Extreme problems for convex curves with given relative Chebyshev radius
Vitor Balestro, Horst Martini, Yurii Nikonorov, Yulia Nikonorova
aa r X i v : . [ m a t h . M G ] O c t EXTREMAL PROBLEMS FOR CONVEX CURVES WITHGIVEN RELATIVE CHEBYSHEV RADIUS
VITOR BALESTRO, HORST MARTINI, YURII NIKONOROV, YULIA NIKONOROVA
Abstract.
The paper is devoted to some extremal problems for convex curves andpolygons in the Euclidean plane referring to the relative Chebyshev radius. In partic-ular, we determine the relative Chebyshev radius for an arbitrary triangle. Moreover,we derive the maximal possible perimeter for convex curves and convex n -gons of agiven relative Chebyshev radius.2010 Mathematical Subject Classification: 52A10, 52A40, 53A04.Key words and phrases: approximation by polytopes, convex curve, convex polygon,relative Chebyshev radius. Introduction
Let (
X, d ) be a bounded metric space. Let us consider the metric invariant δ ( X ) = inf p ∈ X sup q ∈ X d ( p, q ) (1)arising in approximation theory, where it is called the relative Chebyshev radius of X with respect to X itself , see, e.g., [1, p. 119] and [2]. It has also the following obviousgeometric sense for compact X : δ ( X ) is the smallest radius of a ball having its centerin X and covering X . For brevity we will call δ ( X ) just the relative Chebyshev radiusof X .The study of extremal problems for convex curves in the Euclidean plane with a givenrelative Chebyshev radius started with the paper [4] by Rolf Walter. In particular, heconjectured that L (Γ) ≥ π · δ (Γ) for any closed convex curve Γ in the Euclideanplane, where L (Γ) is the length of Γ and d is the standard restricted Euclidean metric.In [4], this conjecture is proved for the case that Γ is a convex curve of class C andall curvature centers of Γ lie in the interior of Γ. It is also shown that the equality L (Γ) = π · δ (Γ) in this case holds if and only if γ is of constant width.It is also proved in [4] that all C -smooth convex curves have good approximationsby polygons in the sense of the relative Chebyshev radius (1). This observation leadsto natural extremal problems for convex polygons. In particular, the following resultgiven in [4] holds. Theorem 1 ([4]) . For each triangle P in the Euclidean plane, one has L ( P ) ≥ √ · δ ( P ) , with equality exactly for equilateral triangles. One of the goals of the present paper is to simplify the proof of this theorem. Forthis aim we apply the explicit expression for the value δ ( P ) obtained in the following. Theorem 2.
Let P be a triangle in the Euclidean plane with side lengths a ≥ b ≥ c and with angles α ≥ β ≥ γ . Then the following formula holds: δ ( P ) = a f or α ≥ π/ ,b sin( γ ) for γ ≥ π/ ,b γ ) for γ ≤ π/ and α ≤ π/ . This theorem is proved in Section 3 of the paper. In the same section, we givea new proof of Theorem 1. We also note that in Section 2 we fix the notation andconsider several auxiliary results. Section 4 is devoted to convex curves (polygons)of maximal perimeter among all curves (among all polygons with a given number ofvertices) with a fixed value of the relative Chebyshev radius (1). The correspondingresults are contained in Theorem 3 and Theorem 4. Finally, we briefly discuss somerelated unsolved problems.2.
Notation and some auxiliary results
We identify the Euclidean plane with R supplied with the standard Euclidean met-ric d , where d ( x, y ) = p ( x − y ) + ( x − y ) .We call Γ a convex curve if it is the boundary of some convex compact set in theEuclidean plane R . Important examples of convex curves are convex polygons ( convexclosed polygonal chains ). A polygon P is called and n -gon if it has exactly n vertices.For n = 2 we get line segments. The perimeter L ( P ) of any 2-gon is defined as thedouble length of the line segment P . Such definition is justified, since it leads to thecontinuity of the perimeter as a functional on the set of convex polygons with respectto the Hausdorff distance.Let Γ ⊂ R be a compact set (in particular, a convex curve). We define the function µ : Γ → R as follows: µ ( x ) = max y ∈ Γ d ( x, y ) . (2)If a point x ∈ Γ is such that µ ( x ) = min y ∈ Γ µ ( y ) = δ (Γ) (see (1)), then we call it extremal (for δ (Γ)), whereas any point x ∈ Γ with d ( x, x ) = µ ( x ) = max y ∈ Γ d ( x, y ) iscalled a footpoint (for x ) and the corresponding chord [ x, x ] is called distinguished .For any polygon P and any given point x ∈ P , max y ∈ P d ( x, y ) is achieved at a vertexof P (see, e.g., Lemma 4.1 in [4]), i.e., µ ( x ) is equal to the maximal distance from x to vertices of P . Note also that the diameter D := max x,y ∈ P d ( x, y ) of a polygon P alwaysconnects two vertices.The following property (monotonicity of the perimeter) of convex curves is wellknown (see, e.g., [3, § Proposition 1.
