aa r X i v : . [ m a t h . M G ] A ug Equisectional equivalence of triangles
Jun O’HaraOctober 12, 2018
Abstract
We study equivalence relation of the set of triangles generated by similarity and operationon a triangle to get a new one by joining division points of three edges with the same ratio.Using the moduli space of similarity classes of triangles introduced by Nakamura and Oguiso,we give characterization of equivalent triangles in terms of circles of Apollonius (or hyperbolicpencil of circles) and properties of special equivalent triangles. We also study rationalityproblem and constructibility problem.
Key words and phrases . Triangle, moduli space, circle of Apollonius, hyperbolic pencil of circles.2010
Mathematics Subject Classification : 51M04.
We study elementary geometric operations on triangles defined as follows. Let △ ABC be a triangle,and q be a real number. Let A ′ , B ′ , and C ′ be division points of the edges BC, CA , and AB by1 − q : q respectively, namely, A ′ = qB + (1 − q ) C, B ′ = qC + (1 − q ) A, C ′ = qA + (1 − q ) B. Let A ′′ ( B ′′ or C ′′ ) be the intersection of the lines AA ′ and BB ′ ( BB ′ and CC ′ or CC ′ and AA ′ respectively). Define equisection operators T q and S q , where S q can be defined when q = 1 /
2, by T q ( △ ABC ) = △ A ′ B ′ C ′ and S q ( △ ABC ) = △ A ′′ B ′′ C ′′ . The operators T q have been studied in articles such as [CD, D, NO, S], et. al. Figure 1: Operator T q Figure 2: Operator S q In this note we study the equivalence relation (denoted by ∼ ) of the set of triangles (denotedby T ) generated by similarity and { T q } q ∈ R , which we shall call equisectional equivalence . Theequivalence relation generated by similarity and { T q } q ∈ Q shall be called rational equisectionalequivalence and denoted by ∼ Q . We say two triangles ∆ and ∆ ′ are equisectionally equivalent (or1 ational equisectionally equivalent ) if ∆ ∼ ∆ ′ (or ∆ ∼ Q ∆ ′ respectively). We remark that we usethe term “similarity” as the equivalence under orientatipon preserving homothetic transformationin this article. We say two triangles are reversely similar if they are equivalent under orientationreversing homothetic transformation.Nakamura and Oguiso introduced the moduli space of similarity classes of triangles in [NO],which is a strong tool for the study of T q and S q . Using their results (explained in Section 3), wegive (projective) geometric characterization of equisectionally equivalent triangles. Namely, twotriangles with a common base, say BC , with the third vertices, say A and b A ′ , in the same side ofthe base are equisectionally equivalent if and only if A and b A ′ are on the same circle of Apolloniuswith foci being two vertices (denoted by D and D ′ ) of regular triangles with the common base BC . Therefore, each equisectional equivalence class with a given base BC corresponds to a circleof Apollonius with foci D and D ′ . It is an element of a hyperbolic pencil of circles defined by D and D ′ from a projective geometric viewpoint.We then study properties of triangles of the following three special types, right triangles, isosce-les triangles, and trianges with sides in arithmetic progression (which shall be denoted by SAP triangles), that appear in the same equisectional equivalence class. There are (at most) two simi-larity classes of such triangles for each type, which are reversely similar in the case of right or SAPtriangles, or the base angles of which satisfy tan θ · tan θ ′ = 3 in the case of isosceles triangles. Foreach type we explicitly give the ratio q such that T q maps one to the other in the same equisectionalequivalence class, which implies that a pair of triangles ∆ and ∆ ′ of one of the above special typeswith rational edges satisfies ∆ ∼ ∆ ′ if and only if ∆ ∼ Q ∆ ′ .We finally study compass and straightedge constructibility of q for a given pair of triangles. Definition 1
