KORFF F-SIGNATURES OF HIRZEBRUCH SURFACES DAISUKE HIROSE AND TADAKAZU SAWADA 7 1 0 2 n Abstract. M.VonKorffintroducedthe F-signatureof a normal a J projective variety, and computed the F-signature of the product of two projective lines P1 P1. In this paper, we compute F- 8 × signatures of Hirzebruch surfaces. ] G A Introduction . h t a Let R be a d-dimensional Noetherian local ring of prime characteris- m tic p with perfect residue field. Let F : R R be the Frobenius map, → [ that is, F(x) = xp for all elements x of R. We denote by FeR the ∗ 1 R-module whose abelian group structure is inherited by R, and whose v R-module structure is given by the e-th iterated Frobenius map Fe. If 5 F R is finitely generated as an R-module, we say that R is F-finite. 0 ∗ 9 Let the R-module FeR be decomposed as FeR = R⊕ae M , where M ∗ ∗ ∼ e e 1 ⊕ does not have free summands. In [HL02], C. Huneke and G. Leuschke 0 . introduced the notion of the F-signature s(R) of R as 1 0 a 7 s(R) = lim e . 1 e→∞ ped : v Xi The existence of the limit lime→∞(ae/ped) is not trivial. In [Tuc12], K. Tucker proved that the limit exists (under the assumption that r a R is F-finite). The notion of the F-signature is generalized by M. Hashimoto and Y. Nakajima in [HN15], and A. Sannai in [San15]. As aglobalanalogue, M. VonKorffintroducedtheF-signatureofanormal projective variety, which we call the Korff F-signature, in [Kor12]. He computed the Korff F-signature of the product of two projective lines P1 P1. In this paper, we compute Korff F-signatures of Hirzebruch × surfaces. In Section1, we review generalities onKorff F-signatures. InSection 2, we compute Korff F-signatures of projective spaces and Hirzebruch surfaces. In Section 3 and 4, we gather together the figures and the source code of wxMaxima in the computation of F-signatures of Hirze- bruch surfaces. 1 2 Daisuke Hirose and Tadakazu Sawada 1. Preliminaries In this section, we review generalities on Korff F-signatures. See [Kor12] for details. Let k be a field of positive characteristic p. Let X be a normal projective variety over k, and let D be a Q-divisor on X. We denote by Sec(X,D) the section ring Γ(X, (nD)) of D. Suppose that n≥0 OX Sec(X,D) is a finitely generated k-algebra of dimension at least two. L If c N is sufficiently divisible, then there exists an ample line bundle ∈ L on X such that Sec(X,cD) is isomorphic to a normal section ring Sec(X,L) = Γ(X,L⊗n) of L. We assume that c is sufficiently n≥0 divisible, and define the Korff F-signature s(X,D) of X along D to be L c s(Sec(X,cD)). (The F-signature of a graded ring is defined in the · same way as the case of local rings.) Let N = Zn be a lattice, and let M be the dual lattice of N. We ∼ denote N R (resp. M R) by N (resp. M ). Let Σ be a fan in Z Z R R ⊗ ⊗ N . We denote by X the toric variety corresponding to Σ. R Σ LetΣbeacompletefaninN withtheprimitivegeneratorsv ,...,v . R 1 n Let D ,...,D be the torus-invariant prime Weil divisors correspond- 1 n n ing to v ,...,v , respectively. Let