The Great Wall of David Shin

Transcription

1 The Great Wall of David Shin Tiankai Liu 115 June 015 On 9 May 010, David Shin posed the following puzzle in a Facebook note: Problem 1. You're blindfolded, disoriented, and standing one mile from the Great Wall of China. How far do you have to walk to guarantee that you will run into the wall? Assume that the wall is innitely long and straight. Though Ian Le posted the correct answer on 7 August, no one posted a proof of optimality. After various unhelpful remarks by the author, such as yan and i convinced each other that we have an convincing argument, the problem seems to have languished for nearly half a decade. Update (19 January 016): Never mind, the proof was given in H. Joris, Le chasseur perdu dans la forêt, Elemente der Mathematik 35 (1), 10 January It is available at 0_0.pdf. 1 The shortest path Before giving the answer to Problem 1, let us slightly reformulate the problem. Let O be the origin in the real plane R, and let be the closed unit disk, = {(x, y) R x + y 1}. Problem 1 asks for the shortest curve C starting at O that (a) meets every line tangent to. We claim this is equivalent to saying (b) the convex hull H of C contains. Proof. To show (a) implies (b): If H misses some point P, then, since H is the intersection of all open half-planes containing C, there must exist some open half-plane R + containing C but not P. Say R + is bounded by the line l ; then let l be the unique line parallel to l, tangent to, and lying outside R +. Then evidently C fails to meet l. To show (b) implies (a): If C fails to meet a line l tangent to at a point P, then C lies entirely in the open half-plane R + containing O and bounded by l. Since R + is convex, the convex hull H also lies in R +, and in particular P / H. 1

2 x = -1 y A (-, 0) O = (0, 0) x B = (-1, -1) (0, -1) Figure 1: The optimal path Let us say a curve C starting at O touches the Great Wall of Shin if it starts at O and satises the equivalent conditions (a) and (b). Problem 1 asks for the minimum possible length of C. The answer is 3 + 7π 6 at (0, 0), and walk straight for 3 1 miles, say, to A := ( 1, + 1 and can be realized by the following curve (Figure 1). Start 3 ). Make a 10 turn and walk 1 straight for 3 mile, say, to ( 1, 3), which lies on. Without making an abrupt turn, walk 7π miles around, to (0, 1). Continue straight for 1 mile, to B := ( 1, 1), and stop. 6 This curve clearly satises condition (b), and therefore touches the Great Wall of Shin. The proof of optimality We would like to show that the curve of length 3 + 7π + 1 in answer to Problem 1 cannot 6 be beaten. For technical reasons, we would like to restrict our attention to contenders that are polygonal paths. 1 Here's why this reduction is justied: Lemma. If C is a nite-length curve that touches the Great Wall of Shin, then for every ɛ > 0 there exists a polygonal path C, also touching the Great Wall of Shin, that has length at most C + ɛ. Proof. Fix some 0 < δ 1. Let O = P 0, P 1,..., P n 1 I'll use path interchangeably with rectiable curve, so paths may intersect themselves. I only feel like doing this because it sounds weird to talk about straight paths as curves.

3 be points on C, spaced less than δ apart in arc length, such that P n is the endpoint of C. Then every point of C is within δ/ of one of the points P i. Therefore, every point of the convex hull of C, being a weighted average of points of C, is also with δ/ of the convex hull H of {P 0,..., P n }. It follows that the disk of radius 1 δ centered at the origin lies within H. Therefore, if we let C be the polygonal path obtained from P 0 P n via a dilation of ratio 1/(1 δ ) from the origin, then C < C 1 δ and C touches the Great Wall of Shin. For suciently small δ depending on C, we can ensure that C C + ɛ, as desired. Therefore, the solution to Problem 1 is completed by the following Proposition: Proposition 3. For any n N and any points P 1,..., P n R, let P denote the polygonal path OP 1 P n, and H the convex hull of {O, P 1,..., P n }. Suppose that P touches the Great Wall of Shin. Then P must have length greater than 3 + 7π Proof. Our proof proceeds by a sequence of reductions. Let us say P is non-redundant if, for any i {1,..., n}, the convex hull of {O, P 1,..., P i,..., P n } does not contain. P is non-self-intersecting if the interiors of the line segments OP 1, P 1 P,..., P n 1 P n are pairwise disjoint. (Caution: a priori we allow the endpoint of one segment to lie in the interior of another segment, so our denition may be a bit dierent from standard terminology.) The violence of P is the number of segments P k 1 P k ( k n) that cut in two. We say P is non-violent if its violence equals zero. Given any polygonal path P that touches the Great Wall of Shin, we will explain how to construct another one, P, of lesser or equal length, that is non-redundant, non-selfintersecting, and non-violent. Then we will prove Proposition 3 for P under these additional assumptions. Step 1: Reduction to non-redundant P. This should be pretty obvious; we just repeatedly throw away one redundant vertex at a time. Henceforth, assume P is non-redundant. Let us pause and introduce some new notation. The convex hull H of {O, P 1,..., P n } is a polygon with exactly n vertices, namely P 1,..., P n. However, a priori, these vertices may not be in order. So let us write A 0,..., A n 1, with indices taken modulo n, to be the vertices of H in some order (either clockwise or counterclockwise). The following lemma is pretty obvious, and I will dispense with the proof. We say that a closed line segment σ cuts a convex region R R in two if R\σ is disconnected. 3