If a convex curve Γ is inside another convex curve Γ in the Euclideanplane, then the perimeter of Γ is less or equal to the perimeter of Γ , and equality holdsif and only if Γ = Γ . The following simple result is very useful.
XTREMAL PROBLEMS FOR CONVEX CURVES WITH . . . 3
Proposition 2.
Let Γ be a convex curve in R such that Γ contains the line segment [ p, q ] with the property Γ ⊂ (cid:8) x ∈ R | d ( x, o ) ≤ d ( p, q ) (cid:9) , where o is the midpoint of [ p, q ] . Then δ (Γ) = d ( p, q ) = µ ( o ) . Moreover, o is a unique extremal point for δ (Γ) . Proof.
Since µ ( o ) = max x ∈ Γ d ( o, x ) ≤ d ( p, q ) and d ( o, p ) = d ( o, q ) = d ( p, q ), then µ ( o ) = d ( p, q ). On the other hand, for any x ∈ Γ we have d ( x, p ) + d ( x, q ) ≥ d ( p, q ),and therefore, µ ( x ) = max y ∈ Γ d ( x, y ) ≥ max { d ( x, p ) , d ( x, q ) } ≥ d ( p, q ). Hence, o is anextremal point for δ (Γ) and δ (Γ) = µ ( o ) = d ( p, q ). It is also clear that o is a uniqueextremal point, since max { d ( x, p ) , d ( x, q ) } > d ( p, q ) for any x = o .3. The relative Chebyshev radius for triangles
In this section we deal with triangles in Euclidean plane R . Lemma 1.
Let P be a triangle KLM such that ∠ KLM ≥ π/ and ∠ LKM ≥ π/ .Then the minimal value of µ ( x ) = max y ∈ P d ( x, y ) for x lying in the line segment [ K, L ] isreached exactly at the point N ∈ [ K, L ] , such that the line M N is orthogonal to the line KL , and d ( M, N ) is equal to the length of the altitude of the triangle KLM throughthe vertex M . Proof.
It is clear that d ( M, N ) ≥ d ( N, K ) and d ( M, N ) ≥ d ( N, L ) (note that ∠ N LM ≥ π/ − ∠ N LM = ∠ LM N and ∠ N KM ≥ π/ − ∠ N KM = ∠ KM N ), seeFig. 1 a). Therefore, µ ( N ) = max { d ( N, K ) , d ( N, L ) , d ( N, M ) } = d ( N, M ) . If x ∈ [ K, L ] and x = N , then d ( M, x ) > d ( M, N ) and, therefore, µ ( x ) = max { d ( x, K ) , d ( x, L ) , d ( x, M ) } > d ( M, N ) , as required.Let ABC be a triangle with vertices
A, B, C . We put a = d ( B, C ), b = d ( A, C ), c = d ( A, B ), where a ≥ b ≥ c , and α = ∠ CAB , β = ∠ ABC , γ = ∠ BCA . It is clearthat α ≥ β ≥ γ .We are going to calculate the value of δ ( P ). Lemma 2.
For a triangle P with α ≥ π/ , we have δ ( P ) = a/ with a unique extremalpoint o , the midpoint of [ B, C ] . Proof.
Since α ≥ π/
2, we have P ⊂ { x ∈ R | d ( x, o ) ≤ a/ } , where o is the mid-point of [ B, C ]. Now it suffices to apply Proposition 2.
Lemma 3.
For a triangle P with γ ≥ π/ , the value δ ( P ) is equal to b sin( γ ) = c sin( β ) ,the length of the altitude of P through the vertex A . Proof.