D = a D be a torus-invariant 1 n i=1 i i Weil divisor on X. The polytope associated to the divisor D is defined P to be the polytope u M u v a for all i , and denoted by R i i { ∈ | · ≥ − } P . We may assume that P contains the origin. Then we denote D D by M the lattice M (P R), where P R is the R-vector sub- D D D ∩ · · space spanned by P in M . We denote by N the dual lattice of M . D R D D We define I to be the subset of 1,...,n such that the hyperplane D { } u u v = a determines a facet of P for i I , and P to be i i D D Σ,D { | · − } ∈ the polytope (v,t) (M Z) 0 (v,t) (v ,a ) < 1 for all i I . D R i i D { ∈ × | ≤ · ∈ } Suppose that P is not full-dimensional. Let π : N N be the D D → naturalprojection map. Let c betherational number such that π(c v ) i i i is the primitive generator for its ray in (N ) . We define P′ to be D R Σ,D thepolytope (v,t) (M Z) 0 (v,t) c (v ,a ) < 1 for alli I . D R i i i D { ∈ × | ≤ · ∈ } Theorem 1.1 ([Kor12], Proposition 4.4.4, Corollary 4.4.7). Let X be a d-dimensional projective toric variety over k, and let Σ be the fan correspondingto X. Let v ,...,v be the primitive generators of Σ, and 1 n let D ,...,D be the torus-invariant prime Weil divisors corresponding 1 n n to v ,...,v , respectively. Let D = a D be a (non-zero) effective 1 n i=1 i i torus-invariant Weil divisor on X. Then: P (1) If P is full-dimensional, then s(X,D) = Vol(P ). D Σ,D (2) If P is not full-dimensional, then s(X,D) = Vol(P′ ). D Σ,D Korff F-signature of Hirzebruch surfaces 3 2. Examples M. Von Korff computed Korff F-signatures of some toric varieties including the product of two projective lines P1 P1 in [Kor12]. In × this section, we compute Korff F-signatures of projective spaces and Hirzebruch surfaces. Example 2.1 (Projective spaces Pn). Let N = Zn with standard basis n e ,...,e . Let e = e , and let Σ be the fan in N consisting 1 n 0 − i=1 i R of the cones generated by all proper subsets of e ,e ,...,e . We see 0 1 n that X is isomorphic Pto the n-dimensional pr{ojective space}Pn. Let Σ D be the torus-invariant prime Weil divisor corresponding to the ray i R e in N for 0 i n. + i R n ≤ ≤ Let D = a D be a (non-zero) effective torus-invariant Weil i=0 i i divisor. We see that P n x a , i=1 i ≤ 0 x a , PD = (x1,...,xn) MR(cid:12) P1 ≥ − 1 . ∈ (cid:12)    (cid:12) ···   (cid:12) xn an  (cid:12) ≥ − (cid:12)    (cid:12)  Since a = 0 for some i, P has an interior point, andthe equations i D (cid:12) n 6 x = a ,x = a ,...,x = a define the facets of P , respec- i=1 i 0 1 − 1 n − n D tively. Hence we have I = 0,1,...,n . In what follows, we compute D P { } the volume of the n-dimensional parallelepipeds n 0 x +a t < 1, ≤ − i=1 i 0 0 x +a t < 1, PΣ,D = (x1,...,xn,t) (MD Z)R(cid:12) ≤ 1P 1 . ∈ × (cid:12)    (cid:12) ···   (cid:12) 0 xn +ant < 1  (cid:12) ≤ (cid:12)   Let a = ni=0ai. The point of the i(cid:12)(cid:12)ntersection of the n+ 1 hyper- n planes defined by x +a t = 1,x +a t = 0,...,x +a t = 0 P − i=1 i 0 1 1 n n is ( a /a,..., a /a,1/a). This is