Is a convex plane body determined by an isoptic?

Is a convex plane body determined by an isoptic?

Arp´´ ad Kurusa^∗

Abstract. We prove that two convex bodies coincide if they have a common isoptic with the same angle ν ∈ (0, π) so that 1−ν/π is an irrational or rational number with even numerator in its lowest terms. In the remaining cases we prove that two convex polygons with a common isoptic coincide, and a common isoptic of different rotationally symmetric bodies is a circle.

1. Introduction

The set of points where a compact convex domainBwith nonempty interior (such domains are called convex bodies, the strictly convex ones are calledovalsor ovoid bodies, and the ones bounded by polygons are called convexpolygonal bodies or simply convex polygonsin this paper) subtends a constant angle ν ∈(0, π) is called the ν-isopticof the convex bodyB. The (π/2)-isoptic is calledorthoptic.

A short list of isoptics of some important, but not necessarily convex plane curves can be found in [10]; further results on isoptics and the strongly related inner isoptics can be found in [7], [5].

This article considers the following question:

(1.1) Is a convex plane body determined by a ν-isoptic of it?

In 1950 Green [1] proved that if theν-isoptic of a convex body Dis a circle, thenDmust be a disc provided 1−ν/πis an irrational or rational number with even numerator in its lowest terms. If the numerator is odd, then there are continuum many bodies not even similar to each other with that circle as theirν-isoptic.

In 1971 W. Wunderlich [9] showed the existence of continuum many bodies in all ellipsesE, which have numerical excentricity near 1, so that theν-isoptic of all AMS Subject Classification(2000): 0052, 0054.

Key words and phrases: visual angle, isoptics, shape recognition, shadow picture, distin- guishability of plane convex bodies.

∗ Supported by the TAMOP-4.2.1/B-09/1/KONV-2010-0005 project.

these bodies are E provided 1−ν/π is rational with odd numerator in its lowest terms.

We provide uniqueness results to question (1.1) for ellipses, for special angles, for convex polygons and for rotationally symmetric convex bodies.

2. Preliminaries

Let the boundary and theν-isoptic of a convex bodyBbe denoted by∂Band byB^ν, respectively. A convex bodyGis said to be aν-ghostof the convex bodyB ifG^ν ≡ B^ν.

Let min_Bαdenote the minimum of the angles at the singular points of∂B. If

∂B does not have singular points, i.e.,B has differentiable border, then we define min_Bα:=π. Ifν ≥min_Bα, then theν-isoptic ofB is a closed curve, but it is not connected if ν <min_Bα.

If min_Bα=π, thenB^π =∂B, hence theν-isoptic is convex ifν is sufficiently close to π. In the other hand theν-isoptics of a “sufficiently depressed” ovals are concave for sufficiently small anglesν [6].

Nevertheless,all isoptics are star-shaped with respect to any point in the con- vex body B, because on every straight line that passes throughB the visual angle ofBis strictly monotone decreasing as the point is moving away from the chord in which the straight line intersectsB. This observation also proves that B^ν is in the interior of the closed domain bounded byB^µ if and only if ν > µ.

The supplementary angle ofν ∈(0, π) is denoted by ¯ν =π−ν, andu(ϕ) is the unit vector (cosϕ,sinϕ). Turning the vectoru(ϕ) anticlockwise byπ/2 gives the unit vector u^⊥(ϕ) :=u(ϕ+π/2) = (−sinϕ,cosϕ).

Letp:S¹→R+ be the support function of a convex bodyBwhose boundary does not contain a segment. Then pis differentiable and the curve

(2.1) r(ϕ) =p(ϕ)u(ϕ) + ˙p(ϕ)u^⊥(ϕ)

is obviously closed and has tangentxcosϕ+ysinϕ=p(ϕ), whence it parametrizes the boundary ∂B ofB.

Lemma 2.1. The ν-isoptic of a convex body B is continuous at its points. It is differentiable at those of its points that are not collinear with any segment in the boundary ∂B.

Proof. Letpbe the support function of the convex bodyB. Any point r^ν_B of the ν-isopticB^ν is the intersection of the tangents

p(ϕ) =xcosϕ+ysinϕ,

p(ϕ+ ¯ν) =xcos(ϕ+ ¯ν) +ysin(ϕ+ ¯ν).

Solving the system of these equations leads to x= p(ϕ) sin(ϕ+ ¯ν)−p(ϕ+ ¯ν) sinϕ

cosϕsin(ϕ+ ¯ν)−sinϕcos(ϕ+ ¯ν) = p(ϕ) sin(ϕ+ ¯ν)−p(ϕ+ ¯ν) sinϕ

sin ¯ν ,

y= p(ϕ) cos(ϕ+ ¯ν)−p(ϕ+ ¯ν) cosϕ

sinϕcos(ϕ+ ¯ν)−cosϕsin(ϕ+ ¯ν) =p(ϕ+ ¯ν) cosϕ−p(ϕ) cos(ϕ+ ¯ν)

sin ¯ν ,

and hence

(2.2) r^ν_B(ϕ) = 1

sin ¯ν(p(ϕ+ ¯ν)u^⊥(ϕ)−p(ϕ)u^⊥(ϕ+ ¯ν)).

