Inscribed triangles of Jordan curves in \mathbb{R}^{n}
aa r X i v : . [ m a t h . M G ] F e b INSCRIBED TRIANGLES OF JORDAN CURVES IN R n ARYAMAN GUPTA AND SIMON RUBINSTEIN-SALZEDO
Abstract.
Nielsen’s theorem states that any triangle can be inscribed in a planar Jordancurve. We prove a generalisation of this theorem, extending to any Jordan curve J embeddedin R n , for a restricted set of triangles. We then conclude by investigating a condition underwhich a given point of J inscribes an equilateral triangle in particular. Introduction A Jordan curve is a continuous image of the unit interval in R n that is injective everywhereexcept the endpoints, which are mapped to the same point. A polygon is inscribed in aJordan curve if the vertices of the polygon lie on the curve. There has been a considerableamount of interest surrounding the inscription of triangles and quadrilaterals in Jordancurves embedded in the plane. A lot of this interest stems from the Toeplitz square pegconjecture, which asks whether any Jordan curve in the plane has an inscribed square.Detailed exposition on this and similar problems can be found in [Mat14] and [Pak10].In the literature, some of these variants have already been resolved. See, for exam-ple, [Mey81] and [Nie92], wherein it is shown, respectively, that a planar Jordan curvenecessarily inscribes a rectangle and any particular triangle. Yet, the original conjectureitself remains unproven, except under certain geometric or topological conditions. See, forinstance, [Str89], [Mat14], [Emc16] and [NW95].Here, instead of squares, we shall consider the inscription of triangles, under geometricconditions on the curve. The motivation for this paper comes from the following two results,proven respectively in in [Nie92] and [Mey80]. Theorem 1.1 (Nielsen) . Let J ⊂ R be a Jordan curve and let △ be any triangle. Theninfinitely many triangles similar to △ can be inscribed in J . Theorem 1.2 (Meyerson) . Let J ⊂ R be a Jordan curve. For every point p ∈ J except atmost two, there exists an inscribed equilateral triangle such that one of its vertices is p . There has also been a smaller, but still significant, amount of interest in higher dimensionalvariants; see for instance [Str89], [Mak16], [AK13], and [NW95]. There are at least twodifficulties in dealing with Jordan curves in at least three dimensions. Firstly, many proofsregarding inscription for planar Jordan curves (including the proof of Theorem 1.1) rely uponthe property that J divides the plane into two disconnected subsets. Since this does not, ofcourse, generalise to higher dimensions, these proofs cannot be generalised in any obviousway. Secondly, since Jordan curves in higher dimensions are able to form knots, they canpotentially be much more pathological than planar Jordan curves.The aim of this paper is to prove that, subject to a certain geometric restriction, anytriangle can be inscribed in a given Jordan curve J embedded in R n .Before stating the main result, we introduce some notation. • J denotes a Jordan curve embedded in R n , defined by γ : [0 , → R n . −→ P −→ Q −→ Q −→ Q - −→ P −→ Q - −→ Ptri ( o, p ) = { q , q } tri ( o, p ) = CC −→ Q −→ Q −→ P R R −→ Q − −→ P −→ Q − −→ P −→ Q −→ Q Q Q p Q p Q Figure 1.
The set tri( o, p ) in R and R respectively. For the former space,this set is the 0-sphere { q , q } . For the latter space, this set is a a 1-sphere. • Let δ ∈ (0 , ). Then, Θ δ : (0 , δ ) × (0 , δ ) → R ≥ denotes the function that maps eachpair ( s, s ′ ) ∈ (0 , δ ) × (0 , δ ) to the angle between oγ ( s ) and −−−→ oγ ( s ′ ). • Θ ′ δ : (1 − δ, × (0 , δ ) → R ≥ denotes the function that maps each pair ( s, s ′ ) ∈ (1 − δ, × (0 , δ ) to the angle between −−−→ oγ ( s ) and −−−→ oγ ( s ′ ). • △ abc denotes the triangle in R n with vertices a, b, c ∈ J .Here is our main result. Theorem 1.3.