given by the following elementary 1 n − − P transformations of matrices: 1 1 1 a 1 0 0 0 a 1 0 − − ··· − ··· 1 0 0 a 0 1 0 0 a 0 1 1  ···   ···  → ··· ···  0 0 1 an 0   0 0 1 an 0   ···   ···   0 0 0 a 1   1 0 0 0 a /a 1 ··· ··· − 1 0 0 0 a /a  ··· − 1   ··· . → → 0 0 1 0 an/a ··· ··· −  0 0 1 0 an/a   0 0 0 1 1/a   ··· −   ···      4 Daisuke Hirose and Tadakazu Sawada Bythesameargument, weseethatthepointoftheintersectionofthe n n+1 hyperplanes defined by x +a t = 0, x +a t = 0 (l = k), − i=1 i 0 l l 6 x +a t = 1is( a /a,...,1 a /a,..., a /a,1/a)foreach1 k n. k k 1 k n Let v = t( −a /a,..., a−/a,P1/a), a−nd ≤ ≤ 0 1 n − − v = t( a /a,...,1 a /a,..., a /a,1/a) k 1 k n − − − for 1 k n. Since P is spanned by v ,v ,...,v , we have Σ,D 0 1 n ≤ ≤ s(Pn,D) = Vol(P ) Σ,D = det(v ,v ,...,v ) 0 1 n | | a /a 1 a /a a /a a /a 1 1 1 1 − − − ··· − a /a a /a 1 a /a a /a (cid:12)  − 2 − 2 − 2 ··· − 2 (cid:12) = (cid:12)det (cid:12) (cid:12) ··· (cid:12) (cid:12)  a /a a /a a /a 1 a /a (cid:12) n n n n (cid:12)  − − − ··· − (cid:12) (cid:12)  1/a 1/a 1/a 1/a (cid:12) (cid:12)  ··· (cid:12) (cid:12)  (cid:12) (cid:12) a1 a a1 a1 a(cid:12)1 (cid:12) − − − ··· − (cid:12) a a a a a (cid:12)  − 2 − 2 − 2 ··· − 2 (cid:12) = 1/an+1(cid:12)det (cid:12) (cid:12) ··· (cid:12) (cid:12)  a a a a a (cid:12) n n n n (cid:12)  − − − ··· − (cid:12) (cid:12)  1 1 1 1 1 (cid:12) (cid:12)  ··· (cid:12) (cid:12)  (cid:12) (cid:12) 0 a 0 0 (cid:12) (cid:12) ··· (cid:12) 0 0 a 0 (cid:12)  ··· (cid:12) = 1/an+1(cid:12)det (cid:12) (cid:12) ··· (cid:12) (cid:12)  0 0 0 a (cid:12) (cid:12)  ··· (cid:12) (cid:12)  1 1 1 1 (cid:12) (cid:12)  ··· (cid:12) = 1/a = 1(cid:12)/(a+a + +a ). (cid:12) (cid:12) 0 1 n (cid:12) ··· (cid:12) (cid:12) The following computation is the main result. Example 2.2 (Hirzebruch surfaces). All figures of this example are in Section 3. Let r be a positive integer, and let N = Z2 with standard basis e and e . Let v = e ,v = e ,v = e , and v = e + re . 1 2 1 1 2 2 3 2 4 1 2 − − We consider a fan Σ in N = R2. The fan Σ consists the four two- r R r dimensional cones v ,v , v ,v , v ,v , and v ,v , and all their 1 2 1 3 2 4 3 4 h i h i h i h i faces. The Hirzebruch surface X = X is defined by the fan Σ (see, Σr r e.g., [CLS11], Example 3.1.16.). Let D be the torus-invariant prime i divisors of X corresponding to v . We have linear equivalents D D i 1 4 ∼ and D D + rD in the divisor class group of X. For an effective 3 2 4 ∼ Q-divisor D = 4 a D , we have i=1 i i D a D +(a +a )D +(ra +a )D P 1 1 2 3 2 3 4 4 ∼ (a +ra +a )D +(a +a )D . 1 3 4 1 2 3 2 ∼ Korff F-signature of Hirzebruch surfaces 5 Hence (D) = ((a + ra + a )D + (a + a )D ). Without loss of ∼ 1 3 4 1 2 3 2 O O generality, we only compute the F-signature s(X,D) of X with respect to a (non-zero) effective Q-divisor D = a D +a D . 