This proves the theorem.

LetBandGbe convex bodies with support functionsp_Bandp_G, respectively.

If theirν-isoptics coincide, that isB^ν ≡ G^ν, then there is a functionξ:S¹→S¹ so that ϕ7→ ϕ+ξ(ϕ) is bijective onS¹ →S¹, and r^ν_G(ϕ) =r^ν_B(ϕ+ξ(ϕ)), that, by (2.2), means

(2.3) pG(ϕ+ ¯ν)u(ϕ)−pG(ϕ)u(ϕ+ ¯ν)

=p_B(ϕ+ξ(ϕ) + ¯ν)u(ϕ+ξ(ϕ))−p_B(ϕ+ξ(ϕ))u(ϕ+ξ(ϕ) + ¯ν).

This equation translates to the system of equations p_G(ϕ+ ¯ν) cos(ϕ)−p_G(ϕ) cos(ϕ+ ¯ν)

=p_B(ϕ+ξ(ϕ) + ¯ν) cos(ϕ+ξ(ϕ))−p_B(ϕ+ξ(ϕ)) cos(ϕ+ξ(ϕ) + ¯ν), p_G(ϕ+ ¯ν) sin(ϕ)−p_G(ϕ) sin(ϕ+ ¯ν)

=p_B(ϕ+ξ(ϕ) + ¯ν) sin(ϕ+ξ(ϕ))−p_B(ϕ+ξ(ϕ)) sin(ϕ+ξ(ϕ) + ¯ν)., which has the only solution

p_G(ϕ) sin(¯ν) =−p_B(ϕ+ξ(ϕ) + ¯ν) sin(ξ(ϕ)) +p_B(ϕ+ξ(ϕ)) sin(ξ(ϕ) + ¯ν), p_G(ϕ+ ¯ν) sin(¯ν) =p_B(ϕ+ξ(ϕ) + ¯ν) sin(¯ν−ξ(ϕ)) +p_B(ϕ+ξ(ϕ)) sin(ξ(ϕ)).

These equations are consistent if and only if (2.4)

p_B(ϕ+ξ(ϕ)) sin(ξ(ϕ))−p_B(ϕ+ξ(ϕ) + ¯ν) sin(ξ(ϕ)−ν¯)

=p_B(ϕ+ ¯ν+ξ(ϕ+ ¯ν)) sin(ξ(ϕ+ ¯ν) + ¯ν)−

−pB(ϕ+ ¯ν+ξ(ϕ+ ¯ν) + ¯ν) sin(ξ(ϕ+ ¯ν)).

With these we have just proved the following result.

Theorem 2.2. A convex body B has a ν-ghost if and only if the functional con- sistency equation (2.4) has a solution ξ such that ϕ 7→ ϕ+ξ(ϕ) is bijective on S¹→S¹, and

(2.5) p(ϕ) := 1

sin(¯ν)(p_B(ϕ+ξ(ϕ)) sin(ξ(ϕ) + ¯ν)−p_B(ϕ+ξ(ϕ) + ¯ν) sin(ξ(ϕ))) is the support function of a convex body.

If the support function of an oval is twice differentiable then the parametriza- tion r^ν_B:S¹ → R² of B^ν as given by (2.2) is also twice differentiable. This allows one to simplify the search for an ovoid ghost of an oval. In fact, exactly this is done in Green’s paper [1] for constant p_B^(2:1), but elementary geometry also offers interesting (π/2)-ghosts (orthoghosts): ellipses are well known to subtend the angle π/2 at the points of their great circle.

However our next result shows that this is the only case of ghosts among ellipses.

Theorem 2.3. If two ellipses have a common isoptic (it may belong to different angles), then either they coincide or that isoptic is a circle.

Proof. As it can be found in [10] (see [4] for a proof), theν-isoptic of the ellipse E_a,b: x²

a² +y² b² = 1 is

(2.6) E_a,b^ν : (x²+y²−a²−b²)²= 4(a²y²+b²x²−b²a²) cot²ν.

(2:1)

A straightforward computation with trigonometric functions shows that ξ(ϕ) = εsin(nϕ) fulfils the conditions of Theorem 2.2 for any smallε >0.

Since thisν-isoptic is symmetric to thex- and y-axis, it may have a further axis of symmetry only through the origin. Suppose that for some α∈(0, π/2) the line y=xtan(α/2) is an axis of symmetry ofE_a,b^ν . Then

((xcosα−ysinα)²+ (xsinα+ycosα)²−a²−b²)²

= 4(a²(xsinα+ycosα)²+b²(xcosα−ysinα)²−b²a²) cot²ν is also an equation ofE_a,b^ν , which is equivalent to

(x²+y²−a²−b²)²

= 4(a²y²+b²x²+ (b²−a²)(y²sin²α−x²sin²α−2xycosαsinα)−b²a²) cot²ν.