Let ∆ = △ ABC be a triangle. Let Π A be a half plane containing A with boundarythe line BC , and D and D be two points ( D ∈ Π A ) such that △ DBC and △ DBC are regulartriangles. Define α (∆) (0 ≤ α (∆) <
1) and β (∆) by α (∆) = | AD | (cid:12)(cid:12) AD (cid:12)(cid:12) , β (∆) = |△ ABC | (cid:12)(cid:12) AD (cid:12)(cid:12) , where |△ ABC | means the area of △ ABC .We remark that both α (∆) and β (∆) are independent of the choice of the base of the triangle.A locus of points X such that | XD | / (cid:12)(cid:12) XD (cid:12)(cid:12) is a given positive constant is a circle, called a circle ofApollonius with foci D and D . Put C A = { X : | XD | / (cid:12)(cid:12) XD (cid:12)(cid:12) = α (∆) } . Note that C A = { A } when ∆ is a regular triangle. The quantity α takes the value 0 if and only if∆ is a regular triangle, and approaches 1 as ∆ becomes thinner and thinner. In that sense, it canbe considered as measuring how far a triangle is from a regular triangle. Theorem 2
Given two triangles ∆ = △ ABC and ∆ ′ = △ A ′ BC . Let b A ′ be a point in Π A suchthat △ b A ′ BC is similar to △ A ′ B ′ C ′ . Then the following conditions are equivalent: (1) ∆ is equisectionally equivalent to ∆ ′ . (2) α (∆) = α (∆ ′ ) , in other words, b A ′ is on the circle of Apollonius with foci D and D that passesthrough A . (3) β (∆) = β (∆ ′ ) . Let A and A be points in Π A such that △ BCA and △ CA B are similar to △ ABC insuch a way that each vertex of △ BCA or △ CA B corresponds to a vertex of △ ABC in thesame sequential order through the similarity (Figure 3). Then b A ′ is on the circle that passesthrough A, A , and A . When ∆ is a regular triangle we agree that the circle through A, A ,and A consists of a single point. Figure 3: Three positions of similar tri-angles with a common base in the samehalf-plane. Figure 4: A hyperbolic pencil consisting of circlesof Apollonius with foci D and D The set of circles of Apollonius with foci D and D is called a hyperbolic pencil of circles definedby D and D (or a Poncelet pencil with limit points (or
Poncelet points ) D and D ). It consists ofcircles that are orthogonal to any circle passing through D and D (Figure 5 left). A set of circlesthrough D and D is called an elliptic pencil (or a pencil of circles with base points ).Let S be the set of similarity classes of triangles and [∆] denote the similarity class of a triangle∆. Nakamura and Oguiso’s result implies that the sets of similarity classes of equisectionallyequivalent triangles form a codimension 1 foliation of S with a unique singularity 0 that correspondsto regular triangles. We study the intersection of each leaf and another codimension 1 subspaceof S which is the set of similarity classes of one of the following three special triangles, isoscelestriangles, right triangles, and SAP triangles (i.e., triangles with sides in arithmetic progression)(the reader is referred to [MOS] for the properties of SAP triangles). Corollary 3
Two right triangles are equisectionally equivalent if and only if they are either similaror reversely similar. Any triangle ∆ is equisectionally equivalent to a right triangle if and only if α (∆) ≥ ( √ − / . Corollary 4
Two SAP triangles are equisectionally equivalent if and only if they are either similaror reversely similar. Any trianle is equisectionally equivalent to such a triangle.