It is clear that min x ∈ P µ ( x ) = min { M , M , M } , where M := min x ∈ [ A,B ] µ ( x ), M := min x ∈ [ A,C ] µ ( x ), and M := min x ∈ [ B,C ] µ ( x ) (see (2)). By Lemma 1, M , M , M areequal to the length of the altitude of the triangle P through the vertices C , B , A respectively. Now it is clear (recall that a ≥ b ≥ c ) that δ ( P ) = min x ∈ P µ ( x ) = M , thelength of the altitude of P through the vertex A . VITOR BALESTRO, HORST MARTINI, YURII NIKONOROV, YULIA NIKONOROVA a) b)
Figure 1.
The pictures for: a) Lemma 1; b) Lemma 4.
Remark 1. If γ ≥ π/
4, then α ≥ β ≥ γ ≥ π/ α ≤ π − β − γ ≤ π/
2. Moreover, α = π/ β = γ = π/ Lemma 4.
For a triangle P with γ ≤ π/ and α ≤ π/ , we have δ ( P ) = b γ ) . Proof.
Let us consider the points E ∈ [ B, C ] and F = [ A, C ] such that ∠ EAC = ∠ F BC = γ , see Fig. 1 b). Note that d ( E, C ) = d ( E, A ) ≥ d ( E, B ) (the midpoint of[
B, C ] is on the line segment [
E, C ] due to the inequality ∠ CAB = α ≤ π/ µ ( E ) = d ( E, C ) = d ( E, A ) = b γ ) (see (2)). Since ∠ CEA = π − γ ≥ π/
2, then wehave d ( x, A ) ≥ d ( E, A ) = µ ( E ) for any point x ∈ [ E, C ]. Moreover, we have d ( x, C ) ≥ d ( E, C ) = µ ( E ) for any x ∈ [ E, B ]. Therefore, N := min x ∈ [ B,C ] µ ( x ) = µ ( E ) = b γ ) .Analogously, d ( F, C ) = d ( F, B ) ≥ d ( E, A ) (the midpoint of [
A, C ] is on the linesegment [
E, C ] due to the inequality ∠ CBA = β ≤ α ≤ π/ µ ( F ) = d ( F, C ) = d ( F, B ) = a γ ) . Since ∠ CF B = π − γ ≥ π/
2, then for any point x ∈ [ F, C ] we have d ( x, B ) ≥ d ( F, B ) = µ ( F ). Moreover, for any x ∈ [ F, A ] we have d ( x, C ) ≥ d ( F, C ) = µ ( F ). Therefore, N := min x ∈ [ A,C ] µ ( x ) = µ ( F ) = a γ ) .Now, we are going to find N := min x ∈ [ A,B ] µ ( x ). Since γ ≤ π/ α ≤ π/
2, we have α ≥ β = π − α − γ ≥ π/
4. Hence, we can apply Lemma 1 for the side [
A, B ] of thetriangle P = ABC . Therefore, N is equal to a sin( β ) = b sin( α ), the length of thealtitude of the triangle P through the vertex C .Note that δ ( P ) = min x ∈ P µ ( x ) = min { N , N , N } . Since a ≥ b , we have N = a γ ) ≥ b γ ) = N , and since γ ≤ π/
4, we get cos( γ ) ≥ cos( π/
4) = 1 / √
2. Since α ≥ π/
3, wehave sin( α ) ≥ sin( π/
3) = √ /
2. Further on, 2 sin( α ) cos( γ ) ≥ · √ · √ = p / > N = b sin( α ) > b γ ) = N . Therefore, we get δ ( P ) = min x ∈ P µ ( x ) = N = b γ ) . Remark 2. If γ ≤ π/ α = π/
2, then β = π/ − γ ≥ π/ δ ( P ) = b γ ) = b β ) = a α ) = a/ the proof of Theorem 2 . Now, weare going to use Theorem 2 in order to get a more simple proof of Theorem 1. XTREMAL PROBLEMS FOR CONVEX CURVES WITH . . . 5
Proof of Theorem 1.
We are going to find all triangles P with maximal value L ( P ) /δ ( P ), where L ( P ) = a + b + c and δ ( P ) is used as shown in Theorem 2. We have L ( P ) = 3 a and δ ( P ) = √ · a for any equilateral triangle P , hence, L ( P ) /δ ( P ) = 2 √ L ( P ) > √ · δ ( P ) for any triangle P that is not equilateral.We consider three cases according to Theorem 2. Case 1.