Let θ v be the angle of some vertex v of △ . If there exists a θ v such that lim sup δ → + Θ δ < θ v < lim inf δ → + Θ ′ δ , then there exist two points p, q ∈ J \ { o } such that △ opq is similar to △ , with vertex v corresponding to o . Note that the hypothesis of Theorem 1.3 is automatically satisfied when o is a smoothpoint.For the proof, our first concern is how we identify when o, p, q ∈ J inscribe △ . Assuming △ opq is similar to △ , let r := k q − o kk p − o k and r ′ := k q − p kk p − o k , where r ≥ o to be γ (0) and assume that it is the origin of R n , and we let p be any element of J − { o } . Then, the set tri( o, p ) of points q such that △ opq is similar to △ is an ( n − o, p ) := ( q ∈ R n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k−→ Q kk−→ P k = r ) ∩ ( q ∈ R n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k−→ Q − −→ P kk−→ P k = r ′ ) , where −→ Q := −→ oq and −→ P := −→ op . Since both sets of the intersection are ( n − o and p respectively, tri( o, p ) is an ( n − p ∈ J − { o } suchthat J ∩ tri( o, p ) = ∅ .Here is a more detailed outline of the path we take to prove our main result. • Setup: In § o, γ ( t )) := S t , a scaled isometry I t such that I t ( S t ) is equal to the same ( n − S n − for all t ∈ (0 , S t is mapped to a constant frame of reference RIANGLES INSCRIBED IN CURVES 3 o S t γ ( t ) (a) At t , S t has notyet crossed through γ .In turn, γ ′ t is not ho-motopic in X to c . S T o γγ ( T ) (b) At T exactly, S T crossesthrough γ . S t o γγ ( t ) d max rd max (c) At t , S t has already crossedthrough γ , when γ ( t ) is farthestfrom o . Recall that r ≥
1. In turn, γ ′ t is homotopic in X to c . Figure 2. As t increases from t to t , and as γ ( t ) goes farther away from o , S T intersects γ for some T ∈ (0 , t ∈ (0 , S t in R n , it will be moreconvenient to reorganise our coordinate system to make S t appear stationary, ratherthan considering a moving complement. In § γ and S t neverintersect. Under this assumption, we prove that I t ( γ ) ≃ I t ′ ( γ ) in X := R n − S n − forall t, t ′ ∈ (0 , ≃ denotes that two loops are freely homotopic in X . • Finding t : In § γ ′ t := I t ( γ ) is homotopicin X to the trivial loop c at I t ( o ) for some t ∈ (0 , • Finding t : In §
4, we prove a series of technical lemmata (namely, Lemmata 4.2to 4.15) leading to a proof of Lemma 4.1, which states that γ ′ t is not homotopic in X to c for some t ∈ (0 , γ ′ t γ ′ t if γ ∩ S t = ∅ for every t ∈ (0 , γ ′ t ≃ γ ′ t if γ ∩ S t = ∅ for every t ∈ (0 , γ ∩ S t = ∅ for every t ∈ (0 , • Inscribing equilateral triangles: In § o ∈ J inscribes an equilateral triangle if J satisfies a certain condition.See Figure 2 for diagrams outlining the sketch of the proof.2. Setup
Translating S t to S n − . We show here that S t can always be mapped onto S n − by ascaled isometry I t , where S n − = ( ( a , . . . , a n ) ∈ R n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − X i =0 a i = 1 , a n = 0 ) . The precise choice of I t is not important; we need only that I • is a continuous map from S to SO( n ) × R > such that I t ( S t ) = S n − for all t , but we provide a concrete descriptionof one such family I • nonetheless. We construct I t by composing a translation function T t : R n → R n , a rotation function R t : R n → R n (whose center of rotation is o ), and ascaling function s t : R n → R n . Here is how we construct each of the three functions. ARYAMAN GUPTA AND SIMON RUBINSTEIN-SALZEDO oS t o ′ T t T t ( S t ) Figure 3.
The translation T t maps o ′ to o .(1) T t : Let o ′ be the centre of S t . We set T t as the translation of R n that maps o ′ to o .(2) R t : Let Π t be the hyperplane that contains S t , let R n := { ( a , a , . . . , a n ) ∈ R n | a n = 0 } , and for any ( n − P ∈ R n , let N ( P ) = ( θ , θ , . . . , θ n − ,
1) denotethe unit normal vector of that plane expressed by its ( n −
1) spherical coordinates.Additionally, let us assume that N ( R n ) = (0 , , . . . , , N (Π t ) = ( φ , φ , . . . , φ n − , R t : R n → R n to be the rotationof R n about o defined by the equation R t (( θ , θ , . . . , θ n − , r )) = ( θ − φ , θ − φ , . . . , θ n − − φ n − , r ) . By this construction, N ◦ R t ◦ T t (Π t ) = N ( R n ). Thus, since every ( n − o is uniquely identified by its unit normal vector, it follows that R t ◦ T t ( S t ) = R n lies completely in R n , and is centred at o . Thus, R t ◦ T t ( S t ) is anonzero scaling of S n − .(3) s t : We set s t as the scaling of R n for which s t ◦ ( R t ◦ T t )( S t ) = I t ( S t ) = S n − .Figures 3, 4a, and 4b illustrate these transformations.Note that whilst I t always maps S t to the same set S n − , it also maps the loop γ to adifferent loop γ ′ t := I t ( γ ) for each t ∈ (0 , A homotopy lemma.Lemma 2.1.
