The shadow picture problem for nonintersecting curves
Arp´´ ad Kurusa
Abstract. We prove that two strictly convex bodies in the plane subtending the same angles at each of the points of two parallel straight lines and a big closed curve, must coincide.
1. Introduction
The problem of reconstructing a plane body from its shadow pictures, S- pictures, was raised in [5] (and for special cases in [3], [6]). The S-picture of a convex body D at a point P ∈ R2 is defined as the angle of the two supporting lines ofDgoing throughP. This angle is called the visual angle too. The question is: What set of S-pictures distinguishes any two convex bodies.
In [4] it is shown that the S-pictures taken from one curve only do not deter- mine the convex bodies in general. However, roughly speaking, they do distinguish the polygons from each other [4]. LetC1,C2,F1andF2be closed convex domains with C2 boundaries. Assume that the strictly convex bodiesF1 andF2 subtend equal visual angles at each point of ∂C1∪∂C2, and F1∪ F2 are in the interior of C1∩ C2. The author showed in [5] that if ∂C1 and ∂C2 intersect each other in non-zero angles, then F1 andF2 must coincide.
In this article we investigate the S-picture problem for curves that do not intersect each other. We prove uniqueness for S-pictures taken from two straight lines, or two hyperbolas. The method, we develop, can also be used for curves having asymptotes at infinity.
The basic idea of our method, coming from [2], is that the S-pictures generate such a measure on the set of straight lines that two domains having equal S-pictures AMS Subject Classification(2000): 0052,0054.
Supported by the Hungarian NSF, OTKA Nr. T4427
have also equal volumes (perhaps infinite) with respect to this measure. For this measure, the finiteness of the volume of the difference of the two convex bodies is proved first. Then an infinite sequence of components of the difference of the bodies is constructed so, that the volumes of the components are the same (or almost the same). This implies that the volume of every component should be zero, hence the two bodies must coincide.
In the third section, we prove with elementary geometry that any two convex bodies can be distinguished by their S-pictures taken on any infinite set of curves.
The author thanks the Soros Foundation for supporting his visit at the De- partment of Mathematics of the MIT, where this research was done. He also thanks the Department of Mathematics of MIT and the Matematische Institut of the Uni- versit¨at Erlangen-N¨urnberg for the support and assistance in finishing this paper.
Thank is also due to the referee for his/her help in improving the form of this paper.
2. Preliminaries
First of all, we redefine the S-picture. LetLbe the Grassman manifold of all the straight lines in the plane. Given a planar compact domain Dwith piecewise C1 boundary, we define ¯Das a domain inLof all the straight lines intersectingD so thatDhas exactly two tangents parallel tol.
The S-picture functionSD of a domainDas above is defined then as SD:R2−→R SD(P) = 1
2 Z
S1
χD¯(l(hω, Pi, ω)) dω,
where χD¯ is the indicator function of ¯D, l(r, ω) means the straight line through rω that is perpendicular to ω, and h., .i is the usual inner product. (The factor 1/2 is needed becausel(r, ω) =l(−r,−ω)). It is easy to see, that this definition is the same as the original one for convex bodies. For simplicity we shall writel(r, β) instead ofl(r, ωβ), whereωβis the unit vector closing angleβwith an appropriately fixed unit vector.
Let g:R → R2 be a C1 curve parameterized by its arclength. Let ξ(s, α) denote the straight line l(r, ωβ) with r = |g(s)| and making hωβ,g(s)i˙ = sinα.
From [7] the invariant measure on Lis
(1) dβdr=|sinα|dαds,
Let D be a strictly convex body with C2 boundary so that the curve g is outside of D. Let a(s) andb(s) be the lengths of the two tangents of D through
g(s);α(s),β(s) are the corresponding angles of these two tangents to ˙g, respectively.
We proved in [5] that
(2) ν˙ =sinβ
b −sinα a , where ν(s) =SD(g(s)).
3. The main results
Theorem 1. D1 and D2 are strictly convex bodies, g1 and g2 are straight lines outside ofD1∪ D2andSD1=SD2 ong1andg2. IfD1andD2 have outer common tangent that intersects the straight lines, then D1 andD2 coincide.
Proof. For intersecting straight lines this has already been proved in [5], hence we can assumeg1kg2. We suppose further that the angle ofg1to the fixed direction is zero, and therefore |cos1β|dβdr= dαdsby (1). From now on we use this measure onL.