We suppose that α ≥ π/ δ ( P ) = a/
2. Since b + c > a , weget L ( P ) = a + b + c > a = 4 · δ ( P ) > √ · δ ( P ). Case 2.
We suppose that γ ≥ π/ δ ( P ) = b sin( γ ). Let D bea point on the line segment [ B, C ], such that [
A, D ] is the altitude of P through A . d ( A, D ) = b sin( γ ) = c sin( β ), d ( B, D ) = c cos( β ), d ( C, D ) = b cos( γ ). Hence, L ( P ) δ ( P ) = a + b + cb sin( γ ) = c cos( β ) + b cos( γ ) + b + cb sin( γ ) = b cos( γ ) + bb sin( γ ) + c cos( β ) + cb sin( γ )= b cos( γ ) + bb sin( γ ) + c cos( β ) + cc sin( β ) = 1 + cos( β )sin( β ) + 1 + cos( γ )sin( γ ) = f ( β ) + f ( γ ) , where f ( x ) = x )sin( x ) . By direct computations we get f ′ ( x ) = − x )sin ( x ) = x )cos ( x ) − = x ) − < f ′′ ( x ) = sin( x )(cos( x ) − > x ∈ (0 , π ).Let us find the minimal value of the function F ( β, γ ) := f ( β )+ f ( γ ) for β ≥ γ ≥ π/ β + γ ≤ π . Since f ′ ( x ) <
0, the minimal value can be achieved only when β + γ = π and γ ∈ [ π/ , π/ h ( x ) := f ( x ) + f (cid:0) π − x (cid:1) . It is convex due to theinequality h ′′ ( x ) = f ′′ ( x ) + f ′′ (cid:0) π − x (cid:1) >
0. Hence, h ′ ( x ) = f ′ ( x ) − f ′ (cid:0) π − x (cid:1) isstrictly increasing and h ′ ( x ) < h ′ ( π/
3) = 0 for x ∈ (0 , π/ h achieves its minimal value on [ π/ , π/
3] exactly at thepoint x = π/
4. This minimal value is h ( π/
3) = 2 f ( π/
3) = 2 √ L ( P ) δ ( P ) = F ( β, γ ) > √ β and γ such that β ≥ γ ≥ π/ β + γ ≤ π ,and ( β, γ ) = ( π/ , π/ β = γ = π/ Case 3.
We suppose that γ ≤ π/ α ≤ π/
2, therefore δ ( P ) = b γ ) . We have β ∈ [ π/ − γ, π/ − γ/
2] due to β + γ = π − α ≥ π/ γ + 2 β ≤ γ + β + α = π ,whereas γ ∈ (0 , π/ L ( P ) = a + b + c = b sin( β ) (cid:16) sin( α ) + sin( β ) + sin( γ ) (cid:17) and L ( P ) δ ( P ) = 2 cos( γ )sin( β ) (cid:16) sin( α ) + sin( β ) + sin( γ ) (cid:17) = 2 cos( γ )sin( β ) (cid:16) sin( β + γ ) + sin( β ) + sin( γ ) (cid:17) = 2 cos( γ )sin( β ) (cid:16) sin( β ) (cid:0) γ ) (cid:1) + sin( γ ) (cid:0) β ) (cid:1)(cid:17) = 2 cos( γ ) (cid:0) γ ) (cid:1) + sin(2 γ ) · β )sin( β ) =: G ( β, γ ) . It is easy to see that ∂G∂β ( β, γ ) = − sin(2 γ ) · β )sin ( β ) = sin(2 γ )cos( β ) − < π ≥ β ≥ γ > G ( β, γ ) for β ∈ [ π/ − γ, π/ − γ/
2] and for a given
VITOR BALESTRO, HORST MARTINI, YURII NIKONOROV, YULIA NIKONOROVA γ ∈ [0 , π/
4] can be achieved only when β = π/ − γ/
2. By direct computations we get G ( π/ − γ/ , γ ) = 2 cos( γ ) (cid:0) γ ) (cid:1) + sin(2 γ ) · γ/ γ/ γ ) + 2 cos ( γ ) + 4 (cid:0) γ/ (cid:1) cos( γ ) sin( γ/ γ ) (cid:0) γ/ (cid:1) = g (cid:0) sin( γ/ (cid:1) , where g ( t ) = 4(1 − t )(1 + t ) = 4(1 + t − t − t ). Since 0 < γ/ ≤ π/
8, we get0 < t = sin( γ/ < sin( π/
8) = q (cid:0) − cos( π/ (cid:1) = √ √ =: t .Since g ′′ ( t ) = − t + 1), g is concave on the interval [0 , ∞ ). The minimal valueof g ( t ) on the interval [0 , t ] is achieved either at t = 0, or at t = t . Since g (0) = 4and h ( t ) = 2 √ p √ > √
3, we get that L ( P ) δ ( P ) = G ( β, γ ) > √ β ∈ [ π/ − γ, π/ − γ/
2] and γ ∈ (0 , π/ Convex curves and polygons of maximal perimeter
A half-disk in the Euclidean plane R is a set which is isometric to HD ( r ) = (cid:8) x = ( x , x ) ∈ R | x ≥ , x + x ≤ r (cid:9) for some fixed r >
