the ellipse.
Keywords:
Integral geometry, geometric probabilities, random con-vex sets, hitting probabilities, intersection probabilities, lattice of tri-angles, Buffon’s problem, Buffon’s needle problem, rectangle, half disc
We consider the random throw of a planar convex body C onto a plane withan unbounded lattice R a, b, c of congruent triangles with sides a , b , c , andopposite angles α , β , γ , respectively. (An example is shown in Fig. 1.) Asusual, we denote the altitudes from the sides a , b , c respectively as h a , h b , h c , and have h a = b sin γ = c sin β , h b = a sin γ = c sin α , h c = a sin β = b sin α . R a, b, c is the union of three families R h a , R h b and R h c of parallel lines, R h a = { ( x, y ) ∈ R | x sin β + y cos β = kh a , k ∈ Z } , R h b = { ( x, y ) ∈ R | x sin α − y cos α = kh b , k ∈ Z } , R h c = { ( x, y ) ∈ R | y = kh c , k ∈ Z } . Every triangle of R a, b, c is congruent to the triangle T := { ( x, y ) ∈ R | ≤ y ≤ h c , y cot α ≤ x ≤ c − y cot β } . Two triangles form a parallelogram that is congruent to the parallelogram Q := { ( x, y ) ∈ R | ≤ y ≤ h c , y cot α ≤ x ≤ c + y cot α } , Q of Q is given by Q = ab sin γ = ac sin β = bc sin α . (1)Let C be provided with a fixed reference point O inside C and a fixed orientedline segment σ starting in O (see Fig. 4), and let φ be the angle between afixed direction in the plane and segment σ . The random throw of C onto R a, b, c is defined as follows: The coordinates x , y of O are random variablesuniformly distributed in [ y cot α, h b csc α + y cot α ] and [0 , h c ], respectively;the angle φ is a random variable uniformly distributed in [0 , π ]. All threerandom variables are stochastically independent. Let u denote the perimeterof C , F the area of C , and s : R → R , φ s ( φ ) the support function of C with reference to the point O . We use s in the following sense (see Fig. 2): C rotates about the origin of a Cartesian x, y -coordinate system with rotationangle φ between the x -axis and segment σ , whereby O coincides with theorigin. The support line, touching C , is perpendicular to the x -axis. Now,the support function s is the distance between the origin and the supportline. In [11, pp. 2-3], the support function, say ˜ s , is defined for fixed C andmoving support line. Clearly, the relation between these two definitions isgiven by s ( φ ) = ˜ s ( − φ ) if the fixed body C is in the position with φ = 0, and O coinciding with the origin. The width of C in the direction φ is given by w : R → R , φ w ( φ ) = s ( φ ) + s ( φ + π ).Fig. 1: Lattice R a, b, c and convex body C Markov [6, pp. 169-173] calculated the probability p (1) that a line segment(needle) of length ℓ hits exactly one triangle of R a, b, c . He found p (1) = 1 + ℓ ( αa + βb + γc )2 πQ − ℓ (4 a + 4 b + 4 c − ℓ )2 πQ . (2)Pettineo [7, pp. 289-294] and B¨ottcher [3, pp. 3-10] calculated the probabil-ity p that the needle hits at least one line of R a, b, c . Clearly, p = 1 − p (1). Al-though giving equal results, the formulae of Markov, Pettineo, and B¨ottcher2re quite different. Santal´o [10, pp. 167-170] (see also [11, pp. 140-141]) cal-culated the probabilities that the needle hits 0, 1, 2 or 3 lines of the lattice R a, a, a of equilateral triangles. Duma and Stoka [4, pp. 16-20] obtained theprobabilities that C hits at least one line, and exactly three lines of R a, b, c .Ren and Zhang [8, pp. 320-321] (see also [9, pp. 72-73]) computed the prob-abilities for the number of hit lines for a bounded convex body C with nosize restriction, and a lattice of parallelograms. B¨asel and Duma [1], [2]calculated the probabilities that C hits exactly 1, 2, 3 or 4 parallelogramsfor two kinds of parallelogram lattices. In the following, we choose φ as the angle between the direction perpendicu-lar to the sides b and segment σ (see Fig. 4). We assume that for every angle φ , 0 ≤ φ < π , there is a position of the reference point O so that C withangle φ , denoted as C φ , is entirely contained in exactly one triangle T of R a, b, c . Therefore, we consider the triangle T ∗ ( φ ) that is similar to T , andevery side of T ∗ ( φ ) touches C φ (see Fig. 3). The condition that C φ is entirelycontained in T is equivalent to the fact that T ∗ ( φ ) is smaller than T . Hencethe length of side c of T is greater than the length of the respective side c ∗ ( φ ) of T ∗ ( φ ). Using Fig. 3, we find c ∗ ( φ ) = s ( φ ) csc α + s ( φ + α + β ) csc β + s ( φ + α + π )(cot α + cot β ) . So we have max ≤ φ< π c ∗ ( φ ) ≤ c , (3)(The equals sign does not influence the calculation of the probabilities.) Theradius ρ of the incircle of T is given by ρ = Q/ ( a + b + c ) [5, p. 6]. Clearly,if max ≤ φ<π w ( φ ) ≤ ρ (cf. [4, p. 16]), then condition (3) always holds true.Fig. 2: Support function s Fig. 3: Triangle T ∗ ( φ )3 heorem 1. If condition (3) holds true, the probabilities p ( i ) that C hitsexactly i triangles of R a, b, c are given by p (1) = 1 − ( a + b + c ) uπQ + ( a + b + c ) J (0)2 πQ + bcf ( α ) + caf ( β ) + abf ( γ ) πQ ,p (2) = ( a + b + c ) uπQ − a + b + c ) J (0)2 πQ − bcf ( α ) + caf ( β ) + abf ( γ ) πQ ,p (3) = 3( a + b + c ) J (0)2 πQ + bcf ( α ) + caf ( β ) + abf ( γ ) πQ ,p (4) = bcJ ( α ) + caJ ( β ) + abJ ( γ ) πQ − ( a + b + c ) J (0)2 πQ − FQ ,p (5) = 0 , p (6) =
FQ , with f ( x ) = I ( x ) − J ( x ) , f ( x ) = 2 I ( x ) − J ( x ) , f ( x ) = I ( x ) − J ( x ) ,I ( x ) = Z π w ( φ ) w ( φ + x ) d φ , J ( x ) = Z π s ( φ ) s ( φ + x ) d φ , and the expectation for the random number Z of hit triangles by E( Z ) = 1 + ( a + b + c ) uπQ + 2 FQ .