1 1 2 2 The polytope P is to be D x a , y a PD = (x,y) ∈ MR y ≥0−, 1 x+≥ry− 20 . (cid:26) (cid:12) ≤ − ≥ (cid:27) (cid:12) By the figure 1, we see that(cid:12)P is a square if and only if ra < a . (cid:12) D 2 1 In that case, I = 1,2,3,4 . On the other hand, if ra a , then P D 2 1 D { } ≥ is a triangle. That implies I = 1,3,4 . D { } (1) Suppose that P is a square, i.e., ra < a . Then I = 1,2,3,4 D 2 1 D { } Therefore 0 x+a t < 1, 1 ≤ 0 y +a t < 1, PΣ,D = (x,y,t) ∈ (MD ×Z)R(cid:12)(cid:12) 0 ≤ y <21, .    (cid:12) ≤ −   (cid:12) 0 x+ry < 1  (cid:12) ≤ − (cid:12)   We define polygons Q, R(x), and S(x)(cid:12)to be  (cid:12) 0 y +a t < 1, Q = (y,t) R2 ≤ 2 ∈ 0 y < 1 (cid:26) (cid:12) ≤ − (cid:27) (cid:12) a t y < 1 a t, = (y,t) R2(cid:12)(cid:12) − 2 ≤ − 2 , ∈ 1 < y 0 (cid:26) (cid:12) − ≤ (cid:27) (cid:12) 0 x+a t < 1, R(x) = (y,t) R2(cid:12)(cid:12) ≤ 1 ∈ 0 x+ry < 1 (cid:26) (cid:12) ≤ − (cid:27) (cid:12) x/a t < x/a +1/a , = (y,t) R2(cid:12)(cid:12) − 1 ≤ − 1 1 , ∈ x/r y < x/r +1/r (cid:26) (cid:12) ≤ (cid:27) (cid:12) (cid:12) (cid:12) and S(x) = Q R(x).(See the figure 2.) ∩ Then Q is a parallelogram, R(x) is a rectangle, and P = (x,y,t) (M Z) x x x ,(y,t) S(x) , Σ,D D R min max { ∈ × | ≤ ≤ ∈ } wherex = min x R S(x) = andx = max x R S(x) = . min max { ∈ | 6 ∅} { ∈ | 6 ∅} Therefore xmax s(X,D) = Vol(P ) = Area(S(x))dx. Σ,D Zxmin We define some points and lines in the ty-plane appearing in our argument. Let A,B,C, and D be the vertices of the rectangle R(x). 6 Daisuke Hirose and Tadakazu Sawada In particular, 1 1 1 1 1 A = x, x ,B = x+ , x , −a r −a a r (cid:18) 1 (cid:19) (cid:18) 1 1 (cid:19) 1 1 1 1 1 1 1 C = x+ , x+ ,and D = x, x+ . −a a r r −a r r (cid:18) 1 1 (cid:19) (cid:18) 1 (cid:19) The point A moves along the line y = a t/r, denoted by l . The 1 1 − points B and D move along the line y = a t/r +1/r, denoted by l . 1 2 − The point C moves along the line y = a t/r+2/r, denoted by l . 1 3 − We denote the vetices of the parallelogram Q except the origin O by E,F, and G. That is, 1 2 1 E = , 1 ,F = , 1 ,and G = ,0 . a − a − a (cid:18) 2 (cid:19) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19) Let H and I be the intersection points of the t-axis with l and l , 2 3 respectively. We denote the intersection points of the line y = 1 with − l , l , and l by J, K, and L, respectively. That is, 1 2 3 1 2 H = ,0 , I = ,0 , a a (cid:18) 1 (cid:19) (cid:18) 1 (cid:19) r r +1 r +2 J = , 1 , K = , 1 , and L = , 1 . a − a − a − (cid:18) 1 (cid:19) (cid:18) 1 (cid:19) (cid:18) 1 (cid:19) For the point A, we denote the t-coordinate of A and y-coordinate of A by A and A , respectively. We use the same notation for all points t y in the ty-plane. For example, B = x/a +1/a and J = 1. t 1 1 y − − The shape of the polygon P depends on how the lines l ,l and Σ,D 1 2 l across the parallelogram Q. We divide into five subcases from (1-1) 3 through (1-5): (1-1) We assume that L E . That is, (r + 2)/a 1/a . This is t t 1 2 ≤ ≤ equivalent to (r + 2)a a . Then 2a a . This is equivalent to 2 1 2 1 ≤ ≤ 2/a 1/a . Hence I G . (See the figure 4.) 