0. The boundary of HD ( r ) is the union of a half-circle of radius r and a line segment of length 2 r . Lemma 5.
Let Γ be the boundary of some half-disk of radius r in R . Then δ (Γ) = r and L (Γ) = (2 + π ) · δ (Γ) , where L (Γ) means the perimeter of Γ . Proof.
The equality δ (Γ) = r follows from Proposition 2. The second statement isobvious. Theorem 3.
For any closed convex curve Γ in the Euclidean plane, one has L (Γ) ≤ (2 + π ) · δ (Γ) , with equality exactly for boundaries of half-disks. Proof.
Let Γ be a convex curve and let o ∈ Γ be an extremal point for δ (Γ). Thismeans that Γ ⊂ { x ∈ R | d ( x, o ) ≤ r } =: D , where r = δ (Γ). Moreover, Γ is a convexcurve, and therefore there is a straight line l through the point o in R such that Γis situated in one of the half-planes determined by l . Let us denote this half-plane by H ( l ). Therefore, Γ is a subset of the half-disk HD := D ∩ H ( l ) of radius r .By the monotonicity of the perimeter of convex curves (see Proposition 1) we getthat the perimeter L (Γ) of Γ is less than or equal to (2 + π ) · r , the perimeter ofthe boundary of the semi-disk HD . Moreover, equality holds if and only if Γ is theboundary of HD . The theorem is proved.Now we are going to find all convex n -gons P ( n ≥
2) of maximal perimeter amongall convex n -gons with the same value of r = δ ( P ). At first, we consider an explicitconstruction of a special family of convex n -gons U n for n ≥ n − P n , inscribed in a circle of radius r > A, B ∈ P n that are opposite vertices of this polygon (i.e., d ( A, B ) = 2 r ).Now, consider one of the two half-planes determined by the straight line AB , say H , XTREMAL PROBLEMS FOR CONVEX CURVES WITH . . . 7 a) b) c)
Figure 2.
The polygons U n for: a) n = 3; b) n = 4; c) n = 8.and consider the union U n of the line segment [ A, B ] with the polygonal line P n ∩ H .This is an n -gon inscribed in the boundary of the half-disk { x ∈ R | d ( x, o ) ≤ r } ∩ H ,where o is the midpoint of the line segment [ A, B ], see Fig. 2. We have δ ( U n ) = r byProposition 2. For n = 2 we see that P = U is a line segment of length 2 r .The perimeter L ( U n ) of U n is equal to λ n · r , where λ n = 2 (cid:16) n −
1) sin (cid:0) π n − (cid:1)(cid:17) for n ≥
2. Note that λ = 4, since L ( U ) is the double length of the line segment U .It is also easy to see that λ = 2(1 + √ λ = 5, and λ n → π as n → ∞ . Lemma 6.
The above defined sequence { λ n } , n ≥ , is strictly increasing. Proof.
Let us consider the function ψ ( x ) = sin( x ) x for x >
0. We have ψ ′ ( x ) = θ ( x ) x ,where θ ( x ) = x cos( x ) − sin( x ). Since θ ′ ( x ) = − x sin( x ) < x ∈ (0 , π ), then θ ( x ) < θ (0) = 0 for all x ∈ (0 , π ). This means that ψ ′ ( x ) < ψ ( x ) strictlydecreases on the interval (0 , π ). In particular, ψ (cid:16) π n − (cid:17) < ψ (cid:0) π n (cid:1) for all n ≥
2, whichis equivalent to λ n < λ n +1 . Theorem 4.