First we prove thatD1andD2have infinitely many common tangents: Lett0 be an assumed common tangent, and suppose t0 intersectsg1andg2 at the points X0 and Y0, respectively. The visual angles of D1 and D2 are equal at Y0, hence there exist an other common tangent t1 throughY0. The tangentt1 intersectsg1 in X1, where the visual angles of D1 and D2 are also equal. Therefore a third common tangent t2 must go through X1. t2 intersects g2 in Y1 and so one can continue the procedure in the same way to get the sequencetiof common tangents.
This sequence of common tangents is certainly infinite, because the intersections Xi=t2i∩g1andYi =t2i∩g2make monotone sequences ong1andg2, respectively.
Further these sequences tend to infinity, because the visual angles ofD1 andD2at their respective limits would be zero otherwise.
The existence of the sequence ti has as first consequence that D1 and D2
have two common tangents limi→∞t2i and limi→∞t2i+1 parallel tog1. Let their distances to g1 behandH, respectively, whereh < H.
We prove next that any common tangent ofD1 andD2 intersects the bodies in a point of∂D1∩∂D2. Let the distances of∂D1∩tiand∂D2∩ti fromti∩g1 be di and ¯di, respectively. Let ei= ¯di−di. Assuminge0≥0 one can easily see from (2) that ei≥0 for alli∈N
Equation (2) gives sinα2i d2i+e2i
− sinα2i−1 d2i−1+e2i−1
=sinα2i d2i
−sinα2i−1 d2i−1
forg1 and
sinα2i
d2i+e2i+sinfα
2i
− sinα2i+1
d2i+1+e2i+1+sinαf
2i+1
= sinα2i
d2i+sinfα
2i
− sinα2i+1
d2i+1+sinαf
2i+1
,
forg2, whereαi is the angle betweenti andg1, andf is the distance ofg1andg2. By an easy rearrangement we have
e2i
sinα2i
d2i(d2i+e2i) =e2i−1
sinα2i−1
d2i−1(d2i−1+e2i−1) and
e2i+1 sinα2i+1
(d2i+1+sinαf
2i+1)(d2i+1+e2i+1+sinαf
2i+1)
=e2i
sinα2i
(d2i+sinαf
2i)(d2i+e2i+sinfα
2i). Therefore
e2i+1
e2i−1 =sinα2i−1 sinα2i+1
(d2i+1+sinαf
2i+1)(d2i+1+e2i+1+sinαf
2i+1) d2i−1(d2i−1+e2i−1) ×
× d2i(d2i+e2i) (d2i+sinαf
2i)(d2i+e2i+sinfα
2i). To find the limit of the right hand side asi→ ∞first we observe that the sequence e2i−1 is bounded and
(3) H = lim
i→∞d2isinα2i and h= lim
i→∞d2i+1sinα2i+1. It is easy to see, that
(4) tanα2i+1∼ H+f
|Yi| , tanα2i−1∼ H
|Xi| and |Yi|
|Xi| ∼ h+f h
where ∼denotes asymptotic equivalence. Therefore
i→∞lim
sinα2i−1 sinα2i+1
= lim
i→∞
tanα2i−1 tanα2i+1
= H
H+f h+f
h .
Taking the above limits into our expression for e2i+1/e2i−1we obtain
i→∞lim e2i+1
e2i−1 = H H+f
h+f h >1.
Sinceeiis obviously bounded, this givese0= 0, i.e. each common tangent intersects D1 andD2 in a point of∂D1∩∂D2. Consequently, there is a common tangent at each point P of ∂D1∩∂D2, because otherwise a common tangent with different points of intersection would there exist.
In sum, we have proved thatD1∩ D2 has nonempty interior, and therefore IntD1\ D2 (resp. IntD2\ D1) is the union of connected components, that are bounded by one arc of ∂D1 and by an other arc of∂D2. These arcs intersect each other in two points, where they have common tangents. It follows from these, that Int( ¯D14D¯2) consists of components inL, too.
Let Φjldenote the straight line throughl∩gj which makes the angleSD1(l∩ gj) = SD2(l∩gj) with l in the appropriate direction. Let ¯R be a component of Int( ¯D14D¯2) and set ΨjR¯ = {Φjl : l ∈ R}. Since Φ¯ jl cuts a component of Int(D14D2) exactly whenl does, ΨjR¯ is a component of Int( ¯D24D¯1).