By comparing this with (2.6), we obtain that

(a) either cotν = 0, i.e.,E_a,b^ν is the circlex²+y²=a²+b² andν =π/2, (b) ora=b, i.e.,Ea,b andE_a,b^ν are the circles with radiusa=banda√

2 =b√ 2, respectively.

This implies that if two ellipses have a common isoptic, then either that isoptic is a circle or the ellipses have the same axes of symmetries.

If the ellipses have the same two axes of symmetries, then the equations E_a,b^ν : (x²+y²−a²−b²)²= 4(a²y²+b²x²−b²a²) cot²ν , E_c,d^µ : (x²+y²−c²−d²)²= 4(c²y²+d²x²−c²d²) cot²µ

of their common isoptic E_a,b^ν ≡ E_c,d^µ are equivalent. Therefore, ifx= 0, then the equations

(y²−b²−a²)²= 4a²(y²−b²) cot²ν and (y²−d²−c²)²= 4c²(y²−d²) cot²µ must have the same solutions fory², which gives

(2.7) b²+a²

1 + 2 cot²ν± q

(1 + 2 cot²ν)²−1

=y²=d²+c²1±cosµ sinµ

2

. In the same way, if y = 0, then the common solutions for x² must be the same, and because of the form of the equations, these solutions have the same form as (2.7), but with aexchanged tob, andcexchanged tod. Therefore we have

b²+a²1±cosν sinν

²

=d²+c²1±cosµ sinµ

² , (2.8)

a²+b²1±cosν sinν

2

=c²+d²1±cosµ sinµ

2

. (2.9)

Ifx²=y², then the equations

(2x²−b²−a²)²= 4((a²+b²)x²−a²b²) cot²ν , (2x²−d²−c²)²= 4((c²+d²)x²−c²d²) cot²µ must have the same solutions forx², that means

(2.10)

1/2 sin²ν

a²+b²± q

(a²+b²)²−((a²+b²)²+ 4a²b²cot²ν) sin⁴ν

=x²= 1/2 sin²µ

c²+d²± |sinµcosµ| |c²−d²| .

Ifa=b orc=d, then the isoptic is a circle, which proves the statement.

If a 6= b and c 6= d, one can eliminate the variables a, b, c, d from equation (2.10) via formulas (2.8) and (2.9), which leads to tan(ν/2) = tan(µ/2), i.e.,ν=µ.

Ifν =µ, equations (2.8) and (2.9) imply that eitherc=aandb=d, i.e., the ellipses coincide, or (1±cosν)²= sin²ν, i.e.,ν=π/2 and so the isoptic is a circle.

The theorem is proved.

We note that ifE_a,b^ν is a circle, i.e.,y²+x² is a constant, then the left-hand side of (2.6) is a constant, hence either cotν= 0, i.e.,ν =π/2, or alsoa²y²+b²x² is a constant. As E_a,b^ν has infinite points, the latter case impliesa² =b², i.e.,Ea,b

is a circle.

3. General uniqueness

To generalize the uniqueness part of Green’s result we introduce the homeo- morphismsχ^±_B,ν:B^ν→ B^ν andh^±_B,ν:S¹→S¹.

Let`<sub>B</sub>(n) be the tangent of the convex bodyBthat is orthogonal to the unit vectornand separates the interior ofBfromn. LetTB(n) be a touching point of the tangent `_B(n), i.e.,T_B(n)∈`_B(n)∩ B^ν.

SinceB^νis star-shaped with respect to each point of the convex bodyB,`B(n) intersects B<sup>ν</sup> in exactly two points. Let the functionP<sub>B,ν</sub><sup>+</sup> :S<sup>1</sup> → B<sup>ν</sup> be such that P<sub>B,ν</sub><sup>+</sup> (n)∈`_B(n), and−−−−−−−−−→

T_B(n)P_B,ν⁺ (n) makes a positive base of the plane withn. By P_B,ν⁻ we mean the same but with negative base.

We define the maps χ^±_B,ν:B^ν → B^ν so that χ^±_B,ν(P) is the point Q 6=P of the tangent `_B(n) that goes through P = P_B,ν^∓ (n). These maps are homeomor- phisms, because each tangent of B meets exactly two points ofB^ν, and P_B,ν^± are

homeomorphisms. We call the homeomorphismsχ_B,ν,±the right (left, respectively) ν-homeomorphism ofBonB^ν. Obviouslyχ⁺_B,ν◦χ⁻_B,ν=χ⁻_B,ν◦χ⁺_B,ν is the identity map onB^ν.