Suppose there is no t ∈ (0 , such that γ ∩ S t = ∅ . Then for any t, t ′ ∈ (0 , ,we have γ ′ t ≃ γ ′ t ′ in X .Proof. Let F t,t ′ ( · , T ) := γ ′ (1 − T ) t +( T ) t ′ . This is a homotopy taking γ ′ t (at T = 0) to γ ′ t ′ (at T = 1). For any T ∈ [0 , F t,t ′ ( · , T ) is simply γ ′ t ′′ for some t ′′ ∈ (0 , γ ′ t ′′ intersects X , F t,t ′ is a homotopy in X . (cid:4) Finding t Notation . Throughout the rest of the paper, we let c denote the constant path at o . RIANGLES INSCRIBED IN CURVES 5 oR t R t ◦ T t ( S t ) T t ( S t ) (a) The rotation R t maps Π t to R n . o s t ◦ R t ◦ T t ( S t ) = S n − can s t R t ◦ T t ( S t ) (b) The scaling function s t maps R t ◦ T t ( S t ) to S n − . Figure 4 S t o γγ ( t ) d max rd max (a) As r ≥ rd max ≥ d max . Thus, S t is so far that it cannot intersect γ as it ishomotoped to c at o . S t o t = 0 t = t = γ (b) γ being homotoped to c by a straight linehomotopy. Figure 5
Let us choose t such that k γ ( t ) − o k := d max is maximal. We shall prove the followinglemma regarding t : Lemma 3.2.
Assume there is no t ∈ (0 , such that S t intersects J . Then γ ′ t is homotopicin X to c . The idea is as follows. Let x denote an arbitrary point in S t . Then we have k x − o k = rd max , since r ≥
1. Then k γ ( t ) − o k ≤ k x − o k for any t ∈ (0 , S t has alreadyslipped out from γ , being far enough from γ that γ can be shrunk to c by a linear homotopy L that does not intersect S t . See Figure 5. Proof of Lemma 3.2.
We first show that the straight-line homotopy L : [0 , × [0 , → R n from γ to c never intersects S t . We consider two cases for L ( s, · ); first, when t = 0, andsecond, when t ∈ (0 , • t = 0: L ( s,
0) = γ does not intersect S t for any s ∈ [0 ,
1] because of our assumptionthat no such intersection occurs.
ARYAMAN GUPTA AND SIMON RUBINSTEIN-SALZEDO • t ∈ (0 , L is a straight-line homotopy to o , k L ( s, · ) − o k must be monotonicallydecreasing over (0 , k L ( s, t ) − o k < d max ≤ k x − o k . That is, the distance of L ( s, t ) from o is always less than that of x from o . Thus, L never intersects S t when t ∈ (0 , L ( s, t ) does not intersect S t for any s, t ∈ [0 , I t induces the desired homotopy I t ◦ L taking I t ◦ γ = γ ′ t to I t ◦ c := c ′ . Since L never intersects S t , and since I t is a bijection, I t ◦ L never intersects I t ( S t ) = S n − .Therefore, γ ′ t ≃ c ′ in X via the homotopy I t ◦ L . (cid:4) Finding t Lemma 4.1.
Let θ v be the angle of the vertex of v the triangle △ to be inscribed. Supposethere is no t ∈ (0 , such that γ intersects S t . If θ v is such that lim sup δ → + Θ δ < θ v < lim inf δ → + Θ ′ δ , then there exists some t ∈ (0 , has the property that γ ′ t c ′ in X . The proof involves proving a series of lemmata (namely, Lemmata 4.2 to 4.15) leading toa proof of Lemma 4.1. We begin by recalling a well-known and simple preliminary result.
Lemma 4.2.