Proposition 5
Two isosceles triangles ∆ and ∆ ′ are equisectionally equivalent if and only if eitherthey are similar or the base angles satisfy tan ∠ θ · tan ∠ θ ′ = 3 . When [∆] = [∆ ′ ] , [ T q (∆)] = [∆ ′ ] ifand only if q = 1 / or / . Any triangle is equisectionally equivalent to an isosceles triangle. Lemma 6
Let ∆ ab denote a triangle with side lengths , a, b in the anti-clockwise order. Then [ T q (∆ ab )] = [∆ ba ] if and only if q = b − a + b − , − a b + 1 − a , or a − b a − b . Theorem 7
Suppose ∆ and ∆ ′ are isosceles (or right or SAP) triangles such that all the sides arerational numbers. If T q (∆) = ∆ ′ then q is also rational, namely, ∆ ∼ ∆ ′ if and only if ∆ ∼ Q ∆ ′ . Nakamura and Oguiso gave a bijection between S and the open unit disc D in C , and showed thatthe set of equisection operators { T q } q ∈ R acts on D as rotations.Let us first introduce the result of Nakamura-Oguiso [NO]. We work in the complex plane C .Let H be the upper half plane { z ∈ C : ℑ m z > } , B and D be the open unit discs in variables Z and w respectively. Put ρ = exp( πi/ λ : H → H , µ : B → B , f : H → B , and g : B → D by λ ( z ) = 11 − z , µ ( Z ) = ρ Z, f ( z ) = ρ − ρzρ + z = − ρ z − ρz − ρ − , g ( Z ) = Z . (1)Then we have the following commutative diagram: H f −−−−→ B g −−−−→ D λ y y µ (cid:13)(cid:13)(cid:13) H f −−−−→ B g −−−−→ D Let us fix the base of a triangle to be [01] so that a triangle can be identified by the vertex z ∈ H . Suppose ∆ = △ ABC is similar to the triangle △ z
01. Since the three choices of the base,
BC, CA , or AB corresponds to z, λ ( z ), or λ ( z ), there is a bijection ϕ : S → D ([NO]) given by ϕ ([ △ z g ◦ f ( z ) = (cid:18) z − ρz − ρ − (cid:19) . (2)Let us call D the Nakamura-Oguiso moduli space of the similarity classes of triangles.From the construction of the moduli space and the property of linear fractional transformations,it follows that the set of isosceles triangles is expressed by a real axis in D (explained in Section4), and reversely similar triangles by complex conjugate numbers, as was pointed out in [NO].The equisection operators and the equisectional equivalence relation on S , denoted by the samesymbols, can be naturally induced from those on T . We shall express the operators on D given by ϕ ◦ T q ◦ ϕ − and ϕ ◦ S q ◦ ϕ − simply by T q and S q respectively. Since our T q and S q are equal to T ,q and T − q,q in [NO] respectively, Theorem 1 of [NO] implies Theorem 8 ([NO])
The operator T q acts on D as a rotation by angle − q ) ρ − (1 − q )) = 6 arg (cid:16) − − q ) √ i (cid:17) , (3) and S q by angle ρ − q ) . Theorem 8 implies that ∼ is in fact an equivalence relation, that ∆ ∼ ∆ ′ if and only if there isa real numbers q such that ∆ ′ is similar to T q (∆), and that the equivalence relation generated bysimilarity and { S q } q ∈ R \{ / } is identical with the equisectional equivalence.The following corollaries can be obtained by simple computation.4 orollary 9 ([NO])(1) T q = Id D if and only if q = 0 , / , or . (2) T q = Id D and T q = Id D if and only if q = 1 / or / . Corollary 10 T q ′ = T q if and only if q ′ = q, q ′ = 2 q − q − (cid:18) q = 13 (cid:19) , or q ′ = q − q − (cid:18) q = 23 (cid:19) . Corollary 11 ([NO]) T q ′ = T q − if and only if q ′ = 1 − q, q ′ = 2 q − q − (cid:18) q = 23 (cid:19) , or q ′ = q q − (cid:18) q = 13 (cid:19) . Remark 12
The six functions of q which appear in the right hand sides in Corollaries 10 and11 form a non-abelian group with the operation of composition, which is isomorphic to the fullpermutation group of three elements. Corollary 13
Given real numbers q and q ′ . T q ◦ T q ′ = T q ′ ◦ T q = T q ′′ if and only if q ′′ = 3 qq ′ − q + q ′ ) + 16 qq ′ − q + q ′ ) + 1 , q ′′ = − q + q ′ − qq ′ − q + q ′ ) + 2 , or q ′′ = 3 qq ′ − ( q + q ′ )3 qq ′ − . Proof.
The formula (3) implies that T q ◦ T q ′ is the rotation by angle6 arg [ ( − q + q ′ )) ρ + ( q + q ′ ) − qq ′ ] . Since 6 arg( Aρ + B ) ≡ (cid:20)(cid:18) − A + BA + 2 B (cid:19) ρ − (cid:18) − A + BA + 2 B (cid:19)(cid:21) (modulo 2 π ) , substitution A = ( q + q ′ ) − B = − qq ′ + ( q + q ′ ) implies that if we put q ′′ = A + BA + 2 B = 3 qq ′ − q + q ′ ) + 16 qq ′ − q + q ′ ) + 1 , then T q ◦ T q ′ = T q ′ ◦ T q = T q ′′ . The other two values for q ′′ can be obtained by Corollary 10. ✷ Corollaries 9 (1), 11, and 13 show that the rational equisectional equivalence ∼ Q is in fact anequivalence relation. In what follows, we restrict ourselves to the case of non-regular triangles.