We define the mapsh^±_B,ν:S¹→S¹ so thath^±_B,ν(n) is the outer normal vector n⁰ 6= n of the tangent`_B(n⁰) that goes through P_B,ν^± (n). These maps are again homeomorphisms, because each point outside Bmeets exactly two tangents of B, and P_B,ν^∓ are homeomorphisms. We call the homeomorphismsh^±_B,ν the right (left, respectively) ν-homeomorphism ofB onS¹. Obviouslyh⁺_B,ν◦h⁻_B,ν=h⁻_B,ν◦h⁺_B,ν is the identity map onS¹.

Theorem 3.1. Letν/π¯ be an irrational or a rational number with even numerator in its lowest terms. If B^ν₁ ≡ B^ν₂ for the convex bodiesB1 andB2, thenB1≡ B2.

Proof. The convex bodiesB1 and B2 have common tangents, because otherwise one of them would contain the other one, and therefore would subtend bigger angle at every points. Furthermore, B1∩ B26=∅ by the following indirect reasoning. If B1∩ B2 =∅, thenB1 andB2 have four common tangents. Two of these common tangents, say`3and`4, meets in a point inside the convex hull ofB1∪B2. The other two common tangents, say`<sub>1</sub>and`₂, meet in a pointP, that may be at the infinity.

Choose the pointQ∈`1∩ B<sup>ν</sup><sub>1</sub> so that the segmentP Qcontains (B1∪ B2)∩`1. One of the convex bodies B1 and B2 is on the same side of`3 as P is, and the other convex body is on the other side of`₃. By renaming, we may assume thatB2is on the same side of `3 as P is, andB1 is on its other side. This immediately implies that B1 subtends bigger angle atQ thanB2 does. This contradicts the fact that Q∈ B₁^ν≡ B^ν₂, henceB1∩ B26=∅.

Assume that ¯ν/π is irrational and consider a common tangent`_B1(n₀) of B₁ andB2. Define the sequencenirecursively byni+1:=h⁺_B

1,ν(ni). Since the tangents

`<sub>B1</sub>(ni+1) and`_B1(ni) intersects each other on B₁^ν ≡ B^ν₂, we obtain by induction that each tangent`_B1(n_i) is a common tangent ofB1 andB2. As ¯ν/π is irrational, the sequenceni is dense inS¹, therefore the support functionspi:S¹→R+ of the convex bodies Bi (i = 1,2) are equal on a dense set. Since the support functions are continuous, we obtain p1=p2, henceB1≡ B2 as stated.

Now assume that ¯ν/π is rational and ¯ν/π = m/n (m, n∈ N) in its lowest terms.

Choose an arbitrary point P0 ∈ B_j^ν (j = 1,2), and define the sequence Pi

recursively by P_i+1 := χ⁺_B

j,ν(P_i). Then we have the sequence n_i ∈ S¹ given by PiPi+1=`_B(ni). Clearly, the angle ofni toni+1 is ¯ν, therefore

nkn=

n₀, ifkm is even, and

−n0, ifkm is odd, wherek∈N.

Ifkm is even, this shows thatP₀P₁ andP_knP_kn+1 are such parallel tangents of Bj that containBj on their same side, henceP0P1=PknPkn+1, i.e.,P0=Pkn.

So, if m is even, then n is odd, therefore n_n = n₀ and P₀ = P_n, hence P0P1· · ·P_n−1 is a polygonP that is inscribed inB^ν_j circumscribed to Bj and has edges of odd numbern∈N.

Now we prove thatP is also circumscribed toB_3−j.

Now choose a pointBinB1∩ B2, and for any pointPkdefineαk,j andβk,j to be the angles P_k−1PkB⁶ andBPkPk+16 , respectively. Also we define α_k,3−j and β_k,3−j as the anglesχ⁻_B

3−j,ν(P_k)P_kB⁶ andBP_kχ⁺_B

3−j,ν(P_k)⁶ , respectively. Clearly, we have αk,i+βk,i =ν for allk∈ {0,1, . . . , n−1}andi= 1,2.

Ifβ0,1 6=β0,2, then eitherβ0,1 < β0,2 or β0,1 > β0,2. Assumeβ0,1 > β0,2. It immediately follows thatα1,1> α1,2, and thereforeβ1,1< β1,2byα1,1+β1,1=ν = α_1,2+β_1,2.Iterating this step leads to 0<(−1)ⁱ(β_i,1−β_i,2) for i∈ {0,1, . . . , n}.

But then

0<(−1)ⁿ(βn,1−βn,2) =β0,2−β0,1, that contradicts the assumption.

If one assumes β_0,1 < β_0,2, then, in the same way, again a contradiction follows, hence we getβ0,1=β0,2.

The equationβ0,1=β0,2means that P is circumscribed also toB3−j. SinceP₀ was chosen arbitrary on B_j^ν (j = 1,2), we have shown that all the tangents of Bj are also tangents ofB3−j, which proves the lemma.

As Green’s examples show, Theorem 3.1 can not be improved in general.