Let
A, B ⊂ R n be two disjoint compact sets. There is some inf( A, B ) > suchthat k a − b k ≥ inf( A, B ) for any a ∈ A and any b ∈ B . For each of the remaining lemmata (from Lemma 4.3 to Lemma 4.15), we assume thesame hypotheses as are assumed for Lemma 4.1, namely that there is no t ∈ (0 ,
1) such that γ intersects S t , and that the angle θ v of some vertex v of △ is such that lim sup δ → + Θ δ <θ v < lim inf δ → + Θ ′ δ . Lemma 4.3.
There exists some ε > such that sup Θ ε < θ v < inf Θ ′ ε .Proof. Since lim sup δ → + Θ δ < θ v < lim inf δ → + Θ ′ δ , there exist arbitrarily small values of ε , ε > ε < θ v < inf Θ ′ ε . Let ε := min( ε , ε ). Then, sup Θ ε < θ v < inf Θ ′ ε . (cid:4) Lemma 4.4.
For a positive real number r , let B ( o, r ) be the closed n -ball of radius r , centredat o . Additionally, let the neighbourhood N := Im γ | (1 − ε,ε ) . Then there exists d min > suchthat J ∩ B ( o, d min ) ⊂ N . Note that (1 − ε, ε ) is taken modulo 1, i.e. (1 − ε, ε ) = [0 , ε ) ∪ (1 − ε, Proof.
Let d min = inf( { o } ,J − N )2 . Since { o } and J − N are disjoint and compact, Lemma 4.2 tellsus that d min >
0. Since d min < inf( { o } , J − N ), it thus follows that ( J − N ) ∩ B ( o, d min ) = ∅ ,and thus that J ∩ B ( o, d min ) ⊂ N . (cid:4) Lemma 4.5.
There exists some t ∈ [0 , such that S t ⊂ ∂B ( o, d min ) .Proof. Let p ∈ N be a point such that k p − o k = d := d min r . We let t ∈ (0 ,
1) be suchthat p = γ ( t ). Then all the points of S t lie at a distance rd = d min from o . Therefore S t ⊂ ∂B ( o, d min ). (cid:4) RIANGLES INSCRIBED IN CURVES 7
Definition 4.6.
We say a point lies in the interior (or exterior) of S t when it lies within the interior (or exterior) of S t when Π t is taken as its ambient space . Lemma 4.7.
The interior of S t lies in B ( o, d min ) .Proof. Since (from Lemma 4.5) S t ⊂ ∂B ( o, d min ), it follows that its interior is a subset of B ( o, d min ) too. (cid:4) Lemma 4.8.
All intersections of J − N with Π t lie in the exterior of S t .Proof. Assume that some p ∈ ( J − N ) ∩ Π t lies in the interior of S t . Since (from Lemma 4.7)the interior of S t is a subset of B ( o, d min ), it follows that p ∈ ( J − N ) ∩ B ( o, d min ). However,from Lemma 4.4, J ∩ B ( o, d min ) ⊂ N , contradicting the assumption that p lies in the interiorof S t . (cid:4) See Figure 6 for an illustration of the preceding Lemmata.Since I t is an isometry, all of the Lemmata above (which apply for γ ) also apply to thecurve γ ′ t = I t ◦ γ —for example, γ ′ t | [1 − ε,ε ] intersects I t (Π t ) = R n only at the exterior of I t ( S t ) = S n − . Definition 4.9.
Let P : R n → R ≥ × R be defined by P (( r , r , . . . , r n )) = ( d ( r ) , r n ) , where d : R n − → R ≥ is defined to be d ( r ) = q r + r + · · · + r n − . Definition 4.10.
For any given p ∈ R \ Im f , let a path f : [0 ,
1] be parametrized to polarform ( r ( t ) , θ ( t )), where r ( t ) = k f ( t ) − p k and θ ( t ) continuously maps t to the angle of thesegment pf ( t ) relative to the positive vertical axis from p . Then, we define the windingnumber η p ( f ) of f relative to p by the equation θ (1) − θ (0)2 π . Lemma 4.11. η (1 , ( P ′ ◦ γ ′ t | [1 − ε,ε ] ) < .Proof. To begin with, note that P ( S n − ) = (1 , ,
1) = u , and let φ ( t ) be the anglethat the segment uP ◦ γ ′ t ( t ) makes relative to the segment uP ◦ γ ′ t (0). See Figure 7.Assume, for the sake of contradiction, that η (1 , ( P ◦ γ ′ t | [1 − ε, ) > ε <θ v , it follows φ ( ε ) < θ v . Also, since inf Θ ′ ε > θ v , it follows that φ (1 − ε ) > θ v .As φ (1 − ε ) > φ ( ε ), if η (1 , ( P ◦ γ ′ t | [1 − ε, − ε ] ) > p ) then φ ( t ) = θ v for some t ∈ [ ε, − ε ], in contradiction to the fact that φ ( t ) ≥ inf Θ ′ ε for all t ∈ [1 − ε, φ ( t ) ≤ sup Θ ε for all t ∈ [0 , ε ]. (cid:4) Definition 4.12.