4 Lemma 4. With A 0,..., A n 1 arising as above from a non-redundant path that touches the Great Wall of Shin, let i and j be two dierent members of Z/nZ. Then there are two mutually exclusive possibilities: (a) j = i ± 1, and the segment A i A j is either disjoint from or tangent to the unit circle. (b) j i ± 1, and the segment A i A j cuts in two. It also cuts H in two, and any line segment joining points of these two components together must meet the segment A i A j. Step : Reduction to non-self-intersecting P. Recall that we are assuming P is nonredundant. For this step, imagine that the set {P 1,..., P n } is xed. There are n! polygonal paths that start at O and visit all the points of {P 1,..., P n } in some order. Each such path is non-redundant and touches the Great Wall of Shin. Therefore, we may assume that, among all n! such paths, P = OP 1 P n happens to be one of minimum length. We claim that this implies that P is non-self-intersecting. If not, there exist 1 i < j n such that the interiors of segments P i 1 P i and P j 1 P j meet, where we temporarily write P 0 for O for the duration of Step. There are two cases to consider. Case 1: The segments P i 1 P i and P j 1 P j meet transversely at a point X. Then by the triangle inequality so P i 1 P i + P j 1 P j = P i 1 X + XP i + P j 1 X + XP j P := P 0 P i 1 }{{} increasing indices = P i 1 X + XP j 1 + P i X + XP j > P i 1 P j 1 + P i P j, decreasing indices {}}{ P j 1 P i P j P n }{{} increasing indices is strictly shorter than P, a contradiction. Case : The segments P i 1 P i and P j 1 P j are collinear. This is an unusual situation: since P 1,..., P n are all vertices of the convex polygon H, it must be that (i, j) = (0, 1), and O P 1 P. In particular, P 1 P cuts in two. Therefore, by Lemma 4, P 1 P also cuts H in two. There must then exist some k {3,..., n} such that P k 1 and P k lie in dierent components of H\P 1 P. But this means that P 1 P and P k 1 P k intersect transversely, as in Case 1 above, which was shown to be impossible. Therefore, Case is also impossible. Henceforth, we may assume that P is non-self-intersecting. Step 3: Reduction to non-violent P. Recall that we are assuming P is non-redundant and non-self-intersecting. Assuming that the violence v of P is positive, we will construct a non-redundant, non-self-intersecting path Q = OQ 1 Q m 4

5 that touches the Great Wall of Shin, has violence equal to v 1, and has length at most the length of P. (Possibly m > n, but this does not matter.) Since v > 0, we may let k {,..., n} be the largest index for which P k 1 P k cuts (and therefore also H) in two. By maximality of k, Lemma 4 implies that P k, P k+1,..., P n must traverse some of the vertices of the convex polygon H in order. So without loss of generality, let us suppose A i = P i for k i n. Since P is non-self-intersecting, Lemma 4(b) tells us that P 1,..., P k must all lie in one component of H\P k 1 P k, and P k+1,..., P n must all lie in the other component. It follows that A 1 = P k 1. And since P touches the Great Wall of Shin, we must have 3 k n 1. Therefore, it makes sense to let R and R + be the two open half-planes, separated by the line P k 1 P k, that contain P k and P n, respectively. Set Q i = P i for i k 1. If it happens that P k 1 P n = A 1 A n A 1 A k = P k 1 P k, (1) then we may simply let Q := O P 1 P k 1 }{{} increasing indices decreasing indices {}}{ P n P k, and be done; Q has violence v 1 because P k 1 P n = A 1 A n no longer cuts H in two, and inequality (1) ensures that Q is not longer than P. Unfortunately, we cannot simply assume that inequality (1) holds, but there is an easy x. Let l be the unique line through P k 1 that is tangent to at a point T R +. Note that the path P k P n must meet l at some point, because P k P k 1 cuts in two but P n P k 1 doesn't. Intuitively, we want to consider the shortest curve γ from P k to anywhere on the line l that lies entirely within the closure of R + \, and replace the path P k P n by a suciently good polygonal approximation P k P m of C, so that by setting Q := O P 1 P k 1 }{{} increasing indices decreasing indices {}}{ P m P k, () we can carry on just as above. To say this rigorously, we claim that there is a unique shortest path γ from P k to l within clos(r + \ ), namely: Case 1: γ is the perpendicular from P k to l, if this perpendicular lies in clos(r + \ ). Case : Otherwise, let X, Y R + be the unique points such that P k X is tangent to, and OY l. Let Z be the foot of the perpendicular from Y to l, so that Y Z is tangent to. Then γ is the union of the segment P k X, the arc XY of in R +, and the segment Y Z. 5