4. Distinguishability of convex polygons

By the inscribed angle theorem, theν-isopticP^νof a convex polygonP is the union of circular arcs.

ν > π/3 ν =π/3 ν < π/3

This implies immediately that a convex bodyBis not polygonal if itsν-isoptic B^ν either contains any small piece of a not circular arc or is differentiable at its every points.

Conjecture 4.1. If B^ν ≡ P^ν for the convex body B and a convex polygonP, then B ≡ P.

This conjecture might be proved via two statements:

(1) IfB^ν≡ P^ν, thenB is polygonal;

(2) IfP₀^ν ≡ P₁^ν for convex polygons P0 andP1, thenP0≡ P1. We can prove the second statement.

Theorem 4.2. Two convex polygons with a common connectedν-isoptic coincide.

Proof. Let Σ denote the set of the circles of circular arcs inP^ν, and let Π be the set of the intersections of these circles. Obviously, the vertices of P are a subset of Π.

letP^ν is the commonν-isoptic of the convex polygons P and P⁰. AsP^ν is connected, we have ν < min(min_Pα,min_P⁰α). Neither of these convex polygons can properly contain the others, because then the one that contains the other would subtend bigger angle at all points outside of it.

Thus, there is a common tangent `<sub>0</sub> of P and P<sup>0</sup>, and therefore there are two further common tangents `M and `N trough the points M 6=N of `0∩ P^ν, respectively.

To prove that`0 intersects P andP⁰ in the same edge or vertex, we have to consider several cases.

(a) IfM is not an intersection of different circles of Σ, then all the straight lines of the edges of the convex polygonsPandP⁰avoidM, hence for some vertices A, B ∈ P and A⁰, B⁰ ∈ P⁰ we have{A}=`<sub>0</sub>∩ P,{A<sup>0</sup>}=`₀∩ P⁰, {B}=`<sub>M</sub> ∩ P and {B<sup>0</sup>}=`M ∩ P⁰.

In the other hand, the segments AB and A⁰B⁰ subtend the same constant visual angle ν on an arc of the circle C ∈ Σ that contains M, hence A, B, A⁰, B⁰ ∈ C follows and therefore `<sub>0</sub>∩ P ={A}=`₀∩ C \ {M}={A⁰}=`₀∩ P⁰, which was to be proved.

(b) IfM is an intersection of different circlesC₋ andC+ of Σ, then there are edges of both convex polygonsP andP⁰ that are collinear toM, and, since`0and

`M are tangent to both convex polygons, these edges are on`0 or`M. By symmetry in logic, we only need to consider three cases:

(b1) AB=`0∩ P,A<sup>0</sup>B<sup>0</sup>=`0∩ P⁰,{C}=`M ∩ Pand{C<sup>0</sup>}=`M∩ P⁰, (b2) AB=`0∩ P,{A<sup>0</sup>}=`0∩ P⁰,{C}=`M∩ P andC<sup>0</sup>D<sup>0</sup>=`M ∩ P⁰, (b3) AB=`0∩ P,A<sup>0</sup>B<sup>0</sup>=`0∩ P⁰,CD=`M∩ P andC<sup>0</sup>D<sup>0</sup>=`M ∩ P⁰. In all these cases there is an open neighbourhoodUP ofM that any tangent line of P through any point ofU_P intersectsP in only one of the pointsAandB. By the same clear reason, there is an open neighbourhoodUP⁰ ofM that any tangent line of P⁰ through any point of U_P⁰ intersects P⁰ in only one of the points C and D.

LetU =UP ∩ UP⁰.

(b1) (b2) (b3)

Observe, that if a tangent`ofP intersects a circular arcC ∩ P<sup>ν</sup> in two points, then the segment of these intersection points contains the touching points`∩ P, because P^ν is star-shaped from any point ofP. Therefore, if a tangent ofPthrough a point P ∈ C₊∩ U cuts P in A, then any tangent that intersects C₋∩ U cuts P in B, hence by renaming the circles C− andC+ of Σ, we may assume that for any point P ∈ C+∩ U the tangent ofP throughP passes A, and the tangents ofP through C−∩ U containB.

If the pointB⁰exists, then by renaming the pointsA⁰ andB⁰, we may assume that for any point P ∈ C+∩ U the tangent of P through P passes A⁰, and the tangents of P throughC₋∩ U containB⁰.

If the pointD exists, then by renaming the pointsCandD, we may assume that for any point P ∈ C+∩ U the tangent of P through P passes C, and the tangents of P throughC₋∩ U containD.

If the pointD⁰exists, then by renaming the pointsC⁰ andD⁰, we may assume that for any point P ∈ C+∩ U the tangent of P through P passes C⁰, and the tangents of P throughC₋∩ U containD⁰.

Thus, the tangents ofP throughC+∩ UcontainA, the tangents ofP⁰ through C+∩ U containA⁰, the tangents ofP throughC₋∩ U containC, and the tangents of P⁰ throughC−∩ U containC⁰.