Proof.
It is sufficient to consider only the cases where O ∈ Q . (It wouldeven be sufficient to consider only the cases where O ∈ T . But considering O ∈ Q is much more convenient.) Since C is convex, it follows that p (5) = 0.For fixed value of φ , C hits exactly i ∈ { , , , , } parallelograms of R a, b, c if O is inside the set with number i (see Fig. 4), where every set i is theunion of all sets with equal shading. In the following, we use the law of totalprobability in the form p ( i ) = Z π p ( i | φ ) d φ π , where p ( i | φ ) denotes the conditional probability of exactly i hits for fixedvalue of φ . We have p ( i | φ ) = F i ( φ ) Q , where F i ( φ ) is the area of set i . By rearranging the parallelogram Q , onegets the parallelogram shown in Fig. 5. With the help of this figure it iseasier to find and calculate the areas F i ( φ ).4ig. 4: Situation in Q for fixed value of the angle φ Fig. 5: Rearrangement of Q φ , F ( φ ) is equal to F since set 6 is a congruent copy of C (seeFig. 4 and 5), hence p (6) = 12 πQ Z π F d φ = F πQ Z π d φ = FQ (cf. [8, p. 321], [9, pp. 73-74]). Now, we determine the probability p (3). C , with direction φ , hits exactly three triangles of R a, b, c if C ∈ T ( φ ) ∪T ( φ ) ∪ . . . ∪ T ( φ ). Every triangle T i ( φ ) is similar to the triangle T . Wedenote by a i ( φ ), b i ( φ ), c i ( φ ) the sides of T i ( φ ) with opposite angles α , β , γ ,respectively. Because of the said similarity, the area of T i ( φ ) is given by a i ( φ ) a Q b i ( φ ) b Q c i ( φ ) c Q , and, therefore, F ( φ ) = Q (cid:18) a ( φ ) a + c ( φ ) c + b ( φ ) b + a ( φ ) a + c ( φ ) c + b ( φ ) b (cid:19) , hence p (3) = 14 π (cid:18) a Z π a ( φ ) d φ + 1 c Z π c ( φ ) d φ + 1 b Z π b ( φ ) d φ + 1 a Z π a ( φ ) d φ + 1 c Z π c ( φ ) d φ + 1 b Z π b ( φ ) d φ (cid:19) . Using the support function s of C , one finds a ( φ ) = s ( φ − β − γ ) csc β + s ( φ ) csc γ − s ( φ − γ )(cot β + cot γ ) . We put d ( β, γ, φ ) := a ( φ ). Due to the symmetry of R a, b, c , c ( φ ) = d ( α, β, φ − γ ) , b ( φ ) = d ( γ, α, φ − γ − β ) ,a ( φ ) = d ( β, γ, φ + α + β + γ ) , c ( φ ) = d ( α, β, φ + α + β ) ,b ( φ ) = d ( γ, α, φ + α ) . Since d ( x, y, · ) with ( x, y ) ∈ { ( α, β ) , ( β, γ ) , ( γ, α ) } is a 2 π -periodic function,we get Z π a ( φ ) d φ = Z π d ( β, γ, φ ) d φ = Z π a ( φ ) d φ , Z π b ( φ ) d φ = Z π d ( γ, α, φ ) d φ = Z π b ( φ ) d φ , Z π c ( φ ) d φ = Z π d ( α, β, φ ) d φ = Z π c ( φ ) d φ , p (3) = 12 π (cid:18) H ( β, γ ) a + H ( γ, α ) b + H ( α, β ) c (cid:19) , where H ( x, y ) := Z π d ( x, y, φ ) d φ . With d ( β, γ, φ + β + γ )= s ( φ ) csc β + s ( φ + β + γ ) csc γ + s ( φ + β )(cot β + cot γ ) + 2 s ( φ ) s ( φ + β + γ ) csc β csc γ − s ( φ ) s ( φ + β ) csc β (cot β + cot γ ) − s ( φ + β + γ ) s ( φ + β ) csc γ (cot β + cot γ ) , and, using the the relations Z π s ( φ + β ) d φ = Z π s ( φ + β + γ ) d φ = Z π s ( φ ) d φ , Z π s ( φ + β ) s ( φ + β + γ ) d φ = Z π s ( φ ) s ( φ + γ ) d φ , Z π s ( φ ) s ( φ + β + γ ) d φ = Z π s ( φ + α ) s ( φ + α + β + γ ) d φ = Z π s ( φ + π ) s ( φ + α ) d φ , that result from the 2 π -periodicity of s , one finds H ( β, γ ) = Z π d ( β, γ, φ ) d φ = Z π d ( β, γ, φ + β + γ ) d φ = (cid:2) csc β + csc γ + (cot β + cot γ ) (cid:3) Z π s ( φ ) d φ + 2 csc β csc γ Z π s ( φ + π ) s ( φ + α ) d φ − β + cot γ ) csc β Z π s ( φ ) s ( φ + β ) d φ − β + cot γ ) csc γ Z π s ( φ ) s ( φ + γ ) d φ . From Z π w ( φ ) w ( φ + α ) d φ = Z π [ s ( φ ) + s ( φ + π )][ s ( φ + α ) + s ( φ + α + π )] d φ Z π [ s ( φ ) s ( φ + α ) + s ( φ ) s ( φ + α + π ) + s ( φ + π ) s ( φ + α )+ s ( φ + π ) s ( φ + α + π )] d φ = 2 Z π s ( φ ) s ( φ + α ) d φ + 2 Z π s ( φ + π ) s ( φ + α ) d φ , we get, with the π -periodicity of w , Z π s ( φ + π ) s ( φ + α ) d φ = 12 Z π w ( φ ) w ( φ + α ) d φ − Z π s ( φ ) s ( φ + α ) d φ = Z π w ( φ ) w ( φ + α ) d φ − Z π s ( φ ) s ( φ + α ) d φ . Therefore, with the abbreviations I ( x ) := Z π w ( φ ) w ( φ + α ) d φ and J ( x ) := Z π s ( φ ) s ( φ + α ) d φ , we find H ( β, γ ) a = 1 a Z π d ( β, γ, φ ) d φ = 1 a Z π d ( β, γ, φ + β + γ ) d φ = (cid:2) csc β + csc γ + (cot β + cot γ ) (cid:3) J (0)+ 2 csc β csc γ [ I ( α ) − J ( α )] − β + cot γ ) csc β J ( β ) − β + cot γ ) csc γ J ( γ )= " c ( ac sin β ) + b ( ab sin γ ) + (cid:18) c cos βac sin β + b cos γab sin γ (cid:19) J (0)+ 2 bcac sin β ab sin γ I ( α ) − bcac sin β ab sin γ J ( α ) − (cid:18) c cos βac sin β + b cos γab sin γ (cid:19) cac sin β J ( β ) − (cid:18) c cos βac sin β + b cos γab sin γ (cid:19) bab sin γ J ( γ )= ( a + b + c ) J (0) Q + 2 bcI ( α ) Q − bcJ ( α ) Q − acJ ( β ) Q − abJ ( γ ) Q .
8t follows that p (3) = 12 πQ (cid:2) ( a + b + c ) J (0) + 2 bcI ( α ) − bcJ ( α ) − caJ ( β ) − abJ ( γ )+ ( b + c + a ) J (0) + 2 caI ( β ) − caJ ( β ) − abJ ( γ ) − bcJ ( α )+ ( c + a + b ) J (0) + 2 abI ( γ ) − abJ ( γ ) − bcJ ( α ) − caJ ( β )]= 3( a + b + c ) J (0)2 πQ + bcI ( α ) + caI ( β ) + abI ( γ ) πQ − bcJ ( α ) + caJ ( β ) + abJ ( γ )] πQ . As the next step, we determine p (4). Using Fig. 5, one sees that F ( φ ) = w ( φ ) w ( φ + α )sin α − a ( φ ) a Q − a ( φ ) a Q − F = bcw ( φ ) w ( φ + α ) Q − a ( φ ) a Q − a ( φ ) a Q − F .
So we find p (4) = 12 πQ (cid:20) bcQ Z π w ( φ ) w ( φ + α ) d φ − Q a (cid:18) Z π a ( φ ) d φ + Z π a ( φ ) d φ (cid:19) − F Z π d φ (cid:21) = 12 πQ (cid:20) bcI ( α ) Q − Qa Z π d ( β, γ, φ ) d φ − πF (cid:21) = bcI ( α ) πQ − H ( β, γ )2 πa − FQ .