6 The rst case is trivial. In the second case, it is easy to deduce from Proposition 5 in the Appendix that the shortest path γ exists and is (in the notation of that Proposition) of the form γ(p k, W ;, R + ) for some W on the ray P k T beyond T. Such a γ comes in three parts: a line segment starting from P k, an arc of, and nally a line segment ending on W. The nal line segment must be perpendicular to l, or else γ could be shortened by replacing a tiny portion of the end with a segment perpendicular to l. Thus, W must be the point Z dened above, as claimed. So indeed we can dene P k P m and Q by equation () above, such that P m is the endpoint of γ. (In case, we can perform the polygonal approximation in such a way that P i 1 P i is tangent to for every i {k + 1,..., m}.) We must verify that P k 1 P k P k 1 P m. To this end, note that, since P is non-self-intersecting, the interiors of segments OP 1 and P k 1 P k are disjoint, so O doesn't lie in R +. Then it is easy to see that the distance to P k 1 decreases monotonically as we travel along γ from P k to P m. So indeed we have replaced P by a shorter, less violent Q. Henceforth, we may assume P is non-violent. Step 4: Solution for non-violent P. Since P is non-redundant and non-violent, the path P 1 P n must avoid int( ) entirely, and therefore Lemma 4 implies that it must wind around in order, so that we may assume P i = A i for all i {1,..., n}. If P n P 1 isn't tangent to, then we can always move P n closer to P n 1 so that it is tangent. Let l = P n P 1, and let R + be the open half-plane bounded by l and containing most of. Then Proposition 5 applies directly, and it suces to show that, if (using the notation of that Proposition) C := OA γ(a, B;, R + ) for two points A, B l on opposite sides of the point l, then then C 3 + 7π Let's impose some coordinates to help with the calculation (see Figure 1). Say l is the line x = 1 and A = ( 1, a) for some a > 0. As in Step 3, our ability to wiggle B along l implies that the nal straight segment of γ is perpendicular to l, so that B = ( 1, 1). By reecting O across l, we see that A is optimal when the ray from (, 0) to A is tangent to, or in other words a = 1 3. Then C is exactly the curve given as the answer in Section 1. This completes the proof of Proposition 3 and the solution to Problem 1. Appendix: Paths avoiding convex bodies Here we prove a basic fact (Proposition 5) about paths with given endpoints that avoid a given convex region of the plane, which was used in steps 3 and 4 of the proof of Proposition 3 above. Let A, B R be two distinct points, and R + R one of the open half-planes bounded by the line AB. Let D R be a compact, strictly 3 convex set such that A, B / D, but D meets both R + and the line segment AB. 3 By strictly we mean every line either splits D in two, or else meets D in at most one point. However, the strictness assumption is not essential, and we include it only to avoid a few technical annoyances. 6