Now we are ready to consider the cases (b1), (b2) and (b3) on by one. In all three cases the use of the reasoning of (a) on C₊ and the pointsA, C, A⁰, C⁰, we conclude A≡A⁰ andC≡C⁰. ConsideringC− in the similar way

(b1) for the pointsB, C, B⁰, C⁰, we inferB≡B⁰ andC≡C⁰,

(b2) for the points B, C, A⁰, D⁰, we obtain B ≡A⁰ and C ≡D⁰, which is a contradiction, becauseA6≡B, showing that case (b2) can not happen, and

(b3) for the pointsB, D, B⁰, D⁰, we getB≡B⁰ andD≡D⁰.

Thus, we conclude the investigations of case (b) by inferring `<sub>0</sub>∩ P=`₀∩ P⁰. As`0 is an arbitrary common tangent ofP andP⁰, we established that each common tangent intersects P and P⁰ in the same edge or vertex.

If the vertex P0 of P is outside of P⁰, and no tangent of P through P0 is tangent to P⁰, then straight lines of both edges of P through P0 avoid P⁰. In this case the vertices P₋₁ andP₁ of P next toP₀ are outside of P⁰, and, by the very same reason as for P0, either one of them is on a common tangent ofP and P⁰, or the vertices P₋₂ andP₂ of P next toP₋₁ andP₁ are again outside of P⁰. Continuing this reasoning and taking into account thatP has finitely many vertices we conclude that either P containsP⁰, which is excluded. or there is a vertexP_±k of P (k ∈N) outside of P⁰, that is on a common tangent of P and P⁰. But this latter case contradicts the fact that any common tangent intersects P and P⁰ in the same edge or vertex.

Therefore no vertex of eitherP or P⁰ can be outside of the other one, which proves the theorem.

5. Distinguishability for rotationally symmetric convex bodies

As a matter of fact our first result is about reconstructibility of convex bodies rotationally symmetric with respect to the angleπ−ν.

Proposition 5.1. For any convex bodyB and any angle ν ∈(0, π)there may be at most one ν-ghost G of B that is rotationally symmetric with respect to the angle π−ν.

Proof. Place the originO at the centre of the rotation ofGand define the function α:S¹→S¹ by the equation u(α(ψ)) =r^ν_B(ψ)/|r^ν_B(ψ)|. Then the support function p_G ofGis periodic with period ¯ν andαis bijective.

As the isoptics are star-shaped, for any point P = r^ν_B(ψ) ∈ B^ν there is a unique angleϕso thatP=r^ν_G(ϕ), hence, by (2.2), we have

|P|²=|r^ν_G(ϕ)|²=p²_G(ϕ)2(1−cos ¯ν)

sin²ν¯ = p²_G(ϕ) cos²(¯ν/2).

This gives p_G(ϕ) = |P|cos(¯ν/2), which means that u(ϕ) closes angle ¯ν/2 to the straight lineOP which is in directionu(α(ψ)). We conclude, thatp_G(α(ψ)−¯ν/2) =

|r^ν_B(ψ)|cos(¯ν/2) for allψ, which, by the bijectivity ofα, determinespG onS¹. This proves the statement.

Now we are relaxing the condition of rotational symmetry.

Theorem 5.2. The convex bodyBhas rotational symmetry with respect to the angle 2(π−ν) if and only if the length |r^ν_B| of the parametrization r^ν_B of B^ν is periodic with periodπ−ν.

Proof. LetBbe a convex body of rotational symmetry with respect to the angle 2¯ν, and choose the origin to be the centre of this rotation. Then its support function p_B is periodic with period 2¯ν and, by (2.2), we have

|r^ν_B(ϕ)|²sin²ν¯=p²_B(ϕ) +p²_B(ϕ+ ¯ν)−2pB(ϕ)pB(ϕ+ ¯ν) cos ¯ν

=p²_B(ϕ+ 2¯ν) +p²_B(ϕ+ ¯ν)−2pB(ϕ+ 2¯ν)p(ϕ+ ¯ν) cos ¯ν

=|r^ν_B(ϕ+ ¯ν)|²sin²ν,¯ hence|r^ν_B(ϕ)|is periodic with period ¯ν.

If|r^ν_B(ϕ)|is periodic with period ¯ν, then|r^ν_B(ϕ)|² =|r^ν_B(ϕ+ ¯ν)|², which, by (2.2), implies

(5.1) 0 = (p_B(ϕ+ 2¯ν)−p_B(ϕ))(p_B(ϕ+ 2¯ν) +p_B(ϕ)−2p_B(ϕ+ ¯ν) cos ¯ν).