Taking into account the symmetry of R a, b, c , we also have p (4) = 13 (cid:20)(cid:18) bcI ( α ) πQ − H ( β, γ )2 πa − FQ (cid:19) + (cid:18) caI ( β ) πQ − H ( γ, α )2 πb − FQ (cid:19) + (cid:18) abI ( γ ) πQ − H ( α, β )2 πc − FQ (cid:19)(cid:21) = bcI ( α ) + caI ( β ) + abI ( γ )3 πQ − p (3) − FQ = bcI ( α ) + caI ( β ) + abI ( γ )3 πQ − (cid:18) a + b + c ) J (0)2 πQ + bcI ( α ) + caI ( β ) + abI ( γ ) πQ − bcJ ( α ) + caJ ( β ) + abJ ( γ )] πQ (cid:19) − FQ = bcJ ( α ) + caJ ( β ) + abJ ( γ ) πQ − ( a + b + c ) J (0)2 πQ − FQ . p (1). One finds F ( φ ) = [ h b − w ( φ )][ h c − w ( φ + α )]sin α − (cid:18) [ h c − w ( φ + α )] w ( φ − γ )sin β − c ( φ ) c Q − c ( φ ) c Q (cid:19) = h b h c sin α − h b w ( φ + α )sin α − h c w ( φ )sin α + w ( φ ) w ( φ + α )sin α − h c w ( φ − γ )sin β + w ( φ + α ) w ( φ − γ )sin β + Q c (cid:2) d ( α, β, φ − γ ) + d ( α, β, φ + α + β ) (cid:3) . Using h b = a sin γ = c sin α , h c = b sin α = a sin β , w ( φ − γ ) = w ( φ + α + β ),and Q = ab sin γ we get F ( φ ) = Q − cw ( φ + α ) − bw ( φ ) + bcQ w ( φ ) w ( φ + α ) − aw ( φ + α + β )+ acQ w ( φ + α ) w ( φ + α + β ) + Q c (cid:2) d ( α, β, φ − γ )+ d ( α, β, φ + α + β ) (cid:3) . This, with Z π w ( φ ) d φ = Z π [ s ( φ ) + s ( φ + π )] d φ = Z π s ( φ ) d φ = u (see [11, p. 3]), and Z π w ( φ + α ) w ( φ + α + β ) d φ = 2 Z π w ( φ + α ) w ( φ + α + β ) d φ = 2 Z π w ( φ ) w ( φ + β ) d φ , gives p (1) = 12 πQ (cid:18) Q Z π d φ − a + b + c ) Z π w ( φ ) d φ + 2 bcQ Z π w ( φ ) w ( φ + α ) d φ + 2 acQ Z π w ( φ ) w ( φ + β ) d φ + Qc Z π d ( α, β, φ ) d φ (cid:19) = 1 − ( a + b + c ) uπQ + bcI ( α ) πQ + caI ( β ) πQ + H ( α, β )2 πc . We now obtain, with the symmetry of R a, b, c , p (1) = 13 (cid:18) − ( a + b + c ) uπQ + bcI ( α ) πQ + caI ( β ) πQ + H ( α, β )2 πc
10 1 − ( b + c + a ) uπQ + caI ( β ) πQ + abI ( γ ) πQ + H ( β, γ )2 πa + 1 − ( c + a + b ) uπQ + abI ( γ ) πQ + bcI ( α ) πQ + H ( γ, α )2 πb (cid:19) = 1 − ( a + b + c ) uπQ + 2[ bcI ( α ) + caI ( β ) + abI ( γ )]3 πQ + 13 p (3)= 1 − ( a + b + c ) uπQ + ( a + b + c ) J (0)2 πQ + bcI ( α ) + caI ( β ) + abI ( γ ) πQ − bcJ ( α ) + caJ ( β ) + abJ ( γ ) πQ . Now we determine the expression for p (2). Let E denote the event that C hits exactly two triangles of R a, b, c ; let further E a , E b , and E c denote theevents that C hits side a , b , and c , respectively. Since the events E ∩ E a , E ∩ E b and E ∩ E c are pairwise disjoint, p (2) = P ( E ) = P ( E ∩ E a ) + P ( E ∩ E b ) + P ( E ∩ E c ) . We have P ( E ∩ E a ) = Z π P ( E ∩ E a | φ ) d φ π , P ( E ∩ E a | φ ) = F , a ( φ ) Q .
For the area F , a ( φ ), we can write F , a ( φ ) = w ( φ − γ )[ h b − w ( φ )]sin γ − b ( φ ) b Q − b ( φ ) b Q h b w ( φ − γ )sin γ − w ( φ − γ ) w ( φ )sin γ − Q b (cid:2) d ( γ, α, φ − γ − β )+ d ( γ, α, φ + α ) (cid:3) = aw ( φ − γ ) − abQ w ( φ − γ ) w ( φ ) − Q b (cid:2) d ( γ, α, φ − γ − β )+ d ( γ, α, φ + α ) (cid:3) . Hence, with R π w ( φ − γ ) w ( φ ) d φ = 2 R π w ( φ ) w ( φ + γ ) d φ , P ( E ∩ E a ) = 12 πQ (cid:20) a Z π w ( φ ) d φ − abQ Z π w ( φ ) w ( φ + γ ) d φ − Qb Z π d ( γ, α, φ ) d φ (cid:21) = auπQ − abI ( γ ) πQ − H ( γ, α )2 πb . Due to the symmetry of R a, b, c , we also have P ( E ∩ E b ) = buπQ − bcI ( α ) πQ − H ( α, β )2 πc ,P ( E ∩ E c ) = cuπQ − caI ( β ) πQ − H ( β, γ )2 πa .
11t follows that p (2) = ( a + b + c ) uπQ − bcI ( α ) + caI ( β ) + abI ( γ ) πQ − p (3)= ( a + b + c ) uπQ − a + b + c ) J (0)2 πQ − bcI ( α ) + caI ( β ) + abI ( γ )] πQ + 3[ bcJ ( α ) + caJ ( β ) + abJ ( γ )] πQ . In order to calculate the expectation E( Z ) = P i =1 i p ( i ) for the randomnumber Z of hit triangles, we put L := ( a + b + c ) J (0)2 πQ , M := bcI ( α ) + caI ( β ) + abI ( γ ) πQ ,N := bcJ ( α ) + caJ ( β ) + abJ ( γ ) πQ . So we have p (1) = 1 − ( a + b + c ) uπQ + L + M − N ,p (2) = ( a + b + c ) uπQ − L − M + 3 N ,p (3) = 3 L + M − N , p (4) = N − L − FQ , p (5) = 0 , p (6) =
FQ , and findE( Z ) = 1 − ( a + b + c ) uπQ + L + M − N + 2 (cid:18) ( a + b + c ) uπQ − L − M + 3 N (cid:19) + 3(3 L + M − N )+ 4 (cid:18) N − L − FQ (cid:19) + 5 · FQ = 1 + ( a + b + c ) uπQ + 2 FQ . w ( φ ) = 2 s ( φ ) For all convex bodies C where the reference point O may be choosen suchthat w ( φ ) = 2 s ( φ ) for every φ ∈ [0 , π ), we have I ( x ) = Z π w ( φ ) w ( φ + x ) d φ = 12 Z π w ( φ ) w ( φ + x ) d φ = 12 Z π s ( φ ) 2 s ( φ + x ) d φ = 2 Z π s ( φ ) s ( φ + x ) d φ = 2 J ( x ) , and, therefore, get the formulas of the following corollary which are evensimpler to compute. 12 orollary 1. If condition (3) holds true and w ( φ ) = 2 s ( φ ) for every φ ∈ [0 , π ) , then p (1) = 1 − ( a + b + c ) uπQ + ( a + b + c ) I (0)4 πQ + bcI ( α ) + caI ( β ) + abI ( γ )2 πQ ,p (2) = ( a + b + c ) uπQ − a + b + c ) I (0)4 πQ − bcI ( α ) + caI ( β ) + abI ( γ )2 πQ ,p (3) = 3( a + b + c ) I (0)4 πQ − bcI ( α ) + caI ( β ) + abI ( γ )2 πQ ,p (4) = bcI ( α ) + caI ( β ) + abI ( γ )2 πQ − ( a + b + c ) I (0)4 πQ − FQ ,p (5) = 0 , p (6) =
FQ .