Let gj(s1) and gj(s2) be the intersections of gj with the tangents at the endpoints ofR. Further, letM andN be the volume of ¯Rand ΨjR, respectively.¯ Then
M = Z
R¯
dβdr
|cosβ| = Z
R¯
dαds= Z s2
s1
SR¯(gj(s)) ds
= Z s2
s1
Z
χR¯(ξ(s, α)) dαds= Z s2
s1
Z
χΨjR¯(Φjξ(s, α)) dαds
= Z s2
s1
Z
χΨjR¯(ξ(s, α)) dαds= Z s2
s1
SΨjR¯(gj(s)) ds
= Z
ΨjR¯
dαds= Z
ΨjR¯
dβdr
|cosβ| =N, hence ¯Rand ΨjR¯ have the same volume.
To see that the volume of ¯D14D¯2 is finite, one needs to consider only the straight lines close to the common tangents parallel to g1, because the measure
dβdr
|cosβ| has singularity only at β = ±π/2. Let us choose the common tangent l(h, π/2) and calculate the volume of Int ¯D1\D¯2 in a smallε >0 neighborhood of l(h, π/2). We have
Z π2+ε
π 2−ε
Z dβdr
|cosβ| = Z π2+ε
π 2−ε
Z
dr dβ
|cosβ| = Z π2+ε
π 2−ε
%(β) dβ
|cosβ| = Z π2+ε
π 2−ε
σ(β) dβ,
where %(β) is the distance of the two tangent of IntD1\ D2 perpendicular to ωβ. σ(β) is the length of the segment these two tangents cut out froml(h, π/2). Since l(h, π/2) meets D1 and D2 in the same point, σ(β) →0 as β →π/2, hence the volume is finite by the last integral.
SinceS
k∈N(Ψ1Ψ2)kR¯ has finite volume and (Ψ1Ψ2)kR¯ is infinite sequence of disjoint sets having constant volume, we conclude that ¯Rand in the same way any other component should be empty. This completes the proof.
Using S-pictures taken from a third curve, we can skip from the condition on the common tangent.
Theorem 2. LetD1andD2be strictly convex bodies and letCbe a compact domain so that D1∪ D2⊂IntC. Letg1 andg2 be straight lines not intersectingD1∪ D2. If the visual angles of D1 and D2 are equal at each point of g1, g2 and ∂C then D1≡ D2.
Proof. Since neither of D1 and D2 can contain the other, they have a common outer tangent t. This cuts∂C, and therefore they have an other common tangent throught∩∂C that is not parallel to t. Now one uses Theorem 1 to conclude the statement.
In the following theorem we substitute the straight linesg1 andg2 with hy- perbolas. A hyperbolahdivides the plane into tree parts, two of which are convex.
We denote the union of these two convex parts by C(h). The proof of the next theorem is very similar to the previous one, therefore we are going into details only where nontrivial differences occur.
Theorem 3. Let the asymptotes of the hyperbola h1 be parallel to the asymptotes of the hyperbola h2. Suppose thath1 does not intersect h2 and the strictly convex bodies D1 and D2 are inIntC(h1)∩IntC(h2). If the visual angles of D1 andD2
are equal at each point of h1 andh2 thenD1≡ D2.
Proof. For easier calculation we assume that the asymptotes ofh1 andh2 are not the same. As the reader will see, a slight modification of our proof can handle that case too (h16≡h2 is necessary of course). We choose the asymptote closest to D1∪ D2 to be thex-axis and parameterize the hyperbolas by arclength. We shall follow the steps of the proof of Theorem 1 using the same notations where possible.
Obviously any common tangent ofD1 and D2 intersects h1 and h2, and so, one can construct an infinite sequence of common tangents, as in the proof of Theorem 1.
To show that each common tangent intersectsD1andD2in a point of∂D1∩
∂D2, we can apply (2) with a calculation very similar to that used in the proof of Theorem 1.
Therefore we have the same situation, i.e. IntD1∩IntD2 is not empty and Int(D14D2) is the union of its components. These components are bounded by one arc of ∂D1 and by an other arc of ∂D2. These arcs have only two common points where D1 andD2 have common tangents.
We define the mappings Φ1, Φ2, Ψ1and Ψ2just as in the proof of Theorem 1.
Again, the sequence (Ψ1Ψ2)kR¯ of components is infinite for any component ¯Rof Int ¯D1\D¯2.
