A translation of Henri Joris' "Le chasseur perdu dans la forêt" (1980)
aa r X i v : . [ m a t h . HO ] O c t A translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980)
Steven Finch
October 1, 2019
Abstract.
This is an English translation of Henri Joris’ article “Lechasseur perdu dans la forˆet (Un probl`eme de g´eom´etrie plane)” that appearedin
Elemente der Mathematik v. 35 (1980) n. 1, 1–14. Given a point P and aline L in the plane, what is the shortest search path to find L , given its distancebut not its direction from P ? The shortest search path was described by Isbell(1957), but a complete and detailed proof was not published until Joris (1980).I am thankful to Natalya Pluzhnikov for her dedicated work and to the SwissMathematical Society for permission to post this translation on the arXiv.
1. A hunter wandered into the woods and became disoriented. After awhile, hefound a sign with information about a road that passes exactly 1 kilometer away.Unfortunately, the tree on which the sign was attached had fallen to the ground, andthe hunter possessed no idea in what direction he should travel to reach the road.He decided to walk 1 kilometer straight in an arbitrary direction and then continuealong a circle with center at the sign location. Therefore, he is certain to find theroad after walking at most 1 + 2 π kilometers.This situation is described in a problem presented at a mathematical competitionfor high school students, which are so popular in the United States. The questionis: Suppose the road is straight. What is the shortest search path that the hunterwould have taken, had he thought about everything a little more carefully?
Theanswer probably is: “Instead of traversing the full circumference of the circle, hewould replace the last quarter by a straight path of length 1 km tangent to the circle(Figure 1), saving π/ − ≈ .
57 km”.However, he can make his way even shorter if he first continues to walk in thedirection of the radius beyond the circle and then return to the circle tangentially(Figure 2). This will give the path
ABCDEF of length1cos α + tan α + ω + 1 = 1cos α + tan α + 32 π − α + 1 = ℓ ( α ) .ℓ ( α ) attains its minimum at α = 30 ◦ = π/ ℓ (cid:16) π (cid:17) = 76 π + 1 + √ . (1)1 translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980)
22. If we disassociate from the hunter and the forest, and rephrase in terms ofplane geometry, we come to the following problem (P):
Given a circle of radius 1, find the shortest path that starts at the center of thecircle and intersects all the tangents to the circle.
The following theorem holds:
Theorem 1 . A solution of (P) is the path of Figure 2 with α = π/ . Any othersolution is obtained from it by rotations and reflections of the plane that leave thecircle in its place. If we remove from (P) the condition about the starting point, we obtain
Theorem 2 . The shortest path that intersects all the tangents of a given circleconsists of a semicircle (subset of the given circle) extended on each side by a segmentof the tangent of length of the radius (Figure 3).(If the radius is of length 1, the length of the path is π + 2.)Although these theorems, especially the second, look quite elementary and plau-sible, I could not find simple and totally elementary proofs of them. This is due tothe fact that one must take into consideration all continuous and rectifiable curvesthat satisfy the conditions, and also that the solutions must be curves composed ofstraight line segments and circular arcs.3. In what follows, we shall prove these two theorems. The idea is to show thateach minimal curve is one of those indicated in the theorems. First we prove theexistence of minimal curves. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) m be the infimum of the lengths of the curves that meet all the tangents, and C n a minimal sequence of such curves; then ℓ ( C n ) = (length of C n ) → m . Each C n can be parametrized by f n : [0 , → E , the parameter being proportional to the arclength. The ℓ ( C n ) are bounded; therefore, all the C n are contained in a closed square,and distance ( f n ( t ) , f n ( t )) ≤ | t − t | · ℓ ( C n ) ≤ | t − t | · M for all t , t , n , where M is a constant. Hence one can apply the Arzela-Ascoli theorem:there is a path C given by f : [0 , → E such that f n → f uniformly. Then ℓ ( C ) ≤ m ,and C meets every tangent. Indeed, assume the contrary. If C does not meet t , thenit is at a positive distance from t , and in view of the uniform convergence f n → f , C n will be at a positive distance from t for n ≥ N , which is impossible. Consequently,there is a minimum for Theorem 2. As for Theorem 1, each C n starts at the centerof the circle, and hence so does C .4. Notation . If
A, B, C, D are points of the plane, then AB will denote the straightsegment between A and B , | AB | the distance between A and B , and ABCD . . . thepath AB ∪ BC ∪ CD ∪ . . . . If A = B , let d ( AB ) be the straight line that connects A and B , and r ( AB ) the half-line of d ( AB ) that starts at A and passes through B . If A = B = C and r ( BA ) is not the half-line opposite to r ( BC ), then a ( ABC ) will bethe closed convex angular region bounded by r ( BA ) and r ( BC ), and ∡ ( ABC ) theangular measure of a ( ABC ). Therefore, 0 ≤ ∡ ( ABC ) < π always.The convex envelope of a set S will be denoted by k ( S ). In particular, if A, B, C, . . . are points, k ( ABC . . . ) will be the smallest convex polygon that contains
A, B, C, . . . ; k ( ABC ) will be the triangle with vertices
A, B, C . translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) K , and its interior and exterior, int( K ) andext( K ).If a path is given by f : [ a, b ] → E and X = f ( t ) , Y = f ( s ), we write X < Y if t < s .5. First, let C be an arbitrary path given by c : [ a, b ] → E . The interval [ a, b ] is aunion of c − ( K ), c − (int( K )) and c − (ext( K )); as we know from topology, c − (int( K ))and c − (ext( K )) are composed of open intervals in [ a, b ], c − ( K ) is composed of closedintervals in [ a, b ], and additionally, if there are infinitely many such intervals, of theaccumulation points. I will denote the paths corresponding to the intervals of c − ( K ), c − (int( K )) and c − (ext( K )) by j , j , j respectively, after having assigned to themtheir initial and terminal points if necessary.6. Consider now a j of a minimal curve C . A j is an arc of K , but not all of K . If the ends of the arc are not the initial and terminal point of j , then a part ofthe arc is traversed twice, which allows for a shortening by a chord. Therefore, for aminimal C the j are arcs of K traversed once. The j , being in the interior of thecircle, where there are no tangents, are obviously chords or segments of chords.7. It remains to consider the j . First let X ∈ ext( K ); let t + and t − be the twotangents of K from X such that, when viewed from X , K is on the left of t + . Let P + ( X ) and P − ( X ) be the points at which t + and t − touch K (Figure 4). For X ∈ K ,we write P + ( X ) = P − ( X ) = X . Thus P + and P − are continuous maps of K ∪ ext( K )onto K . Each P + ( j ) and P − ( j ) is a closed arc on K , and if j has a point on K ,then P + ( j ) ∪ P − ( j ) is a closed arc on K . Now let X ∈ C ∩ ext( K ) and let t be atangent that properly separates K from X so that X / ∈ t . Suppose for simplicitythat X is not initial or terminal on C . There are Y, Z on C , Y < X < Z , such thatfor Y ≤ X ≤ Z , X is also properly separated from K by t . Let X , X , X , X befour points, Y ≤ X ≤ X ≤ X ≤ X ≤ Z , for which the minimum and maximumof P + and P − on the path { X ∈ C : Y ≤ X ≤ Z } are attained. Then the polygonalpath Y X X X X Z meets the same tangents and is strictly shorter than any otherpath that contains Y, X , X , X , X , Z in the same order. It follows that each j isa succession of segments. We call AB a maximal segment of C if AB is not part of asegment A ′ B ′ of C that properly contains AB .8. It is useful to consider the problem in a different way. Let D be a convexcompact domain on the plane E . The support lines of D are the lines ℓ that intersect D in such a way that D is contained completely in one of the two half-planes definedby ℓ . Let S be a connected set. Then D belongs to k ( S ) if and only if S intersectsall the support lines of D . In particular, if D is bounded by a smooth curve, thenthe support lines are the tangents of the boundary curve. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) C such that K ⊆ k ( C ).We recall that k ( S ∪ S ) is a union of segments XY with X ∈ k ( S ) and Y ∈ k ( S ).We say that an arc S of a path C is repeating if either K ⊆ k ( C r S ) or C r S intersects all the tangents to K . In particular, S is repeating if S ⊆ k ( C r S ). Arepeating arc is always a straight segment traversed once.9. Let P ∈ K ∩ C , let t be a tangent at P , and let Q ∈ C be properly separatedfrom K by t , that is, Q belongs to the interior of the half-plane defined by t that doesnot contain K . Then C is a straight segment near P . Indeed, if S is an arc of C thatcontains P and is contained in the interior of the triangle k ( P + ( Q ) QP − ( Q )) (Figure5), then there must be points X, Y in the dashed regions with
X, Y ∈ k ( C r S ).Therefore, S ⊆ k ( Y QX ) ⊆ k ( C r S ) hence S is repeating, and hence a segment.10. The arguments of § AB and CD that intersect at an interior pointof at least one of the segments AB and CD . Indeed, in Figure 6, k ( AB ∪ DC ) = k ( DB ∪ AC ). We obtain a shortening by replacing AB and CD with CA and DB and by changing the direction of a part of C .Imp. b): Two consecutive segments ABC form an angle such that the oppositeangular region contains a point X of K or of k ( C r ABC ), X = B . Indeed, in Figure7, let S = ( C r ABC ) ∪ A ∪ C , C = ABC ∪ S . There exists Q ∈ k ( S ) such that X ∈ P Q , P ∈ k ( ABC ); then
ABC ⊆ k ( QAC ), ABC is repeating and must bereplaced with the segment AC , which is shorter than ABC . translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) AC such that C is the initial or terminal point of C andsuch that r ( AC ) intersects K beyond C . The only exception: C is the initial pointimposed by Problem (P).Imp. d): Two coinciding segments at least one of which is followed by a segment,in a different direction (Figure 8).Imp. e): Two consecutive segments A ′ BC ′ such that K is contained in a ( A ′ BC ′ ).Indeed, if A and C are in the interior of A ′ B and C ′ B respectively, sufficiently closeto B , then it is easily seen that d ( AC ) properly separates B from K and that ABC can be replaced with AC , which is shorter.Imp. f): Two consecutive segments ABC such that K is contained in one of theangular regions supplementary to a ( ABC ), respectively, the half-planes defined by d ( AB ) if r ( BA ) = r ( BC ). The only exception is the case that A = C ∈ K and d ( AB )is a tangent to K . We consider here the nondegenerate case where r ( AB ) = r ( BC )and ABC ⊆ ext( K ) (Figure 9). Suppose that BA and BC are maximal and that d ( BA ) separates K from BC . Let the tangents t and t from A and B be as in thefigure. For C we have the possibilities C , C , C . BC is repeating, hence C = C .If C = C , then BA ′ is repeating, and hence BA is not maximal. Therefore, C = C .If C is the terminal point of C , then BC is superfluous (because the only imposedterminal point of C is the center of K ). Consider the continuation of C beyond C . Ifthis is a segment CD with D properly separated from K by t , then BA cannot bemaximal. If, on the other hand, the continuation is CD with D in the same closedhalf-plane (determined by t ) as K , then BCD is repeating, which is impossible. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) j can have. Let AB be a maximalsegment of j such that d ( AB ) does not intersect K . In view of Imp. b), e), and f),either B is the terminal point or C passes after B along a segment directed towards K . The same is true for A . If AB ⊆ j is such that d ( AB ) intersects K [say, r ( AB )intersects K ], then by Imp. c) and b), B must be on K . If, moreover, d ( AB ) is notthe tangent to K from B , then the segment AB must continue to int( K ) (in view of § B . A can be terminal. If A is not terminal, C continues beyond A along AC , C ∈ K , or along AC such that d ( AC ) ⊆ ext( K ), which is the case consideredabove.Thus, we found for j the seven possibilities indicated in Figure 10.The continuations to the interior of K are indicated. Identical angles are markedby identical letters. Also the right angles are indicated.For example, the equality of the two angles β in case V is due to the followingelementary fact: if X , Y are two points on the same side of a straight line d , thenthe shortest path between X and Y that passes through d consists of two segments XZ , Y Z with Z ∈ d that form the same angle with d . (The explanation of the rightangles is even simpler.) In case IV we obtain ε = 0 as a limit case. It is easily seenthat α ≤ β + γ, β ≤ α + γ. (2)It can also be shown that 2 β + 2 α + ε ≥ π , but we will not use this inequality.12. We have shown that j , j , and j are the simplest arcs of C ; it remains to translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) j , j , j on K ∩ C . First we prove the following.If AB is a j of C , that is, a chord or part of a chord of K , and B ∈ K , then B isnot the terminal point of C , and C continues to a segment BC ; C ∈ ext( K ), B ∈ AC ,that is, C continues straight ahead to d ( AB ). First of all, if B were terminal, let A ′ ∈ AB ∩ int( K ), S = ( C r A ′ B ) ∪ A ′ . S meets all the tangents, except perhapsthose from B . But since S is closed, it meets also those from B by continuity, andhence A ′ B would be superfluous. In the same way, the rest of C is not formed byanother j . The arguments that follow refer to Figure 11: t is the tangent from B that we take to be horizontal. If there is a Y ∈ C that is properly separated from K by t , then the assertion follows from §
9. Otherwise there will be a
Z > B suchthat S = { X ∈ C : Z ≥ X ≥ B } lies in the rectangle k ( P QRW ), where
R, W ∈ K and P, Q ∈ t , with small | QB | and | P B | . Suppose ∡ ( A ′ BP ) ≤ π/
2. To pass from k ( QBT R ) to k ( P BT W ), S must pass through B , since S cannot cross A ′ B inside,in view of Imp. a). If S passes the two sides of A ′ B infinitely many times, it musthave infinitely many loops b departing from and arriving at B , in k ( P BT W ). Let b be such a loop, with horizontal elongation | BS | . For a sufficiently small BP , thepath A ′ SB is shorter than A ′ B and b together, since | A ′ S | + | BS | ≤ | A ′ B | cos γ + | BS | ≤ | A ′ B | cos γ − | BS | + length( b ) ≤ | A ′ B | cos γ − | A ′ B | sin γ + length( b ) < | A ′ B | + length( b )if γ is sufficiently small. Therefore, if b is a loop with maximum horizontal elongation,then we can replace A ′ B with A ′ SB and omit all the loops in k ( P BT W ). Therefore,we may assume that S is entirely on one side of A ′ B , say in k ( QBT R ). However,in this case S ⊆ k ( QBR ). Indeed, if X is a point in the triangle k ( BT R ) without RB , then X is on a chord j that must intersect RW or BT in the interior, unless S ⊆ j , the case which has already been excluded. Now let U ∈ S be the leftmost forall of S . We replace A ′ B ∪ { X ∈ S : U ≥ X ≥ B } with A ′ V U , which meets all thetangents met by A ′ B ∪ { X ∈ S : U ≥ X ≥ B } and has the length ≥ | A ′ B | + | BV | ,whereas for ω = ∡ ( A ′ BV ), | AA ′ V | + | V U | = | A ′ B | cos δ + | V B | cos( π − ω − δ ) + | V U |≤ | A ′ B | cos δ + | V B | | cos( ω + δ ) | + | V B | tan α < | A ′ B | + | V B | if α and δ are sufficiently small. This proves the assertion.13. Each j = j or j “covers” a set of tangents whose intersection points with K form an arc P + ( j ) ∪ P − ( j ) on the circle