By convexity we have

(5.2)

p_B(ϕ+ 2¯ν) +p_B(ϕ)

=hu(ϕ+ 2¯ν),r_B(ϕ+ 2¯ν)i+hu(ϕ),r_B(ϕ)i

≥ hu(ϕ+ 2¯ν),rB(ϕ+ ¯ν)i+hu(ϕ),rB(ϕ+ ¯ν)i

=hu(ϕ+ 2¯ν) +u(ϕ), pB(ϕ+ ¯ν)u(ϕ+ ¯ν) + ˙pB(ϕ+ ¯ν)u^⊥(ϕ+ ¯ν)i

= 2pB(ϕ+ ¯ν) cos ¯ν,

which is an equality if and only if

(5.3) hu(ϕ+ 2¯ν),r_B(ϕ+ 2¯ν)−r_B(ϕ+ ¯ν)i= 0 and hu(ϕ),r_B(ϕ+ ¯ν)−r_B(ϕ)i= 0.

Condition (5.3) means that the point r_B(ϕ+ ¯ν) is the intersection of the tangents with normal vectors u(ϕ+ 2¯ν) andu(ϕ). The number of such points is at most π/(2¯ν), therefore inequality (5.2) is strict except for finitely manyϕ. Taking into account that the support functionp_Bis continuous, this impliesp_B(ϕ+2¯ν)≡p_B(ϕ) by (5.1).

If an isopticB^ν is a circle, thenr^ν_B is periodic with periodπ−ν, hence the convex bodyBhas rotational symmetry with respect to the angle 2(π−ν)^(5:2)by our Theorem 5.2.

Lemma 5.3. Let the convex body B be rotationally symmetric with respect to the angle2(π−ν). If the polygonP is circumscribed toBand inscribed inB^ν, then the vertices of P are on a circle centred at the rotational centre of B and the lengths of the edges ofP are alternating with two values.

Proof. Letr^ν_B(ϕ) be a vertex ofP. Thenr^ν_B(ϕ+k¯ν) are the vertices ofP, where k∈Z. Theorem 5.2 gives then that the vertices are on the circleCof radius|r^ν_B(ϕ)|

centred to the origin.

The edges ofP are the chords ofC, so the lengths of them are determined by their distances from the origin, which arep_B(ϕ+k¯ν). This alternates according to the parity ofk, because the period ofp_B is 2¯ν.

Theorem 5.4. If the convex bodies BandG have only finitely many common tan- gents, both are rotationally symmetric with respect to the angle 2(π−ν), and if they have common ν-isopticB^ν ≡ G^ν, thenB^ν is a circle.

Proof. Since neitherG norBcan contain the other, there are common tangents of G andB. If a sequence of common tangents has a limit`(that is an accumulation point of the set of common tangents), then ` is a common tangent of G and B.

Therefore the setL of common tangents is closed.

Let`be a common tangent ofGandBthat is isolated inL, and intersectsB<sup>ν</sup> in the points {P, Q}=`∩ B^ν. Assume, that B is in the halfplaneS<sub>`</sub><sup>−</sup> of `, which is to the left of −−→

P Q.

(5:2)

A direct proof of this comes from ¯ν= arccos(p_B(ϕ)/%) + arccos(p_B(ϕ+ ¯ν)/%),where

% >0 is the radius of the circleB^ν.

Choose a pointP₀∈ B^ν in the halfplaneS<sub>`</sub><sup>+</sup>of`, which is to the right of−−→ P Q, so thatP0is so near toP that no straight line inLintersects a small neighbourhood of the arcB^ν₀ ofB^ν that connectsP to P₀ (except of course`).

Let P1 be the only such point of B^ν that P0P1 is a tangent line of B that separates G ∪ B from Q. If such point does not exist, then there must exist a unique point P₁ ∈ B^ν so that P₀P₁ is a tangent line of G that separates G ∪ B from Q.

For simplicity we assume the first case, that is,P₁ is the only such point of B^ν thatP0P1 is a tangent line ofB,P0P1is outside ofG andBis on the left-hand side of−−−→

P₀P₁.

LetP2be the only such point ofB^ν that P1P2 is a tangent line ofG andGis on the right-hand side of−−−→

P1P2. Then, obviously,P1P2 intersectsB.

Given the pointP2k onB^ν for some k∈N, letP2k+1be the only such point of B^ν that P2kP2k+1 is a tangent line of B and B is on the left-hand side of −−−−−−→

P2kP2k+1. Then, obviously, P2kP2k+1 is outside ofG. Next, letP2k+2 be the only such point of B^ν that P_2k+1P_2k+2 is a tangent line of G and G is on the right-hand side of

−−−−−−−−→

P2k+1P2k+2. Then, again obviously,P2k+1P2k+2intersects B.

This gives an infinite sequence{P_i:i∈N}of points onB^ν so that the points of even indexes are in S_{`</sub><sup>+</sup> and the points of odd indexes are inS<sub>`}⁻.

Since bothBand Gare rotationally symmetric with respect to the angle 2¯ν, Lemma 5.3 implies |P2i| =|P2i+1| and |P2i+1| = |P2i+2|, respectively. Thus, we obtain|P2i|=|P2i+1|=|P| for alli∈N.