In the present case it is no longer true that a component and its images by Ψ1 and Ψ2have the same volume. It is true, however, that if the component ¯Ris not empty, then the volume ofS∞
j=1(Ψ2Ψ1)jR¯ is infinite w.r.t. the measure |cosdβdrβ|. To show this, let us first observe that all the components (Ψ2Ψ1)jR¯ are dis- joint. Let ¯Rj = (Ψ2Ψ1)jR¯ and the volume of ¯Rj be Mj. We shall estimate the ratioMj+1/Mj whenj→ ∞.
Let the angle between ˙hi(s) and thex-axis beγi(s). Thenβ=αi−γi−π2, i.e.
cosβ= sin(αi−γi), where (si, αi) is the parameter of the straight lineξi(si, αi) on hi. By (1) we have dβdr= sinαidαidsi, hence |cosdβdrβ| = sin(αsinα1
1−γ1)dα1ds1 that implies
(5) Mj+1= Z
R¯j+1
sinα2
sin(α2−γ2)dα2ds2= Z
Ψ1R¯j
sin(α2−ν2)
sin(α2−ν2−γ2)dα2ds2, where νi(s) =SD1(hi(s)) =SD2(hi(s)) is the visual angle athi(s) and the second equality follows from the definition Ψ2Ψ1R¯ ={Φ2l:l ∈Ψ1R}. By (1) dα¯ 2ds2 =
sinα1
sinα2dα1ds1 forξ1(s1, α1) =ξ2(s2, α2) therefore Mj+1=
Z
Ψ1R¯j
sin(α2−ν2) sinα1 sin(α2−ν2−γ2) sinα2
dα1ds1, Substituting Φ1lagain, by Ψ1R¯ ={Φ1l:l∈R}¯ we arrive to
Mj+1= Z
R¯j
sin(α2−ν2) sin(α1+ν1) sin(α2−ν2−γ2) sinα2
dα1ds1,
where α2, ν2 and γ2 are the h2-parameters of the straight line ξ1(s1, α1+ν1).
Substitutingα2=α1+ν1−γ1+γ2and combining the result with Mj=
Z
R¯j
sinα1
sin(α1−γ1)dα1ds1
one gets
(6) Mj+1−Mj =
Z
R¯j
µ(s1, α1) sinα1
sin(α1−γ1)dα1ds1, where
µ(s1, α1) = sin(α1+ν1−ν2−γ1+γ2) sin(α1+ν1) sin(α1−γ1) sin(α1+ν1−ν2−γ1) sin(α1+ν1−γ1+γ2) sinα1 −1.
Since all the indicated angles tend to zero as s1→ ∞, we have
s1lim→∞s1µ(s1, α1) = lim
s1→∞s1
(α1+ν1−ν2−γ1+γ2)(α1+ν1)(α1−γ1) (α1+ν1−ν2−γ1)(α1+ν1−γ1+γ2)α1
−1
= lim
s1→∞s1 α1ν2γ2−ν2(α1+ν1−ν2−γ1+γ2)γ1
(α1+ν1−ν2−γ1)(α1+ν1−γ1+γ2)α1.
Elementary calculation gives lims1→∞s21sinγi = constant. By this and (4) the termsα1,ν1andν2are asymptotically equivalent toc/s1for appropriate constants c. Therefore lims1→∞s1µ(s1, α1) =kfor some constantk.
The component ¯Rjhas two common tangents ofD1andD2. They intersecth1
in two points, say, inh1(σj) andh1(¯σj) so thatσj < s1<σ¯jfor allξ1(s1, α1)∈R¯j. Therefore the integral in (6) can be estimated as
|Mj+1−Mj| ≤ |k|+ε σj Mj
for j big enough and ε > 0 small enough. In more suitable equivalent form this means
1−|k|+ε σj
≤Mj+1 Mj
≤1 +|k|+ε σj
.
We shall show that σj grows exponentially with j → ∞. This proves limj→∞Mj =constant >0 that givesP∞
j=1Mj =∞as we stated.
Let the common tangent t ∈ Ψ1R¯j through h1(σj) cuth2 in h2(τj). Then the common tangent Φ2t is in ¯Rj+1 and therefore cutsh1 in h1(σj+1). Let f be the distance of the two asymptotes parallel to thex-axis. LethandH denote the distances from the x-axis of the two common tangents parallel to the x-axis, so that h < H. Obviously,
j→∞lim τj
σj
=h+f
h and lim
j→∞
σj+1
τj
= H
H+f,
hence
j→∞lim σj+1
σj
= H
H+f h+f
h >1, which was to be shown.