Let a n : R n → R be the last-coordinate map, i.e. a n (( r , r , . . . , r n )) = r n . Lemma 4.13. a ◦ P ◦ γ ′ t ( ε ) > .Proof. Assume that this is not true. Then, it would follow that k o − γ ′ t ( ε ) k < k o − u k , whichimplies that γ ′ t ( ε ) lies in the interior of B ( o, d min ). This would, in turn, also imply that J N (in whose closure γ ′ t ( ε ) belongs) has a nonempty intersection with B ( o, d min ). However, thiscontradicts the definition of N in Lemma 4.4. (cid:4) ARYAMAN GUPTA AND SIMON RUBINSTEIN-SALZEDO op ′ θS t Np d min B ( o,d min ) b J − NJ − N b ′ inf Θ ′ ε Figure 6.
Here, the angle between the blue lines is sup Θ ε . The angle be-tween the left-facing blue line and the left-facing black line is inf Θ ′ ε . Sincesup Θ ε < θ v < inf Θ ′ ε , it follows S t is wide enough such that N always inter-sects Π t at the interior of S t . Here, p ′ denotes one such point of intersection. P ◦ γ ′ t (0) (1 , ε inf Θ ′ ε γ ′ t ( ε ) γ ′ t (1 − ε ) P ◦ γ ′ t | [ ε, − ε ] Figure 7.
The path P ◦ γ ′ t | [1 − ε,ε ] . This path cannot have a negative windingnumber, since that would contradict the fact that φ ( t ) ≥ inf Θ ′ ε for all t ∈ [1 − ε,
0] and φ ( t ) ≤ sup Θ ε for all t ∈ [0 , ε ]. RIANGLES INSCRIBED IN CURVES 9 (1 , P ◦ γ ′ t ( ε ) P ◦ γ ′ t (1 − ε ) η (1 , ( P ◦ γ ′ t | ε, − ε ) < η (1 , ( P ◦ γ ′ t | ε, − ε ) > Figure 8. η (1 , ( P ◦ γ ′ t | ε, − ε ) > ,
1) by theleft, whilst η (1 , ( P ◦ γ ′ t | ε, − ε ) < Lemma 4.14. η (1 , ( P ◦ γ ′ t | [ ε, − ε ] ) < .Proof. Let p = ( a, ≤ a <
1. Then, the angle that up makes against thepositive vertical axis from u is π . From Lemma 4.13, we infer that φ ( ε ) < π , and thus that φ ( ε ) − π <
0. See Figure 8.Since we also know that π − φ (1 − ε ) <
0, it follows that if η (1 , ( P ◦ γ ′ t | [1 − ε,ε ] ) > t ′ ∈ [ ε, − ε ] such that φ ( t ′ ) = π , i.e. that P ◦ γ ′ t ( t ′ ) = ( a,
0) for some0 ≤ a <
1. However, it would then follow that γ ′ t ( t ′ ) ∈ γ ′ t | [ ε, − ε ] = J − N lies within theinterior of S n − can , which is in contradiction to Lemma 4.8. (cid:4) Lemma 4.15. P ◦ γ ′ t P ◦ c ′ := c ′′ in P ( X ) = R − (1 , .Proof. Since η (1 , ( P ◦ γ ′ t ) = η (1 , ( P ◦ γ ′ t | [1 − ε,ε ] )+ η (1 , ( P ◦ γ ′ t | [ ε, − ε ] ), Lemmata 4.11 and 4.14imply that η (1 , ( P ◦ γ ′ t ) = 0, which is true if and only if this curve is not nullhomotopic. (cid:4) Proof of Lemma 4.1.
Assume that γ ′ t ≃ c ′ . Let H : [0 , × [0 , → R ≥ × R be a homotopysuch that H (0 , · ) = γ ′ t and H (1 , · ) = c ′ . Then, the homotopy P ◦ H takes P ◦ H (0 , · ) = P ◦ γ ′ t to the constant path P ◦ H (1 , · ) = c ′′ . However, from Lemma 4.15, we know that this isimpossible. (cid:4) Therefore, Lemmata 3.2 and 4.1 immediately imply Theorem 1.3.As a corollary of independent interest of this theorem, we get the following.