In the following, we give some examples.
Corollary 2.
Let R a, b, c be a lattice of acute or right triangles. Let C be arectangle with side lengths g and h , and statisfying condition (3) . Then p (1) = 1 − ( a + b + c ) uπQ + ( αa + βb + γc )( g + h )2 πQ + 3( g + h )2 πQ + ( a + b + c ) FπQ + FQ ,p (2) = ( a + b + c ) uπQ − ( a + b + c )( g + h )4 Q − ( αa + βb + γc )( g + h )2 πQ − g + h )2 πQ − a + b + c ) FπQ − FQ ,p (3) = ( a + b + c )( g + h )2 Q − ( αa + βb + γc )( g + h )2 πQ − g + h )2 πQ + ( a + b + c ) FπQ − FQ ,p (4) = 3( g + h )2 πQ − ( a + b + c )( g + h )4 Q + ( αa + βb + γc )( g + h )2 πQ ,p (5) = 0 , p (6) = FQ , with u = 2( g + h ) and F = gh .Proof. The width of the rectangle in the direction φ is given by w ( φ ) = g | cos φ | + h | sin φ | . has the restriction w ( φ ) | ≤ φ< π = w ( φ ) if 0 ≤ φ < π/ ,w ( φ ) if π/ ≤ φ < π ,w ( φ ) if π ≤ φ < π/ ,w ( φ ) if 3 π/ < φ < π , with w ( φ ) = g cos φ + h sin φ ,w ( φ ) = − g cos φ + h sin φ ,w ( φ ) = − g cos φ − h sin φ ,w ( φ ) = g cos φ − h sin φ . For the calculation of I ( x ), x ∈ { α, β, γ } , we have to distinguish the cases0 ≤ φ < π/ , π/ ≤ φ ≤ π , and x ≤ φ + x < π/ , π/ ≤ φ + x < π , π ≤ φ + x ≤ π + x . Since 0 < x ≤ π/
2, this yields0 ≤ φ < π/ − x , π/ − x ≤ φ < π/ , π/ ≤ φ < π − x , π − x ≤ φ ≤ π ;therefore I ( x ) = Z π/ − x + Z π/ π/ − x + Z π − xπ/ + Z ππ − x ! w ( φ ) w ( φ + x ) d φ = Z π/ − x w ( φ ) w ( φ + x ) d φ + Z π/ π/ − x w ( φ ) w ( φ + x ) d φ + Z π − xπ/ w ( φ ) w ( φ + x ) d φ + Z ππ − x w ( φ ) w ( φ + x ) d φ = 12 (cid:8)(cid:2) ( π − x ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos x + 2 (cid:0) g + h + 2 xgh (cid:1) sin x (cid:9) . This, with (1), F = gh , and2 bc cos α = b + c − a , ac cos β = a + c − b , ab cos γ = a + b − c , gives bcI ( α ) + caI ( β ) + abI ( γ )= ( bc/ (cid:2) ( π − α ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos α + bc (cid:0) g + h + 2 αgh (cid:1) sin α + ( ca/ (cid:2) ( π − β ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos β + ca (cid:0) g + h + 2 βgh (cid:1) sin β + ( ab/ (cid:2) ( π − γ ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos γ + ab (cid:0) g + h + 2 γgh (cid:1) sin γ
14 (1 / (cid:2) ( π − α ) (cid:0) g + h (cid:1) + 4 F (cid:3) (cid:0) b + c − a (cid:1) + (cid:0) g + h + 2 αF (cid:1) Q + (1 / (cid:2) ( π − β ) (cid:0) g + h (cid:1) + 4 F (cid:3) (cid:0) c + a − b (cid:1) + (cid:0) g + h + 2 βF (cid:1) Q + (1 / (cid:2) ( π − γ ) (cid:0) g + h (cid:1) + 4 F (cid:3) (cid:0) a + b − c (cid:1) + (cid:0) g + h + 2 γF (cid:1) Q = (1 / π − α ) (cid:0) g + h (cid:1) (cid:0) b + c − a (cid:1) + (1 / π − β ) (cid:0) g + h (cid:1) (cid:0) c + a − b (cid:1) + (1 / π − γ ) (cid:0) g + h (cid:1) (cid:0) a + b − c (cid:1) + (cid:0) a + b + c (cid:1) F + 3 (cid:0) g + h (cid:1) Q + 2 πF Q = − ( α/ (cid:0) g + h (cid:1) (cid:0) a + b + c − a (cid:1) − ( β/ (cid:0) g + h (cid:1) (cid:0) a + b + c − b (cid:1) − ( γ/ (cid:0) g + h (cid:1) (cid:0) a + b + c − c (cid:1) + ( π/ (cid:0) a + b + c (cid:1) (cid:0) g + h (cid:1) + (cid:0) a + b + c (cid:1) F + 3 (cid:0) g + h (cid:1) Q + 2 πF Q = (cid:0) αa + βb + γc (cid:1) (cid:0) g + h (cid:1) − ( π/ (cid:0) a + b + c (cid:1) (cid:0) g + h (cid:1) + (cid:0) a + b + c (cid:1) F + 3 (cid:0) g + h (cid:1) Q + 2 πF Q . Furthermore, one finds I (0) = 12 (cid:2) π (cid:0) g + h (cid:1) + 4 gh (cid:3) = 12 (cid:2) π (cid:0) g + h (cid:1) + 4 F (cid:3) . With Corollary 1, the result follows.