7 Let A denote the unique point on D R + such that AA D = {A }. Dene B D R + likewise with respect to B. Let γ(a, B; D, R + ) denote the path from A to B that consists of the line segment AA, the arc along D from A to B inside R +, and the line segment B B. Proposition 5. With the above notation and assumptions, γ := γ(a, B; D, R + ) is the shortest path from A to B that lies in R + (except at its endpoints) and avoids the interior of D. Proof. Suppose for the sake of contradiction that there were a shorter alternative β to γ. Step 1: Reduction to the case where β is polygonal. any point O D AB, and let { ɛ := min dist({a, B}, D), Choose a sequence of points } η 3 dist(o, β), η 6 γ > 0. A = P 0, P 1,..., P n, P n+1 = B Fix η := 1 β γ > 0. Choose along β, spaced less than ɛ apart in arc length along β. For each i {1,..., n}, let Q i be the point on the ray OP i lying a distance 3ɛ beyond P i. Let β be the polygonal path AQ 1 Q n B. We claim that β avoids D. Since ɛ dist({a, B}, D) and, by assumption, AP 0, P n B < ɛ, it is easy to see that the segments AQ 1 and Q n B avoid D. We claim that all segments Q i Q i+1 avoid D as well. Indeed, suppose for the sake of contradiction that X D Q i Q i+1. By the law of cosines, OP i + OP i+1 OP i OP i+1 cos P i OP i+1 = P i P i+1, so Moreover, 0 1 cos P i OP i+1 = P ip i+1 ( OP i OP i+1 ) OP i OP i+1 P i P i+1 OP i OP i+1. Q i Q i+1 = OQ i + OQ i+1 OQ i OQ i+1 cos Q i OQ i+1 = (3ɛ + OP i ) + (3ɛ + OP i+1 ) (3ɛ + OP i )(3ɛ + OP i+1 ) cos P i OP i+1 = P i P i+1 + 6ɛ(3ɛ + OP i + OP i+1 )(1 cos P i OP i+1 ), Since OP i, OP i+1 dist(o, β) 3ɛ η, 7

8 we therefore have and In particular, and 1 Q iq i+1 P i P i ɛ(3ɛ + OP i + OP i+1 ) OP i OP i+1 1 Q iq i+1 P i P i+1 Q i X Q i Q i+1 (1 + η) P i P i+1 < ɛ, 1 + η + η + η = (1 + η), 1 + η. (3) OX OQ i Q i X > OP i + 3ɛ ɛ = OP i + ɛ. Now between P i and P i+1, by assumption, the arc of β has length less than ɛ, and yet this arc must meet the ray OX at some point Y outside int(d). So P i Y OY OP i OX OP i > ɛ, by the previous inequality. This is a contradiction, so our assumption of the existence of X must have been false. So indeed the polygonal path β is disjoint from D. Moreover, inequality (3) implies that β < P 1 Q 1 + (1 + η) β + P n Q n = (1 + η) β + 6ɛ (1 + η)(1 η) γ + η γ = γ. We have thus reduced to the case where β is a polygonal path, which henceforth we will call where P 0 = A and P n+1 = B. P := P 0 P n+1, Step : Reduction to the case where each segment P i P i+1 meets D at one point. Let us induct primarily on n (the number of intermediate vertices in the polygonal path P), and, in cases of equal n, on the number of segments P i P i+1 (0 i n) disjoint from D. If, for some pair of indices (i, j) (0, n+1) with j i+, the segment P i P j fails to cut in two, then throw away P i+1,..., P j 1. The resulting polygonal path P := P 0 P i P j P n is no longer than the original P, so we are done by induction on n. Therefore, we may assume that no such pair (i, j) exists. If there exists a segment of the form P i P i+1 disjoint from D, then either i 0 or i n. Without loss of generality, assume the former. Then, as either (i 1, i + 1) = (0, n) or P i 1 P i+1 cuts D in two, there must exist a unique point P i strictly between P i and P i+1 such that P i 1 P i meets D in exactly one point. Then replacing P i by P i, we are done by induction, since n stays the same, and the polygonal path P := P 0 P i 1 P i P i+1 P n+1 is shorter than P and has fewer segments disjoint from D. So we are reduced to the case where P i P i+1 meets D at a unique point W i, for every i {0,..., n}. Then W 0 = A and W n = B. Moreover, since all segments of the form P i P i+ cut D in two, we must have W i W i+1. It follows that W 0,..., W n are in order along the arc of D in R +. 8

9 Step 3: a local calculation. It remains to show that, for i {1,..., n}, W i 1 W i D < W i 1 P i + P i W i, where W i 1 W i D denotes the length of the arc between W i 1 and W i along D in R +. To this end, let us introduce some coordinates. Suppose that P i = (0, 0) R, and the y-axis bisects W i 1 P i W i, so that W i 1 = ( a, ta) and W i = (b, tb) for some a, b, t > 0. The arc of D in question is the graph of a convex function f : [ a, b] R 0 with f( a) = ta and f(b) = tb; this graph remains above the graph of for all x [ a, b]. Then if f is dierentiable, convexity says so that W i 1 W i D = ˆ b a x t x t f (a) f (x) f(b) t, 1 + f (x) dx (a + b) 1 + t = W i 1 P i + P i W i, as needed. If f is not dierentiable, the same idea works by taking piecewise linear approximations to f. 9