Theorem 4.16. If o ∈ J is a differentiable point, then any triangle △ can be inscribed in J .Proof. When o is differentiable, lim sup δ → + Θ δ = 0 and lim inf δ → + Θ ′ δ = π . Then, the angle θ v of any vertex of △ always satisfies lim sup δ → + Θ δ < θ v < lim inf δ → + Θ ′ δ . Thus, △ alwayssatisfies the requirements of Theorem 1.3 to be inscribed in J . (cid:4) Generalising theorem 1.2
We now prove that subject to certain restrictions, Theorem 1.2 can be generalised to anygiven Jordan curve J ⊂ R n . We state this condition before stating our generalisation. γ (1 − ε ) γ ( ε ) op (a) a p | (1 − ε,ε ) is mono-tone for any p ∈ γ | (1 − ε,ε ) − o . γ (1 − ε ) γ ( ε ) op (b) a p | (1 − ε, is monotone de-creasing but a p | [0 ,ε ) is mono-tone increasing, so a p | (1 − ε,ε ) can’t be monotone for any p ∈ γ | (1 − ε,ε ) − o . Figure 9. J is strongly locally monotone at o in Figure 9a, and is not inFigure 9b. Definition 5.1.
For any given o ∈ J , and any p ∈ J − o , let a p : [0 , → R be defined as a p ( t ) = oγ ( t ) k oγ ( t ) k · op. If there is some ε > a p | (1 − ε,ε ) is monotone for each p ∈ Im γ | (1 − ε,ε ) − o , wesay that J is strongly monotone at o . Here, γ | (1 − ε,ε ) is called a strongly locally monotoneneighbourhood of o .Examples wherein similar conditions have been used to prove inscription theorems canbe found in, for example, [Str89]. See Figure 9 for examples of neighbourhoods that bothsatisfy and fall outside this condition. Theorem 5.2. If J is strongly locally monotone at o , then the point o ∈ J has an inscribedinscribed equilateral triangle. Note that, in comparison to Theorem 1.3 for an equilateral triangle, this theorem isstronger, since it allows for an unaccountably infinite number of equilateral triangles to beinscribed at the strongly monotone neighbourhood containing the point o , since any point ofthat neighbourhood can be chosen as the first point for the equilateral triangle. By contrast,finding a suitable neighbourhood that satisfies the hypothesis of Theorem 1.3 only ensuresat least a single triangle of the desired specifications can be inscribed in it.For the proof, our first concern is how we identify when o inscribes an equilateral triangle.Let φ s : [0 , → [0 , s ] be defined by φ s ( t ) = st . For all s ∈ (0 , r ,s : [0 , → R ≥ and r ,s : [0 , → R ≥ be respectively defined by r ,s ( t ) = k γ ◦ φ s ( t ) − o kk γ ( s ) − o k and r ,s ( t ) = k γ ◦ φ s ( t ) − γ ( s ) kk γ ( s ) − o k , where, for any t ∈ [0 , r ,s ( t ) and r ,s ( t ) respectively represent the ratio of the lengths ofthe sides oγ ◦ φ s ( t ) and γ ( s ) γ ◦ φ s ( t ) to the side oγ ( s ). Then, the ratio path R s : [0 , → R RIANGLES INSCRIBED IN CURVES 11 for any s ∈ (0 ,
1) is R s ( t ) = ( r ,s ( t ) − , r ,s ( t ) − . To prove Theorem 5.2, then, it suffices to show that, for some s ′ , t ′ ∈ (0 , r ,s ′ ( t ′ ) = r ,s ′ ( t ′ ) = 1, and thus that R s ′ ( t ′ ) = (0 , R s (0) = ( − ,
0) and R s (1) = (0 , − Definition 5.3.
For any pair of paths f, g , let f ∗ g denote the concatenation of the paths f, g . Definition 5.4.
For any path f , let ¯ f denote the inverse of f , defined by ¯ f ( t ) = f (1 − t ). Definition 5.5.