Corollary 3.
Let R a, b, c be a lattice of obtuse or right triangles with π/ ≤ α < π . Let C be a rectangle with side lengths g and h , and statisfyingcondition (3) . Then p (1) = 1 − ( a + b + c ) uπQ + ( αa + βb + γc )( g + h )2 πQ + 3( g + h )2 πQ + 2 a FπQ + 2 FQ − αFπQ ,p (2) = ( a + b + c ) uπQ − ( a + b + c )( g + h )4 Q − ( αa + βb + γc )( g + h )2 πQ − g + h )2 πQ − (3 a + b + c ) FπQ − FQ + 2 αFπQ ,p (3) = ( a + b + c )( g + h )2 Q − ( αa + βb + γc )( g + h )2 πQ − g + h )2 πQ + 2( b + c ) FπQ − FQ + 2 αFπQ , (4) = 3( g + h )2 πQ − ( a + b + c )( g + h )4 Q + ( αa + βb + γc )( g + h )2 πQ − ( b + c − a ) FπQ + FQ − αFπQ ,p (5) = 0 , p (6) = FQ .
Proof.
We have I (0) = 12 (cid:2) π (cid:0) g + h (cid:1) + 4 gh (cid:3) ,I ( β ) = 12 (cid:8)(cid:2) ( π − β ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos β + 2 (cid:0) g + h + 2 βgh (cid:1) sin β (cid:9) ,I ( γ ) = 12 (cid:8)(cid:2) ( π − γ ) (cid:0) g + h (cid:1) + 4 gh (cid:3) cos γ + 2 (cid:0) g + h + 2 γgh (cid:1) sin γ (cid:9) , as in the case 0 < α ≤ π/
2. For the calculation of I ( α ), we have to distin-guish the cases 0 ≤ φ < π/ , π/ ≤ φ ≤ π , and α ≤ φ + α < π , π ≤ φ + α < π/ , π/ ≤ φ + α ≤ π + α . This gives0 ≤ φ < π − α , π − α ≤ φ < π/ , π/ ≤ φ < π/ − α , π/ − α ≤ φ ≤ π ;therefore I ( α ) = Z π − α + Z π/ π − α + Z π/ − απ/ + Z π π/ − α ! w ( φ ) w ( φ + α ) d φ = Z π − α w ( φ ) w ( φ + α ) d φ + Z π/ π − α w ( φ ) w ( φ + α ) d φ + Z π/ − απ/ w ( φ ) w ( φ + α ) d φ + Z π π/ − α w ( φ ) w ( φ + α ) d φ = 12 (cid:8)(cid:2) ( π − α ) (cid:0) g + h (cid:1) − gh (cid:3) cos α + 2 (cid:2) g + h + 2( π − α ) gh (cid:3) sin α (cid:9) . Now, we denote the formula of I ( α ) for 0 < α ≤ π/ I ( α ), and find I ( α ) − ˜ I ( α ) = 2( π − α ) gh sin α − gh cos α . (Of course, for α = π/ I ( α ) − ˜ I ( α ) = 0.) With bc sin α = Q and bc cos α = 12 (cid:0) b + c − a (cid:1) ,
16e get bc (cid:2) I ( α ) − ˜ I ( α ) (cid:3) πQ = 2( π − α ) ghQ πQ − gh (cid:0) b + c − a (cid:1) πQ = ( π − α ) FπQ − ( b + c − a ) FπQ = FQ − αFπQ − ( b + c − a ) FπQ , hence bcI ( α )2 πQ = bc ˜ I ( α )2 πQ + FQ − αFπQ − ( b + c − a ) FπQ . We denote the probabilities for 0 < α ≤ π/ p ( i ). From Corollary 1, itfollows that p (1) = ˜ p (1) + FQ − αFπQ − ( b + c − a ) FπQ ,p (2) = ˜ p (2) − FQ + 2 αFπQ + ( b + c − a ) FπQ ,p (3) = ˜ p (3) − FQ + 2 αFπQ + ( b + c − a ) FπQ ,p (4) = ˜ p (4) + FQ − αFπQ − ( b + c − a ) FπQ ,p (5) = ˜ p (5) = 0 , p (6) = ˜ p (6) = FQ .