For any convex n -gon P ⊂ R , n ≥ , one has L (Γ) ≤ (cid:18) n −
1) sin (cid:18) π n − (cid:19)(cid:19) · δ (Γ) , with equality exactly for the n -gon U n defined above. Proof.
For r > n ≥
2, we denote by K n ( r ) the set of all convex polygons P inthe Euclidean plane with δ ( P ) = r and having no more than n vertices. In particular,polygons with 2 vertices are line segments, and we repeat that the perimeter of sucha 2-gon is assumed to be equal to its double length (and the perimeter is continuousfunctional on K n ( r ) for every n ≥ M n ( r ) be the supremum of all perimetersof polygons from the set K n ( r ). Let us prove that there is a polygon in K n ( r ) whoseperimeter equals M n ( r ).Take a sequence { P k } , k ∈ N , P k ∈ K n ( r ), such that L ( P k ) → M n ( r ) as k → ∞ .Without loss of generality, we may suppose that all P k are situated in some ball of radius r (we can move each P k in such a way that an extremal point for δ ( P k ) coincides withthe center of a given ball). Using compactness arguments, we may assume that P k → P (in the Hausdorff distance) for some P ∈ K n ( r ) as k → ∞ , hence L ( P ) = M n ( r ).Now we are going to prove that P is isometric to the n -gon U n as above. For thisgoal we will use induction on n . For n = 2, P is a line segment of length 2 r , that is U . Suppose that the assertion is true for all k < n and prove it for n .First note that P is an n -gon. Indeed, if P is an m -gon for some m < n , then λ m = M m ( r ) ≥ L ( P ) by the induction assumptions. On the other hand, we know that VITOR BALESTRO, HORST MARTINI, YURII NIKONOROV, YULIA NIKONOROVA M n ( r ) ≥ L ( U n ) = λ n , but λ n > λ m by Lemma 6. Therefore, we get M n ( r ) > L ( P ),and this contradiction proves that P is an n -gon.Now we are going to show that P is isometric to U n . Let o ∈ P be an extremal pointfor δ ( P ). This means that P ⊂ { x ∈ R | d ( x, o ) ≤ r } =: D . Moreover, P is a convex n -gon, and therefore we have a straight line l through the point o in R such that P is situated in one of the half-planes determined by l . Let us denote this half-plane by H ( l ). Therefore, P is a subset of the half-disk HD := D ∩ H ( l ) of radius r .Note that all vertices of P are situated in the boundary ∂ ( HD ) of the half-disk HD .Indeed, suppose the contrary. Take a point S inside of P and consider the rays SA i , i = 1 , ..., n , where A i are the vertices of P . If some point A i is not in ∂ ( HD ) then wecan modify P into a polygon P ′ , replacing the vertex A i by some point A ′ i lying on theray SA i such that d ( S, A ′ i ) > d ( S, A i ) and A ′ i is taken so close to A i that A ′ i ∈ HD and P ′ is convex. By the monotonicity of the perimeter of convex curves (Proposition 1),we get that L ( P ′ ) > L ( P ) = M n ( r ), which is not true. This contradiction shows thatall vertices of P are situated in the boundary ∂ ( HD ) of HD .Now let us consider vertices of P that are in the line l . Since o ∈ l ∩ P , there is atleast one such vertex. On the other hand, it is impossible to have three or more suchvertices (in this case P is an m -gon with some m < n ). Hence we have two possibilities:1) only one vertex of P (that should coincide with the point o ) is in l , 2) exactly twovertices of P are in l . Fortunately, the first possibilities could be easily reduced to thesecond one. Indeed, one can rotate the polygon P around the point o so that it stillremains in the half-disk HD , but its two vertices will already be on the line l . Hence,without loss of generality, we may assume that exactly two vertices (say A and A )of P are in l . Let us prove that d ( A , A ) = 2 r , i.e., d ( o, A ) = d ( o, A ) = r . Supposethe contrary and modify the polygon P into a polygon P ′ , replacing the points A and A with two points A ′ , A ′ ∈ l such that d ( A ′ , o ) = d ( A ′ , o ) = r . It is clear that