Let 0< γ_i < π/2 be the angle of the straight line `<sub>i</sub> = P<sub>i</sub>P<sub>i+1</sub> to `. It is an immediate consequence of the construction of the points {Pi : i ∈ N}, that the sequence γ_i is strictly decreasing, hence there exists 0 ≤γ_∞ = lim_i→∞γ_i <

π/2. Since every straight line`i intersects the segmentP Q, we infer the existence of a common tangent `_∞ = lim_i→∞`i. But `0 was chosen so that B^ν₀ is not intersected by any common tangent, so`∞=`follows, hence limi→∞P2i=P and lim_i→∞P2i+1=Q.

We have proved thatB^ν is a circular arc near any endpoint of any isolated common tangent and the radii of the circles of these arcs are equal.

SinceLis finite, every common tangents ofBandG are isolated.

Take a maximal arcB^ν_maxofB^ν that is not intersected by any common tangent ofBandG. LetPandP⁰ be the two closing points ofB^ν_max. Then there are unique common tangents`, `⁰∈ LthroughPandP⁰, respectively, so that`and`⁰separate P⁰andP, respectively, fromB∪G. We define the pointsQandQ⁰by`∩B^ν ={P, Q}

and `⁰∩ B^ν={P⁰, Q⁰}.

LetP₀ be any point of B^ν_max, and let P_i be the sequence of points generated from P0 as above so thatP2iP2i+1 separatesB ∪ Gfrom {Q, P⁰}. Then the angles 0 < γi < π/2 of the straight lines P2iP2i+1 to P Q decreases monotonously again by the construction, so the sequence of the straight lines P_2iP_2i+1 converges, and therefore it converges toP Q. Thus any point ofB_max^ν is on the circle of the circular arc ofB^ν nearP, henceB^ν_maxis a circular arc.

This proves the theorem completely.

An obvious, but interesting statement follows.

Theorem 5.5. Two convex bodies with rotational symmetry with respect to the angle 2(π−ν) having the same noncircularν-isoptic coincide.

According to W. Wunderlich’s result mentioned already in the introduction, the condition of rotational symmetry is crucial in this theorem. However Theo- rem 2.3 shows that the result may be valid with some other conditions as well.

6. More isoptics

Referring to Green’s result, Klamkin in 1988 raised the question “If two isoptic curves of a convex domain are concentric circles, is it then also a concentric circle?”

(see [2]), which was affirmatively answered by J.C.C. Nitsche [8] two years later.

It is most natural to broaden Klamkin’s question as follows:

(6.1) How many isoptics of a convex body are necessary to permit its exact reconstruction?

To the best knowledge of the author, not any answer to this question — except, of course, the ones of Green and Nitsche — was published by now. The following specialized version of [3, Theorem 5] gives the best answer the author could give by now.

Theorem 6.1. If {ν_i}_i∈N is a strictly decreasing sequence of angles in(0, π), and the νi-isoptics of two convex bodies coincide for all i, then the convex bodies also coincide.

Nitsche’s result and the additional conditions imposable by Theorem 3.1 offer hope for improvement of Theorem 6.1, but the author sees nothing that would support a better result in general.

References

[1] J. W. GREEN, Sets subtending a constant angle on a circle, Duke Math. J., 17 (1950), 263–267.

[2] M. S. KLAMKIN, Conjectured isoptic characterization of a circle, Amer. Math.

Monthly, 95(1988), 845.

[3] ´A. KURUSA, The shadow picture problem for nonintersecting curves, Geom. Dedi- cata, 59(1996), 113–125.

[4] ´A. KURUSA, K´upszeletek izoptikusai, Polygon, 19:1(2011), 27–46 (in Hungarian).

[5] H. MARTINIand W. MOZGAWA, An integral formula related to inner isoptics, Ren- dic. Sem. Mat. Univ. Padova, 125(2011), 39–49.

[6] A. MIERNOWSKIand W. MOZGAWA, On some geometric condition for convexity of isoptics, Rend. Sem. Mat. Univ. Pol. Torino, 55(1997), 93–98.

[7] W. MOZGAWA, Integral formulas related to ovals, Beitr¨age Algebra Geom., 50 (2009), 555–561.

[8] J. C. C. NITSCHE, Isoptic characterization of a circle, (Proof of a conjecture of M.S.

Klamkin), Amer. Math. Monthly, 97(1990), 45–47.

[9] W. WUNDERLICH, Kurven mit isoptischen Ellipse, Monatshefte Math., 75(1971), 346–362.

[10] R. C. YATES,A Handbook on Curves and their Properties, J. W. Edwards, Ann.

Arbor, 1947, pp. 138–140.

A. K´ URUSA, Bolyai Institute, University of Szeged, Aradi v´ertan´uk tere 1., H-6720 Szeged, Hungary; e-mail: kurusa@math.u-szeged.hu