Proof of Theorem 2.
Theorem 8 shows that a set of equisectionally equivalent trianglescorresponds to a circle with center 0 in D . Therefore, the formula (2) implies[ △ z ∼ [ △ z ′ ⇐⇒ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) z − ρz − ρ − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) z ′ − ρz ′ − ρ − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⇐⇒ | z − ρ || z − ρ − | = | z ′ − ρ || z ′ − ρ − | , which proves the equivalence between (1) and (2).Since △ ABC , △ A BC , and △ A BC are similar, the above argument implies that a circle ofApollonius with foci ρ and ρ − that passes through A also passes through A and A , which provesthe equivalence of (1) and (4). 5he equivalence between (2) and (3) follows from computation. Since | z − ρ | | z − ρ − | = 1 − √ ℑ m z | z − ρ − | , (4) | z − ρ || z − ρ − | = | z ′ − ρ || z ′ − ρ − | if and only if ℑ m z | z − ρ − | = ℑ m z ′ | z ′ − ρ − | , which, translated to a scale-invariant statement, is equivalent to (3). We remark that the formula(4) implies β (∆) = 1 − α (∆) √ (cid:3) Let us give projective geometric explanation of the equivalence between (1) and (2). Theset of lines through 0, which are considered as circles through 0 and ∞ , is an elliptic pencil ofcircles defined by 0 and ∞ , and the set of concentric circles with center 0 is a hyperbolic pencil ofcircles defined by 0 and ∞ . They are mutually orthogonal. A linear fractional transformation isa conformal map (i.e., it preserves the angles) that maps circles (which include lines that can beconsidered as circles through ∞ ) to circles, and hence it maps an elliptic pencil (or a hyperbolicpencil) of circles defined by a pair of points to an elliptic pencil (or a hyperbolic pencil) definedby a pair of corresponding points.Since f − is a linear fractional transformation which maps 0 and ∞ to ρ and ρ − , it mapsthe set of lines through 0 to an elliptic pencil consisting of the circles through ρ and ρ − , and theset of concentric circles with center 0 to a hyperbolic pencil defined by ρ and ρ − , which consistsof circles of Apollonius with foci ρ and ρ − (Figure 5). Now it follows from the construction ofNakamura-Oguiso moduli space that the vertices z of equisectionally equivalent triangles △ z ρ and ρ − , which proves the equivalence between (1) and (2).Figure 5: A pair of a hyperbolic pencil (left, black) and an elliptic pencil (left, blue) of circlesdefined by ρ and ρ − , and a pair of a hyperbolic pencil (right, black) and an elliptic pencil (right,blue) of circles defined by 0 and ∞ . The former is mapped to the latter by a linear fractionaltransformation that maps ρ and ρ − to 0 and ∞ respectively. In each case, a circle in a hyperbolicpencil is orthogonal to a circle in an elliptic pencil.In what follows, it sometimes makes things easier to work in a a fundamental domain of λ ,Ω = { z ∈ H : | z | < , | z − | ≤ } ∪ { ρ } , (5)6hich corresponds to studying triangles with the longest edge (one of the longest edges) beingfixed to [0 , − ,
1) in the real axis in the Nakamura-Oguiso moduli space D . Definition 14
Let Γ( ξ, η, ζ ) be a circle through three points ξ, η , and ζ . When one of the threepoints is ∞ , Γ( ξ, η, ζ ) is a line. We assume that Γ( ξ, η, ζ ) is oriented by the cyclic order of ξ, η ,and ζ .In Ω, a vertex of an isosceles triangle lies either on the line ℜ e z = 1 / | z − | = 1.Since f is a linear fractional transformation with f : ρ , , ρ , ρ −
7→ ∞ , and ∞ 7→ − ρ, it maps a circle | z − | = 1 (Γ( ρ, , ρ − )) to the real axis (Γ(0 , , ∞ )), another circle | z | = 1(Γ( ρ, , ρ − )) to a line joining 0 and ρ (Γ(0 , ρ , ∞ )), a line ℜ e z = 1 / ρ, ρ − , ∞ )) to a linesegment joining 0 and ρ (Γ(0 , ∞ , − ρ )), the real axis (Γ(0 , , ∞ )) to the unit circle (Γ(1 , ρ , − ρ )),and Ω to one third of the open unit disc { Z ∈ B : | Z | < , ≤ arg Z < π/ } . Therefore, theimages of ϕ of Ω ∩ { z : | z − | = 1 } and Ω ∩ { z : ℜ e z = 1 / } are [0 ,
1) and ( − ,
0] respectively.