The formulae of the corollary follow easily.As special case of Corollaries 2 and 3, respectively, we get the probabilitiesfor a needle of length ℓ := g with h = 0, u = 2 ℓ and F = 0. We find p (1) = 1 − a + b + c ) ℓπQ + ( αa + βb + γc ) ℓ πQ + 3 ℓ πQ ,p (2) = 2( a + b + c ) ℓπQ − ( a + b + c ) ℓ Q − ( αa + βb + γc ) ℓ πQ − ℓ πQ ,p (3) = ( a + b + c ) ℓ Q − ( αa + βb + γc ) ℓ πQ − ℓ πQ ,p (4) = 3 ℓ πQ − ( a + b + c ) ℓ Q + ( αa + βb + γc ) ℓ πQ ,p (5) = 0 , p (6) = 0 .p (1) is the result obtained by Markov (see (2)).17or the random throw of a needle of length ℓ ≤ √ a/ R a, a, a of equilateral triangles we have α = β = γ = π/ Q = √ a /
2, andtherefore p (1) = 1 − √ π ℓa + √ π + 23 ! ℓ a , p (2) = 4 √ π ℓa − √ π + 53 ! ℓ a ,p (3) = − √ π ! ℓ a , p (4) = √ π − ! ℓ a , p (5) = 0 , p (6) = 0 . This is the result obtained by Santal´o [10, pp. 167-170] (see also [11, pp. 140-141]).
Let g and h be the lengths of the major axis and the minor axis, respectively.We have F = πgh/ w ( φ ) = q g cos φ + h sin φ = g q − µ sin φ , µ = 1 − (cid:18) hg (cid:19) , hence u = Z π w ( φ ) d φ = 2 Z π/ w ( φ ) d φ = 2 gE ( µ ) , where E ( µ ) = Z π/ q − µ sin φ d φ is the complete elliptic integral of the second kind. Furthermore, one has I ( x ) = g Z π q (1 − µ sin φ )(1 − µ sin ( φ + x )) d φ . Hence, if the inequality condition (3) with s ( φ ) = g q − µ sin φ is fulfilled, we have found all hitting probabilities for ellipses and R a, b, c .For an ellipse with h = 0 we have µ = 1, hence u = 2 g Z π/ q − sin φ d φ = 2 g Z π/ | cos φ | d φ = 2 g Z π/ cos φ d φ = 2 g , and I ( x ) = g Z π q (1 − sin φ )(1 − sin ( φ + x )) d φ = g Z π p cos φ cos ( φ + x ) d φ = g Z π | cos φ cos( φ + x ) | d φ g Z π/ − x − Z π/ π/ − x + Z ππ/ ! cos φ cos( φ + x ) d x = g Z π/ − x − Z π/ π/ − x + Z ππ/ ! (cid:18)
12 cos x + 12 cos x cos 2 φ −
12 sin x sin 2 φ (cid:19) d φ = g (cid:18)
12 ( π − x ) cos x + sin x (cid:19) . These formulas are equal to the respective formulas of the rectangle for h = 0. Therefore, as was to be expected, we get needles also as special casesof ellipses. Now, we calculate the hitting probabilities in the case that C is a half discof radius r . Corollary 4.
Let R a, b, c be a lattice of acute or right triangles. Let C be ahalf disc with radius r , and satisfying condition (3) . Then p (1) = 1 − ( a + b + c ) uπQ + ( a + b + c ) r Q + 4( ab + bc + ca ) r πQ − ( αa + βb + γc ) r πQ + ( αbc + βca + γab ) r πQ − r πQ ,p (2) = ( a + b + c ) uπQ − a + b + c ) r Q − (8 − π )( ab + bc + ca ) r πQ + 5( αa + βb + γc ) r πQ − αbc + βca + γab ) r πQ + 33 r πQ ,p (3) = 7( a + b + c ) r Q + (4 − π )( ab + bc + ca ) r πQ − αa + βb + γc ) r πQ + 3( αbc + βca + γab ) r πQ − r πQ ,p (4) = ( ab + bc + ca ) r Q − a + b + c ) r Q + 3( αa + βb + γc ) r πQ − ( αbc + βca + γab ) r πQ + 15 r πQ − FQ ,p (5) = 0 , p (6) =
FQ , with u = ( π + 2) r and F = πr / . roof. The width of the half disc in the direction φ is given by w ( φ ) = r (1 + | cos φ | ) , see Fig. 6.Fig. 6: Width function w and support function s for a half disc w has the restriction w ( φ ) | ≤ φ< π = w ( φ ) if 0 ≤ φ < π/ ,w ( φ ) if π/ ≤ φ < π/ ,w ( φ ) if 3 π/ ≤ φ < π , with w ( φ ) = r (1 + cos φ ) , w ( φ ) = r (1 − cos φ ) , w ( φ ) = r (1 + cos φ ) . For the calculation of I ( x ), x ∈ { α, β, γ } , we have to distinguish the cases0 ≤ φ < π/ , π/ ≤ φ ≤ π , and x ≤ φ + x < π/ , π/ ≤ φ + x ≤ π + x . Since 0 < x ≤ π/