P isinside of P ′ , hence (by Proposition 1) we have L ( P ′ ) > L ( P ) = M n ( r ), which is nottrue. Hence, d ( A , A ) = 2 r .The last step is to prove that the rays emanating from the point o successively to allthe vertices of the polygon P divide the semicircle { x ∈ R | d ( x, o ) = r } ∩ H ( l ) intoequal arcs. Indeed, this will imply that P is isometric to U n .Now it suffices to prove the equality d ( A i − , A i ) = d ( A i +1 , A i ) for three successivevertices A i − , A i , A i +1 of P . Let us take a point B on the arc between A i − and A i +1 and consider the polygon P ′ which is obtained from P by replacing the point A i with B . By construction of P we have L ( P ) ≥ L ( P ′ ), equivalent to the inequality d ( A i − , A i ) + d ( A i +1 , A i ) ≥ d ( A i − , B ) + d ( A i +1 , B ). Hence A i is the point that givesthe maximal value of the function B d ( A i − , B ) + d ( A i +1 , B ).Put ϕ := ∠ A i − oA i +1 and ψ := ∠ A i − oB . Then d ( A i − , B ) + d ( A i +1 , B ) = 2 r (cid:16) cos( ψ ) + cos( ϕ − ψ ) (cid:17) . Note that ϕ ∈ (0 , π/
2] and ψ ∈ [0 , ϕ ]. The minimal value of the function h ( ψ ) =cos( ψ ) + cos( ϕ − ψ ) on the interval [0 , ϕ ] is attained at the point ψ = ϕ/
2, since h ′ ( ψ ) = sin( ϕ − ψ ) − sin( ψ ) > ( < ) 0 for ψ ∈ [0 , ϕ/
2) (for ψ ∈ ( ϕ/ , ϕ ]). This meansthat d ( A i − , A i ) = d ( A i +1 , A i ). The theorem is completely proved. XTREMAL PROBLEMS FOR CONVEX CURVES WITH . . . 9 Final remarks
Note that there is no result for n -gons, n ≥
4, similar to Theorem 1 for triangles.We do not even know quadrangles that have the smallest perimeter among all convexquadrangles with a given value of (1). It is easy to check that squares have no suchproperty. In [4], it is conjectured that L ( P ) ≥ p √ · δ ( P ) for any convexquadrangle P ⊂ R . Note that this inequality becomes an equality for quadrangles P called “magic kites” (This notion is taken from [4] and means convex quadrangles whichare hypothetically extreme with respect to the Chebyshev radius.) Up to similarity,such a quadrangle could be represented by its vertices, that are as follows:( − , , (1 , , , √ q √ ! , (cid:18) , − q √ − (cid:19) . Note also that we have L ( P ) = √ · δ ( P ) for a square P and √ > p √ ≈ , Problem 1.
Given real number l > n ≥
3, determine thebest possible constant C ( n, l ) such that the inequality L ( P ) ≥ C ( n, l ) holds for everyconvex polygon P with n vertices and with the following property: for the endpoints A and B of every side of P , there is a point C ∈ P such that d ( A, C ) = d ( B, C ) ≥ l .It is easy to show that C (3 , l ) = 3 l . On the other hand, the answer for n ≥ L ( P ), the area S ( P ) of the polygon P . References [1] D. Amir,
Characterizations of Inner Product Spaces,
Operator Theory: Advances and Ap-plications Series Profile, Vol. 20. Basel–Boston–Stuttgart: Birkh¨auser– Verlag. Basel, 1986, MR Zbl.
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Theory of Convex Bodies , BCS Associates, Moscow, ID, 1987. Trans-lated from the German and edited by L. Boron, C. Christenson and B. Smith., MR Zbl.
On a minimax problem for ovals,
Minimax Theory Appl. 2(2) (2017), 285–318, MR Zbl.
Vitor BalestroInstituto de Matem´atica e Estat´ısticaUniversidade Federal Fluminense24210201 Niter´oiBrazil
Email address : [email protected] Horst MartiniFakult¨at f¨ur MathematikTechnische Universit¨at Chemnitz09107 ChemnitzGermany
Email address : [email protected] Yurii NikonorovSouthern Mathematical Institute ofthe Vladikavkaz Scientific Center ofthe Russian Academy of Sciences,Vladikavkaz, Markus st., 22,362027, Russia
Email address : [email protected] Yulia NikonorovaVolgodonsk Engineering Technical Institute the branchof National Research Nuclear University “MEPhI”,Rostov region, Volgodonsk, Lenin st., 73/94,347360, Russia
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