To complete the proof we only have to note that the volume of ¯D14D¯2 is finite by the same reason as in the case of the straight lines.
The above method can be used for a large class of curves. These curves should have two important properties: asymptotes in the infinity and something that ensures that the sequence of common tangents gets onto the asymptotic part.
We state this observation in the following theorem.
Theorem 4. Let h1 andh2 be convex curves so that 1) hi (i= 1,2) has an asymptotic straight linegi,
2) g1 is parallel to g2 and the intersection of the convex envelopes of h1 andh2
is not empty,
3) the angleγi(s)between gi andh˙i(s) satisfieslims→∞sγi(s) = 0.
Let D1 and D2 be strictly convex bodies so that D1∪ D2 is in the interior of the intersection of the convex envelopes of h1 and h2. If the visual angles of D1 and D2 are equal at each point ofh1 andh2 thenD1≡ D2.
4. Infinitely many arbitrary curves
Although the results above and in [5] cover many cases, we still do not know any general result for S-pictures taken from two concentric circles, the first case considered in the literature. We offer the following theorem as a first step in this direction.
Theorem 5. Let Ci (i∈N),D1 andD2 be convex bodies so thatD1∪ D2⊂IntC0 andCi⊂IntCi+1 for(i∈N). If the S-pictures ofD1 andD2are equal at all points of each ∂Ci, thenD1≡ D2.
Proof. Lett0 be a common supporting line ofD1 and D2. It intersects each ∂Ci
in two points, Pi and Qi. Let T1 =t0∩ D1 and T2 =t0∩ D2. T1 and T2 divide the straight linet0into parts. On one of the two infinite parts are the intersection pointsPi and on the other one are the intersection pointsQi (i= 0,1,2, . . .).
LetP be a limit point of{Pi}, which may be the infinity. At each point Pi
there must be an other common supporting line of D1andD2, sayti. SinceP is a
limit point, the sequence ti must have a limit straight linetP throughP which is a common supporting lines. SincetP is a limit of common supporting line, it must meet the bodies in a point of∂D1∩∂D2.
The pointP must also be a limit point of the intersection points tP ∩∂Ci. Through these intersection points there must be other common supporting lines tPi . Obviouslyt0is a limit straight line of these common supporting linestPi , hence T1 should coincide with T2, i.e. every common supporting line intersects the two bodies in∂D1∩∂D2.
Also every point of∂D1∩∂D2is an intersection point of a common supporting line, because otherwise a common supporting line bridging the point of∂D1∩∂D2
would exist, which intersects ∂D1 and∂D2 in different points.
Let ˜Pi∈C˜ibe a monotone subsequence ofPitending toP. Then the sequence C˜i is also monotone with respect to the inclusion relation, i.e. either ˜Ci⊂C˜i+1 for eachior ˜Ci+1⊂C˜ifor eachi. Let ˜Qi∈(t0∩∂C˜i) be different from ˜Pi. The sequence Q˜i is monotone, because ˜Ciis monotone. Moreover, ifI denotes the intersection of t0 and ∂D1∩∂D2, then the distance d( ˜Pi, I) is decreasing or increasing together withd( ˜Qi, I). LetQbe the limit of ˜Qi, which may be the infinity. As in the case of P we obtain a new common supporting line throughQ, saytQ. Qis a limit point of the intersection pointstQ∩∂C˜i, which implies a sequence of common supporting lines tQi through these points tending tot0. Obviously, the intersections oftQi and tPj with∂D1∩∂D2are on the same side oft0and their distances toItends to zero.
Therefore a common supporting line is a limit line of convergent sequences of common supporting lines from both clockwise and anti-clockwise directions. The normals of the common supporting lines then constitutes a closed subsetN of the unit circle S1, so that any point of N has convergent sequence in N from both clockwise and anti-clockwise directions. Let ωbe a unit vector. SinceN is closed, there is a pointα∈ N closest toω. Both arcs ofωαinS1contain sequences inN convergent to α, thereforeω=α, henceN =S1.
So,D1 andD2 have the same set of supporting lines, hence they coincide.
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A. K´ URUSA, Bolyai Institute, Aradi v´ertan´uk tere 1., H–6720 Szeged, Hungary; e-mail:
kurusa@math.u-szeged