Proof of Proposition 5.
We work in the Nakamura-Oguis moduli space D . The isoscelestriangles correspond to the real axis in D . Each circle with center 0, which corresponds to a setof equisectionally equivalent non-regular triangles, intersects the real axis in two points. It provesthe third statement. It also proves that if T q satisfies T q (∆) = ∆ ′ with two isosceles triangles ∆and ∆ ′ then either [∆] = [∆ ′ ] or T q is a rotation by π on D .Theorem 8 shows that T q is a rotation by angle (2 n + 1) π for some n ∈ Z if and only if q = 1 / /
3, which proves the second statement.The equation tan ∠ θ · tan ∠ θ ′ = 3 follows directly from the fact that [∆ ′ ] = T / ([∆]) as isillustrated in Figure 6. (cid:3) Figure 6: tan θ − · tan θ = 3 Proof of Corollary 3.
In the fundamental domain Ω, a vertex z of a right triangle △ z
01 lieson the upper half hemi-circle | z − / | = 1 /
2, which intersects any circle of Apollonius in at mosttwo points, which are symmetric in the line { x = 1 / } . The extremal value of α is given by a rightisosceles triangle. (cid:3) Proof of Corollary 4.
In the fundamental domain Ω, a non-regular triangle such that the ratioof the edge lengths is 1 : 1 − d : 1 − d corresponds to P ± ( d ) = ± (2 d − d )2 , √ p (1 − d )(1 − d )(1 − d ) ! (cid:18) < d < (cid:19) . Let Γ ± be a curve Γ ± = { P ± ( d ) } 0) on the real axis, it mustintersect C A . Secondly, if we put Ω ∓ = Ω ∪ {ℜ e z ≷ / } , then Γ + ⊂ Ω − , and hence Γ + ∩ C A =Γ + ∩ ( C A ∩ Ω − ). Let P + ( d ) = ( x ( d ) , y ( d )), then, as x ( d ) is an increasing function and y ( d ) adecreasing function of d , we have dy/dx < + , whereas C A ∩ Ω − can be expressed as a graphof an increasing function. Therefore Γ + intersects C A in at most one point.We remark that the statement can also be proved by computation showing that | ϕ ( P ± ( d )) | isa monotonely increasing function of d with | ϕ ( P ± (0)) | = 0 and lim d → / | ϕ ( P ± ( d )) | = 1. (cid:3) Proof of Lemma 6. Suppose ∆ ba is expressed by a complex number z = z ba = x + yi ( x, y ∈ R )in the fundamental domain Ω. Since | z | = a and | z − | = b , x and y satisfy x = 12 ( a − b + 1) ,y = 14 (cid:0) b − ( a − (cid:1) (cid:0) ( a + 1) − b (cid:1) . (6)Let w = f ( z ) and θ = arg w . Thentan θ = ℑ m (cid:16) − ρ z − ρz − ρ − (cid:17) ℜ e (cid:16) − ρ z − ρz − ρ − (cid:17) = √ x + y − xx + y + 2 x − . Substitution of (6) gives tan θ = √ b − a − b − . Since ∆ ab is a mirror image of ∆ ba , we have ϕ ( z ab ) = ϕ ( z ba ), and since ϕ ( z ) = ( f ( z )) , we havearg ϕ ( z ab ) = arg ϕ ( z ba ) + 6 θ modulo 2 π. On the other hand, T q acts on D as a rotation by 6 τ q , where τ q = arg (cid:0) − − q ) √ i (cid:1) .Suppose − (1 − q ) √ √ b − a − b − . (7)It means tan τ q = tan θ , which implies τ q ≡ θ modulo π and hence 6 τ q ≡ θ modulo 2 π , whichimplies T q ( ϕ ( z ab )) = ϕ ( z ba ), i.e., T q ([∆ ab ]) = [∆ ba ].The equation (7) gives q = 1 − a b + 1 − a . The other two values follow from the above by Corollary 10. (cid:3) Proof of Theorem 7. When [∆] = [∆ ′ ] the conclusion follows from