K . We denote this arc P ( j ). We claim that translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) P ( j ) does not have common interior points with P ( j ′ ) if j = j ′ . This is clear if j or j ′ is a j , and hence an arc of K , since a part of this j can be replaced by a shorterchord. Thus, assume that j and j ′ are j . Obviously, P ( j ) P ( j ′ ); otherwise j isrepeating. Therefore, we are in a situation of Figure 12. If j is a j of type V (Figure10), we can shorten the path by cutting through the angle formed by C at C . If j is oftype IV (Figure 10), we can go down from C along the left tangent, which will ensurea shortening unless β = α + γ , and so forth. We arrive at the following situation asthe only possibility (Figure 13): j contains the segment b = BD , with β ≤ π/
2, and j ′ contains the segment a = AC with α ≤ π/
2. In the same way as in the proofs of §
10, if S = C r a r b , we find points X, Y, W ∈ k ( S ), in the indicated angular regionsfor which R, Q, P , respectively, are in k ( C ). But then a, b ⊆ k ( ABW XY ), and a r A and b r B are repeating, as well as their continuations up to the tangents t and t ,respectively, which would enforce a forbidden crossing. We note that this is the firsttime we used the fact that K is a circle, or rather that the normals to the tangentsat Q and P meet at Z ∈ int( K ). Up until this moment, everything was applicablefor smooth and convex K s that do not contain straight segments.14. It is easily seen from the above that if j = j or j , then C r j is entirelyon one side of K with respect to each tangent that touches K at P ( j ), that is, inthe non-dashed part of the plane shown in Figure 14. It follows that if AB ⊆ C is amaximal segment that contains a chord P Q of K , then neither A nor B is terminaland the continuations of C from A and from B do not go in the same half-planedetermined by d ( AB ). In other words, we have the situation of Figure 15. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) C with the tangent t k AB must be in the dashed region, which C can reach neither from X nor from Y without crossing AB .15. We show that C consists of finitely many j , j , j . Assume the contrary. Thenthere exists an X ∈ K ∩ C such that for all Y, Z ∈ C , Y < X < Z , there are infinitelymany j , j , j between Y and Z , say between Y and X . Suppose there are no chords j among them. Then the j must be of type II’ in Figure 10. But each of those haslength 2, and so there are only finitely many of them; therefore, if Y is close enoughto X , there are only j ’s, whence { W ∈ C : Y ≤ W ≤ X } is an arc on K and belongsto a single j . Consequently, there are infinitely many chords j approaching X . Inview of §
14, they must zigzag, as in Figure 16. The j have exactly the same directionas the tangent t at X . To each j we attach a j of the form III, IV, or V (Figure 10).For a j of the form V we have, in Figure 17, 2 π = ε + δ + π − α + π − δ ; therefore α + β = ε + δ . But α + β = π − δ , and hence δ = π/ − ε/
2; for | ε | small, δ ≈ π/
2, and P ( j ) has a length ≈ π/
2, which is too large. In the same way we have contradictionsfor j of the types III and IV. Thus we have found that C is a finite union of paths j , j , j .16. We can now prove Theorems 1 and 2. In Theorem 1 we have one free terminalpoint; in Theorem 2, both terminal points are free. We see that an arc j on K cannotbe a terminal path of C . We already know that the j are not terminal except for thefixed terminal point in Theorem 1. In view of §
14, the j of type I are not terminal.Therefore, only the j of type II or VI remain candidates for a free terminal path. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) j of type II, and hence a tangent oflength 1. It may be followed by a j of type III or by an arc j . In the first casewe have Figure 18. But here we see that ABC can be replaced with