For any s, s ′ ∈ (0 , L s,s ′ be defined by the equation L s,s ′ := R s ∗ ¯ R s ′ .The approach is as follows. Let c ′′ denote the constant path at R s (0) = ( − , L s,s ′ ≃ c ′′ in R − (0 , s, s ′ ∈ (0 ,
1) if R s doesn’t contain (0 ,
1) for any s ∈ (0 , L s ,s c ′′ if R s doesn’t contain (0 ,
0) for any s ∈ (0 , L s ,s ≃ c ′′ if R s doesn’t contain (0 ,
0) for any s ∈ (0 , R s doesn’t contain (0 ,
0) forany s ∈ (0 , Lemma 5.6.
Assume R s ( t ) does not contain (0 , for any s ∈ (0 , . Then, R s ≃ R s ′ in R − (0 ,
0) := X ′ for any s, s ′ ∈ (0 , .Proof. Let F ′ s,s ′ ( · , T ) := R s (1 − T )+ s ′ T . This is a homotopy taking R s (at T = 0) to R s ′ ( T = 1).Then, for each T ∈ [0 , F ′ s,s ′ ( · , T ) is simply R s ′′ for some s ′′ ∈ [ s, s ′ ]. Thus, if no curve R s ′′ intersects (0 , F s,s ′ is the desired homotopy taking R s to R s ′ in X ′ . (cid:4) Lemma 5.7.
Assume R s not contain (0 , for any s ∈ (0 , . Then L s,s ′ ≃ c ′′ in X ′ for any s, s ′ ∈ (0 , .Proof. From Lemma 5.6, R s ≃ R s ′ if R s does not contain (0 ,
0) for any s ∈ (0 , L s,s ′ = R s ∗ ¯ R s ′ ≃ R s ′ ∗ ¯ R s ′ ≃ c ′′ . (cid:4) Lemma 5.8.
There exists some s ∈ (0 , such that R s ⊂ R ≤ × R .Proof. Let s ∈ (0 ,
1) be such that k γ ( s ) − o k is maximal. Then, by definition, k γ ( t ) − o k ≤k γ ( s ) − o k for all t ∈ [0 , r ,s ( t ) = k γ ◦ φ s ( t ) − o kk γ ( s ) − o k ≤
1. Let a : R → R map eachpoint to its first component. Then, a ◦ R s ( t ) = r ,s ( t ) − ≤
0, and thus R s ⊂ R ≤ × R . (cid:4) Lemma 5.9.
Let J be strongly locally monotone at o . There exists some s ∈ (0 , suchthat R s ∩ ( R < × R < ) = ∅ .Proof. Let ε > γ | (1 − ε,ε ) := U ⊂ J is a strongly locally monotone neighbour-hood of o . Let d := inf( { o } ,J − U )3 , and let s ∈ (0 , ε ) be such that k γ ( s ) − o k = d . Note that J − U ⊂ γ | [0 ,s ] . Also, let B ′ ( o, r ) denote the open n -ball of radius r centred at o .Then, R s ( t ) ∈ R < × R < if and only if k γ ( t ) − o k < d and k γ ( t ) − γ ( s ) k < d for all t ∈ [0 , γ ◦ φ s ∩ B = ∅ , where B := B ′ ( o, d ) ∩ B ′ ( γ ( s ) , d ). Assuch, it shall suffice to show that γ ◦ φ s ∩ B = ∅ to complete the proof. We consider threeparts of γ ◦ φ s separately: γ ◦ φ s | [0 , εs ] , γ ◦ φ s | [ εs , − εs ] = J − U , and γ ◦ φ s | [ − εs , . oγ ◦ φ s (1) = γ ( s ) B γ | [ s , B ′ ( o, d ) B ′ ( γ ( s ) , d ) Figure 10.