Corollary 9. When[∆] = [∆ ′ ] the conclusion follows from Proposition 5 for isosceles triangles and from Corollaries 3,4 and Lemma 6 for right and SAP triangles. (cid:3) Proposition 15 Given two triangles ∆ = △ ABC and ∆ ′ = △ A ′ B ′ C ′ . Whether ∆ is equisection-ally equivalent to ∆ ′ or not can be determined using a straightedge and compass, and if the answeris affirmative, a real number q such that T q ([∆]) = [∆ ′ ] is compass-and-straightedge constructible. Proof. One can construct the following with a compass and straightedge in the following order: The image of Ω + (or Ω − ) of ϕ is the intersection of D and the upper half plane (or the lower half planerespectively). 81) Two points D and D ( D ∈ Π A ) such that both △ DBC and DBC are regular triangles.(2) A vertex b A ′ in Π A such that [ △ b A ′ BC ] = [ △ A ′ B ′ C ′ ].(3) An oriented circle Γ( D, A, D ) and another oriented circle Γ( D, b A ′ , D ) (see Definition 14).(4) A circle of Apollonius C A with foci D and D that passes through A , since the center is theintersection of a bisector of the edge BC and a tangent line of Γ( D, D, A ) at point A .(5) A decision whether ∆ ∼ ∆ ′ or not, since the answer is affirmative if and only if b A ′ ∈ C A .Assume ∆ ∼ ∆ ′ in what follows.(6) The signed angle θ ( − π/ < θ < π/ 3) at point D from the oriented circle Γ( D, A, D ) toΓ( D, b A ′ , D ).(7) At least one n ∈ Z such that θ nπ ∈ (cid:18) π , π (cid:19) modulo 2 π. (8) The value q which is given by q = 12 (cid:20) √ (cid:18) θ nπ (cid:19)(cid:21) , (8)where n is given by the proceeding step.We explain the process (6),(7),(8). By working in the fundamental domain Ω, we may assumethat B = 0 , C = 1 , A = z , and b A ′ = z ′ . By formulae (2) and (3), we want a real number q suchthat 6 arg (cid:16) − − q ) √ i (cid:17) = 3 (cid:18) arg (cid:18) z ′ − ρz ′ − ρ − (cid:19) − arg (cid:18) z − ρz − ρ − (cid:19)(cid:19) modulo 2 π. (9)Since the right hand side divided by 3 is equal to the singed angle ˆ θ at 0 from the oriented lineΓ(0 , ( z − ρ ) / ( z − ρ − ) , ∞ ) to the oriented line Γ(0 , ( z ′ − ρ ) / ( z ′ − ρ − ) , ∞ ), and a linear fractionaltransformation f − maps 0 , ( z − ρ ) / ( z − ρ − ) , ( z ′ − ρ ) / ( z ′ − ρ − ), and ∞ to ρ, z, z ′ , and ρ − respectively, the signed angle ˆ θ is equal to the signed angle θ at point ρ from the oriented circleΓ( ρ, z, ρ − ) to the oriented circle Γ( ρ, z ′ , ρ − ).If q is given by (8), then − − q ) √ i = − − i tan (cid:18) θ nπ (cid:19) , and therefore, 6 arg (cid:0) − − q ) √ i (cid:1) ≡ θ (modulo 2 π ), which means (9). ✷ References [CD] G. Chang and P. J. Davis, Iterative Processes in Elementary Geometry , Amer. Math.Monthly 90 (1983) 421–431.[D] P. J. Davis, Cyclic transformations of polygons and the generalized inverse , Can. J. Math.29 (1977), 756–770.[MOS] K. Mikami, J. O’Hara and K. Sugawara, Triangles with sides in arithmetic progression , toappear in Elem. Math. 9NO] H. Nakamura and K. Oguiso, Elementary moduli space of triangles and iterative processes ,J. Math. Sci. Univ. Tokyo 10 (2003), 209–224.[S] D. B. Shapiro,