BAC , which isshorter and has the same convex envelope. The same argument is valid if C ends witha j of type VI. Consequently, the only possibility is a tangent of length 1 followed byan arc. We show that C does not admit entire chords. Indeed, assume the contraryand take the first such chord after the arc. We obtain Figure 19. Let α + ω ≤ π/
2. If | RE | < π − − π − , we verify that the circle centered at E of radius | ER | + ( π/ − V (on the diameter RM ), and hence | ED | > | ER | + π − . (3)This enables us to replace the path ABCDE (where BC is an arc) with DCBRE ,where
CBR is an arc on K . The length of the first is 1 + ω + tan α + | DE | > ω + tan α + | ER | + ( π/ − ω + tan α + ( π/
2) + | ER | which is the length ofthe second. If | RE | > π − − π − , translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) | AB | + | BE | > | RE | ≥ π − − π − > π + 2 , and hence it is longer than the path in the theorem.For Theorem 1, the curve must return to the center from E . Therefore, the pathwill be longer than | AB | + | BE | + | EZ | > | RE | ) > π , and hence longerthan the path of Theorem 1.Now let ω + α > π/
2. For Theorem 1, we see that DE is an obstacle on the wayof C back to the center. For Theorem 2, we start from the other terminal point, butwe must have ω ′ + α ′ < π/ §
14, which reduces to the previous case.Thus, we see that the single j can only be the semichord leading to the center,for Theorem 1, and so we cannot obtain anything different from the paths describedin the theorems. Remarks (a) The problem considered can be largely generalized by replacing, for example, K with an arbitrary convex domain, or the tangent lines with tangent circles, etc.The general statement of the problem is this: In an arcwise connected metric space,given a family F of closed F and a connected compact set K that intersects each F ∈ F , find the shortest path that intersects all the F .(b) In higher dimensions, if E n is n -dimensional Euclidean space, S n − the unitsphere, and ℓ n the length of the shortest path that meets all hyperplanes tangent to translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) S n − , or the shortest path C with S n − ⊆ k ( C ), then by induction ℓ ≥ s(cid:18) √ π (cid:19) + 4 ≈ . ,ℓ n ≥ const +2 n. By constructing particular paths we find that ℓ n ≤ const · n / ,ℓ ≤ √ · · π ≈ . . The upper bounds seem to me to be the closest to reality.(c) In an infinite-dimensional Hilbert space, say H = l , the circumstances areslightly different. If S is the unit sphere, there is no path of finite length, nor acompact path C such that S ⊆ k ( C ), since k ( C ) is compact, but S is not. Therefore, wemust consider compact convex sets K ⊆ H . In this case there surely exists a compactpath C (which can be easily constructed) such that k ( C ) ⊇ K . [In view of a theoremby Hahn and Mazurkiewicz, there even exists a continuous map f : [0 , → H withIm f = K (see [1]).] However, this curve in general is not of finite length. If K = ( ( x , x , . . . ) : ∞ X j =1 j x j ≤ ) , then K is compact and convex, but any curve C with k ( C ) ⊇ K has infinite length.If K = ( ( x , x , . . . ) : ∞ X j =1 j x j ≤ ) , then there exists C of finite length with k ( C ) ⊇ K . Moreover, it can be shown that if K is compact and convex and C has finite length and k ( C ) ⊇ K , then there exists aminimal curve. This can be done as in §
3. It must simply be shown that if { C n } is aminimal chain of curves, then S ∞ n =1 C n is relatively compact.Written by Henri Joris, Gen`eveTranslated by Natalya Pluzhnikov Addendum
The only citation to the literature given in Joris (1980) is [1].I have included [2, 3, 4, 5, 6, 7, 8, 9, 10] for the sake of completeness.John Wetzel provided valuable comments as I prepared this draft for posting. translation of Henri Joris’ “Le chasseur perdu dans la forˆet” (1980) References [1] M. H. A. Newman,
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Geom. Dedicata196(2018) 123–143; MR3853631.Steven FinchMIT Sloan School of ManagementCambridge, MA, USA