The Shortest Planar Arc of Width 1
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The Shortest Planar Arc of Width 1 Ani Adhikari Department of Statistics Sequoia Hall Stanford University Stanford, CA 94305 and Jim Pitman Department of Statistics University of California Berkeley, CA 94720 Technical Report No. 113 September 1987 Department of Statistics University of California Berkeley, California 1.1 THE SHORTEST PLANAR ARC OF WIDTH 1 ANI ADHIKARI AND JIM PITMAN Departrnent of Statistics, Sequoia Hall, Stanford University, Stanford, CA 94305 and Department of Statistics, University of California, Berkeley, CA 94720 1. Introduction. A floor is ruled with parallel lines spaced unit distance apart. You are given a piece of wire of length I which you are free to bend but not stretch. Can you bend the wire so that if dropped on the floor the bent piece of wire is certain to cross at least one of the lines, no matter how it falls? In this article we find the least l such that this can be done, and show how the wire should be bent. More formally, let a: [0, 1] -4 R2 be a continuous rectifiable arc, of length 1(a). For 0.9.< r, define the width of a between parallels at angle 9 by w0(a) = distance between supporting parallel lines at an angle 9 to the x-axis. FIG. 1. The width between parallels at angle 0 And define the width of a by 1.2 w (a) = inf wQ(x) Our problem is to identify an arc of minimal length among all arcs of width at least 1. A cir- cular arc of diameter 1 has length it= 3.14* h.Tree sides of the unit square reduces the length to 3. Two sides of an equilateral triangle of altitude I further reduces the length to 4/t3= 2.309401 . The minimal length tums out only slightly less, namely 2.27829 * - *, and is achieved by bending the wire into a shape which we call the calliper, shown in Fig. 2. FIG. 2. The calliper. In Section 2 we identify the calliper of length 2.27829 - as the shortest convex arc of width 1. In Sections 3 and 4 we show that the calliper is in fact shortest among all arcs, con- vex and non-convex. For the most part, our arguments are geometric, in the spirit of Kazar- inoff (1961), Yaglom and Boltyanski (1961) and Niven (1982). Our problem is a deterministic variation of Buffon's famous needle problem: in case you do not bend the wire at all, but leave it as a straight needle of length 1, what is the chance that the needle crosses at least one line if dropped at random on the floor? As observed by Bertrand in the last century, and argued in detail by Barbier (1860), under natural assumptions of random- ness the expected number of crossings of the grid by a needle bent into a planar curve is a constant c times the length 1, regardless of the shape of the curve. A closed convex curve of constant width 1 must cross the lines exactly twice, no matter how it falls. Since a circle of diameter I has constant width 1 and length it, it follows that 1.3 (i) the constant is c = 2/xt, (ii) every closed convex curve of constant width 1 has the same length it, and (iii) for I . 1 the probability that a randomly dropped straight needle crosses the lines is (2/i)l. For other questions about randomly dropped curves, and further references, see DeTemple and Robertson (1980). The problem solved in this article is a special case of a stochastic geometry problem which we do not know how to solve for all l. Given a wire of length l, how should it be bent to max- imize the probability that it crosses at least one line when tossed at random on the ruled floor? Here we just find the least length L such that this maximum probability is one. For 0 < I < 1 the best strategy is easily shown to be leaving the wire straight, with crossing probability (2/7i)1. But for 1 <1 < L we do not know how to bend the wire to maximize the crossing probability. Some remarks on this and other problems in the same vein are mentioned in the final section. Notation. Since rotation, translation, and reflection of an arc in the plane affect neither its length nor its width, we will often use these operations to reduce calculations to arcs with some convenient orientation. A generic arc a will often be denoted instead by A-B to indi- A cate that it starts at A and ends at B. To -indicate the arc is convex we write A B instead of A-B. For points A and B in the plane we use the following notation: AB for the straight line through A and B, AB for the segment of this line between A and B, I AB I for the length of this line segment. Constructions of arcs may be indicated by notation such as this: if A B is some arc under consideration, then A BA is the closed arc obtained by following A-B from A to B, then returning to A along a line segment. In such constructions the precise paramaterization of the arc will be irrelevant, though the order in which points are visited may be important 2.1 2. The Shortest Convex Arc of Width One. Suppose now that the arc A B is convex, A meaning that the closed arc A BA is the boundary of a convex subset of the plane: A B A B A FIG. 3. Some convex arcs. A A convex arc A B must lie on one side of the line AB. Without loss of generality, we take AB to be horizontal, and suppose A B lies above AB. We propose to find the shortest such A arc of width one. In case A B extends to the left of A or to the right of B, a shorter arc with greater width is obtained by dropping a perpendicular to AB: A 3 FIG. 4. Reduction to arcs over an interval. So we may assume that d = AB 2 1, and confine our search to arcs A B lying in the strip above AB, such as the leftmost arc in Fig. 3. In this case the width condition can be reformu- lated more simply as follows, with reference to the horizontal segment A'B' at unit height above AB, and open discs Disc(A) and Disc(B) centered at A and B, with radii 1, as shown in the next figure: Al . ;; _-- Ds(B)"C I,A A JB FIG. 5. An arc of width greatera an 1. LEMMA. For a convex arc A B above AB with l AB > 1 width (A B) > 1 if and only if the following three conditions all hold: A (i) A B intersects A'B' at some point F A (ii) A F does not intersect Disc (B) A (iii) F B does not intersect Disc (A) Proof of Sufficiency. Suppose the three conditions hold. Width at least 1 in the horizontal and vertical directions is clear. For the width between down-sloping parallels, consider two paral- lels distance 1 apart, one through A, the other above it tangent to Disc (A ). The upper paral- lel must hit the arc between F and B by (i) and (iii). Similarly, width at least 1 between up- sloping parallels is implied by (i) and (ii). Proof of Necessity. If (i) fails, the vertical width is less than 1. The only remaining case is if A (i) holds but either (ii) or (iii) fails. If say (ii) fails, there exist points C and D on A B as in the following diagram such that C and D lie on Circle (B) bounding Disc (B), and apart from these points arc C D lies inside the open Disc (B). defined in tems of CD and isc (B ). Th upe-uprigl'e fABprle oC A'.A By convexity of A B, the arc lies inside the shaded region in the diagram, with boundary d d e f n c . ppersupportinglineofA 2.3 must therefore intersect Disc (B ), giving a width less than 1. According to the lemma, for fixed A and B at distance d . 1, the shortest convex arc A B of width 1 lying above AB is the shortest path from A to B via some F on A'B', not entering Disc (B ) on the way from A to F, and not entering Disc (A ) on the way from F to B. In case d > 2/43, this shortest path is obvious. For F the midpoint of A'B' the straight lines AF and FB do not meet the discs, so AFB is the shortest, with length V1Id. For d =2h13, the length is 2.30940... .. FIG. 7. Two sides of an equilateral triangle: d =P3 If d < 2/43, the constraint of the discs is felt. For any F on A'B', it is easy to see that the shortest way to get from A to F without entering Disc (B) is to go along AF, if this can be done without entering Disc (B); else to go from A to Circle (B ) along a tangent, stay on the circle awhile, and leave for F along another tangent. And the shortest way to return from F to B without entering Disc (A) can be described similarly, with Circle (A) replacing Circle (B). A ' .;.F A.:.