2, this yields0 ≤ φ < π/ − x , π/ − x ≤ φ < π/ , π/ ≤ φ < π ;therefore I ( x ) = Z π/ − x + Z π/ π/ − x + Z ππ/ ! w ( φ ) w ( φ + x ) d φ = Z π/ − x w ( φ ) w ( φ + x ) d φ + Z π/ π/ − x w ( φ ) w ( φ + x ) d φ + Z ππ/ w ( φ ) w ( φ + x ) d φ = ( π + 4) r + 12 ( π − x ) r cos x + r sin x . bc cos α = b + c − a , ac cos β = a + c − b , ab cos γ = a + b − c , yields bcI ( α ) + caI ( β ) + abI ( γ )= ( π + 4)( ab + bc + ca ) r + (1 / bc ( π − α ) r cos α + bcr sin α + (1 / ca ( π − β ) r cos β + car sin β + (1 / ab ( π − γ ) r cos γ + abr sin γ = ( π + 4)( ab + bc + ca ) r + (1 / π − α )( b + c − a ) r + Qr + (1 / π − β )( c + a − b ) r + Qr + (1 / π − γ )( a + b − c ) r + Qr = ( π + 4)( ab + bc + ca ) r + ( π/ a + b + c ) r + 3 Qr − ( α/ a + b + c − a ) r − ( β/ a + b + c − b ) r − ( γ/ a + b + c − c ) r = ( π + 4)( ab + bc + ca ) r + ( π/ a + b + c ) r + 3 Qr − ( π/ a + b + c ) r + ( αa + βb + γc ) r = ( π + 4)( ab + bc + ca ) r − ( π/ a + b + c ) r + ( αa + βb + γc ) r + 3 Qr . The support function s has the restriction s ( φ ) | ≤ φ< π = s ( φ ) if 0 ≤ φ < π/ ,s ( φ ) if π/ ≤ φ < π ,s ( φ ) if π ≤ φ < π ,s ( φ ) if 2 π ≤ φ < π/ ,s ( φ ) if 5 π/ ≤ φ < π , with s ( φ ) = r cos φ , s ( φ ) = − r cos φ , s ( φ ) = r . For the calculation of J ( x ), x ∈ { , α, β, γ } , we have to distinguish the cases0 ≤ φ < π/ , π/ ≤ φ < π , π ≤ φ ≤ π and x ≤ φ + x < π/ , π/ ≤ φ + x < π , π ≤ φ + x < π , π ≤ φ + x ≤ π + x . ≤ x ≤ π/
2, this yields0 ≤ φ < π/ − x , π/ − x ≤ φ < π/ , π/ ≤ φ < π − x ,π − x ≤ φ < π , π ≤ φ < π − x , π − x ≤ φ ≤ π ;therefore J ( x ) = (cid:18) Z π/ − x + Z π/ π/ − x + Z π − xπ/ + Z ππ − x + Z π − xπ + Z π π − x (cid:19) s ( φ ) s ( φ + x ) d φ = Z π/ − x s ( φ ) s ( φ + x ) d φ + Z π/ π/ − x s ( φ ) s ( φ + x ) d φ + Z π − xπ/ s ( φ ) s ( φ + x ) d φ + Z ππ − x s ( φ ) s ( φ + x ) d φ + Z π − xπ s ( φ ) s ( φ + x ) d φ + Z π π − x s ( φ ) s ( φ + x ) d φ = ( π − x ) r + 12 ( π − x ) r cos x + 52 r sin x . This, with (1) and2 bc cos α = b + c − a , ac cos β = a + c − b , ab cos γ = a + b − c , gives bcJ ( α ) + caJ ( β ) + abJ ( γ )= bc ( π − α ) r + (1 / bc ( π − α ) r cos α + (5 / bcr sin α + ca ( π − β ) r + (1 / ca ( π − β ) r cos β + (5 / car sin β + ab ( π − γ ) r + (1 / ab ( π − γ ) r cos γ + (5 / abr sin γ = π ( ab + bc + ca ) r − ( αbc + βca + γab ) r + (1 / π − α )( b + c − a ) r + (1 / π − β )( c + a − b ) r + (1 / π − γ )( a + b − c ) r + (15 / Qr = π ( ab + bc + ca ) r − ( αbc + βca + γab ) r + ( π/ a + b + c ) r − (3 α/ a + b + c − a ) r − (3 β/ a + b + c − b ) r − (3 γ/ a + b + c − c ) r + (15 / Qr = π ( ab + bc + ca ) r − ( αbc + βca + γab ) r + ( π/ a + b + c ) r − (3 π/ a + b + c ) r + (3 / αa + βb + γc ) r + (15 / Qr = π ( ab + bc + ca ) r − ( αbc + βca + γab ) r − ( π/ a + b + c ) r + (3 / αa + βb + γc ) r + (15 / Qr .
22e have J (0) = 3 πr /
2. The assertation follows by inserting the formulasfor bcI ( α ) + caI ( β ) + abI ( γ ), bcJ ( α ) + caJ ( β ) + abJ ( γ ), and J (0) in theformulas for p ( i ) at the end of the proof of Theorem 1. Knowing the width function w and the support function s of any planeconvex body C , it is possible to calculate the hitting probabilities for C and R a, b, c with Theorem 1 and, if w ( φ ) = 2 s ( φ ) for every φ ∈ [0 , π ), withCorollary 1, respectively. w and s for a parallelogram and a triangle are tobe found in [8, pp. 319/320].In particular, numerical values for p ( i ) are easily obtained by numericalintegration of the integrals I ( α ), I ( β ), I ( γ ), J (0), J ( α ), J ( β ), and J ( γ ). References [1] U. B¨asel, A. Duma: Intersection probabilities for random convex bodiesand lattices of parallelograms,
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Tˆohoku Math. J. (1940), 159-171.2311] L. A. Santal´o: Integral Geometry and Geometric Probability , Addison-Wesley, London, 1976. Uwe B¨aselHTWK Leipzig,Fakult¨at f¨ur Maschinenbau und Energietechnik,PF 30 11 66, 04251 Leipzig, Germany [email protected]@htwk-leipzig.de