No point of γ ◦ φ s | [0 , εs ] lies low enough to intersect B , and nopoint of γ ◦ φ s | [ − εs , is high enough to intersect B . • γ ◦ φ s | [0 , εs ] : Since (without a loss of generality) a γ ( s ) is monotonically increasing,the starting point o of γ ◦ φ s | [0 , εs ] is its lowest point. Thus, since a γ ( s ) ( o ) > a γ ( s ) ( b )for any b ∈ B , no point of γ ◦ φ s | [0 , εs ] lies close enough to the axis defined by oγ ( s )to intersect B . See Figure 10 for a diagram that illustrates this. • γ ◦ φ s | [ εs , − εs ] = J − U : Since d < inf( { o } , J − U ), it follows that ( J − U ) ∩ B ′ ( o, d ) = ∅ .Theninf( { γ ( s ) } , J − U ) ≥ k inf( { o } , J − U ) k − k γ ( s ) − o k = 3 d − d = 2 d. Since d < d ≤ inf( γ ( s ) , J − U ), it follows that ( J − U ) ∩ B ′ ( γ ( s ) , d ) = ∅ . Therefore, J − U intersects neither B ′ ( o, d ) nor B ′ ( γ ( s ) , d ), and in particular, ( J − U ) ∩ B = ∅ . • γ ◦ φ s | [ − εs , : Since (without a loss of generality) a γ ( s ) is monotonically increasing,the end point γ ◦ φ s (1) = γ ( s ) of γ ◦ φ s | [ − εs , is its “highest” point. That is tosay, a s ◦ γ ( s ) > a γ ( s ) ◦ γ ◦ φ s | [ − εs , ( t ) for any t . Thus, since a γ ( s ) ( o ) < a γ ( s ) ( b )for any b ∈ B and any t , a γ s ◦ γ ◦ φ s | [0 , εs ] ( t ) < a γ ( s ) ( b ). In other words, no pointof γ ◦ φ s | [0 , εs ] lies high enough (relative to the axis defined by oγ ( s )) to intersect B . Again, see Figure 10.Therefore, γ ◦ φ s ∩ B = ∅ , and thus R s ∩ ( R < × R < ) = ∅ . (cid:4) Lemma 5.10. If R s doesn’t contain (0 , for any s ∈ (0 , , then L s ,s c ′′ in X .Proof. Since neither R s nor R s contain the origin, L s ,s has a well-defined winding numberaround (0 , η ( L s ,s ) of L s ,s around (0 ,
0) is 1. As η ( c ′′ ) = 0, and as the winding number of a curve is a homotopy invariant, demonstratingthis shall be sufficient for this proof. RIANGLES INSCRIBED IN CURVES 13 ( − ,
0) (0 , − L s ,s | [0 , ] = R s (a) θ ( ) − θ (0) = − π only if L s ,s | [0 , ] passes through R > × R . ( − ,
0) (0 , − L s ,s | [ , = ¯ R s (b) θ (1) − θ ( ) = − π only if L s ,s | [ , passes through R < × R < . For any t ∈ [0 , θ ( t ) be equal to the angle of L s ,s ( t ) relative to the x -axis. Tocompute η ( L s ,s ), we first find θ ( ) − θ (0) and θ (1) − θ ( ). • θ ( ) − θ (0): Since L s ,s | [0 , ] = R s ⊂ R ≤ × R (from Lemma 5.8), k θ ( ) − θ (0) k 6≥ L s ,s to pass through all 4 quadrants.Then, as R s goes from ( − ,
0) to (0 , − θ ( ) − θ (0) can either be π or − π .However θ ( ) − θ (0) = − π , only if L s ,s | [0 , ] passes through R > × R , in contradictionto Lemma 5.8. Therefore, θ ( ) − θ (0) = − π , and thus θ ( ) − θ (0) = π . See Figure 11afor a diagram illustrating this. • θ (1) − θ ( ): Since L s ,s | [ , = ¯ R s R < × R < (from Lemma 5.9), k θ (1) − θ ( ) k 6≥ L s ,s | [0 , ] to pass through all 4 quadrants.Then, as L s ,s | [0 , ] goes from (0 , −
1) to ( − , θ (1) − θ ( ) can either be π or − π . However, θ (1) − θ ( ) = − π only if L s ,s | [ , travels through R < × R < , incontradiction to Lemma 5.9. Therefore, θ (1) − θ ( ) = − π , and thus θ (1) − θ ( ) = π .See Figure 11b for a diagram illustrating this.Hence, η ( L s ,s ) = θ (1) − θ (0)2 π = ( θ (1) − θ ( )) + ( θ ( ) − θ (0))2 π = π + π π = 1 . Then, as discussed above, it follows that L s ,s c ′′ in X ′ . (cid:4) We now conclude with the proof of Theorem 5.2.
Proof of Theorem 5.2. If R s doesn’t contains (0 ,
0) for any s ∈ (0 , L s ,s ≃ c ′′ in X ′ and that L s ,s c ′′ , which is a contra-diction. Therefore, our assumption for these lemmata is false, so there exists some s ′ and t ′ such that R s ′ ( t ′ ) = (0 , γ ( s ′ ) and γ ( t ′ )—that, by theconstruction of R s , inscribe an equilateral triangle such that one of its vertices is o . (cid:4) Acknowledgements
We would like to thank Rachana Madhukara for helpful comments.
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