1 2 t t ≤ ≤ We denote the intersection points of the line OE with l and l by 2 3 T and U, respectively. Then 1 a 2 2a 2 2 T = , , and U = , . a ra −a ra a ra −a ra (cid:18) 1 − 2 1 − 2(cid:19) (cid:18) 1 − 2 1 − 2(cid:19) If C = U, then ( x/a +1/a ,x/r+1/r) = (2/(a ra ), 2a /(a 1 1 1 2 2 1 − − − − ra )). Hence x = (a +ra )/(a ra ). If D = T, then ( x/a ,x/r+ 2 1 2 1 2 1 − − − 1/r) = (1/(a ra ), a /(a ra )). Hence x = a /(a ra ). If 1 2 2 1 2 1 1 2 − − − − − B = T, then ( x/a + 1/a ,x/r) = (1/(a ra ), a /(a ra )). 1 1 1 2 2 1 2 − − − − Hence x = ra /(a ra ). If C = I and D = H, then ( x/a + 2 1 2 1 − − − Korff F-signature of Hirzebruch surfaces 7 1/a ,x/r + 1/r) = (2/a ,0). Hence x = 1. If A = O and B = H, 1 1 − then ( x/a ,x/r) = (0,0). Hence x = 0. 1 − Since 0 < ra < a , we have ra < a < a +ra and a ra > 0. 2 1 2 1 1 2 1 2 − Hence (a + ra )/(a ra ) < a /(a ra ) < ra /(a ra ). 1 2 1 2 1 1 2 2 1 2 − − − − − − Since a ra < a , we have a /(a ra ) < 1. If 2ra a , then 1 2 1 1 1 2 2 1 − − − − ≤ 1 ra /(a ra ). Therefore 2 1 2 − ≤ − − a +ra a ra 2 2 1 2 < < 1 < < 0. −a ra −a ra − −a ra 1 2 1 2 1 2 − − − On the other hand, if 2ra > a , then ra /(a ra ) < 1. Therefore 2 1 2 1 2 − − − a +ra a ra 2 2 1 2 < < < 1 < 0. −a ra −a ra −a ra − 1 2 1 2 1 2 − − − Forvariousvaluesofx, weconsider shapesofS(x)andareasofthem. If (a + ra )/(a ra ) < x < a /(a ra ), then S(x) is a 2 2 1 2 1 1 2 − − − − triangle as in the figure 5. Hence 1 1 1 a a 1 1 1 1 2 2 Area(S(x)) = x+ x x+ x 2 r r − a − a · −a a − −ra − ra (cid:26)(cid:18) (cid:19) (cid:18) 1 1(cid:19)(cid:27) (cid:26)(cid:18) 1 1(cid:19) (cid:18) 2 2(cid:19)(cid:27) 1 = (a ra )x+(a +ra ) 2. 2a2a r2 { 1 − 2 1 2 } 1 2 (1-1-1) Suppose that 2ra a . 2 1 ≤ If a /(a ra ) < x < 1, then S(x) is a trapezoid as in the figure 1 1 2 − − − 6. Hence 1 1 1 a 1 1 a a 1 2 2 2 Area(S(x)) = x+ x + x+ x+ 2 r r − a r r − a a · a (cid:26)(cid:18) 1 (cid:19) (cid:18) 1 1(cid:19)(cid:27) 1 1 = 2(a ra )x+2a +ra . 2a2r { 1 − 2 1 2} 1 We denote this area by V . 1 If 1 < x < ra /(a ra ), then S(x) is a tropezoid as in the 2 1 2 − − − figure 7. Hence 1 a a a 1 2 2 2 Area(S(x)) = x x+ 2 −a − a a · a (cid:18) 1 1 1(cid:19) 1 a 2 = (1 2x). 2a2 − 1 We denote this area by V . 2 8 Daisuke Hirose and Tadakazu Sawada If ra /(a ra ) < x < 0, then S(x) is a pentagon as in the figure 2 1 2 − − 8. Hence 1 1 1 a 1 1 1 2 Area(S(x)) = x x x x x −r · a − 2 a − r · −ra − −a 1 (cid:18) 1 (cid:19) (cid:26) 2 (cid:18) 1 (cid:19)(cid:27) 1 (a ra )2 = x 1 − 2 x2. −ra − 2r2a2a 1 1 2 We denote this area by V . 3 Therefore s(X,D) = Vol(P ) Σ,D = −a1−a1ra2 1 (a ra )x+(a +ra ) 2dx Z−aa12−+rraa22 2a21a2r2 { 1 − 2 1 2 } −1 − ra2 0 a1−ra2 + V dx+ V dx+ V dx 1 2 3 Z−a1−a1ra2 Z−1 Z−a1r−ar2a2 a 2 = . a (a a r) 1 1 2 − In the above calculation, we use wxMaxima (see Section 4 (%o1)). (1-1-2) Suppose that 2ra > a . If a /(a ra ) < x < ra /(a 2 1 1 1 2 2 1 − − − − ra ), then S(x) is a trapezoid as in the figure 9, which is the same 2 shape as that in the case where a /(a ra ) < x < 1 on (1-1-1). 1 1 2 − − − Then Area(S(x)) = V . 1 If ra /(a ra ) < x < 1, then S(x) is a pentagon as in the 2 1 2 − − − figure 10. Hence 1 1 1 1 1 a 1 1 1 2 Area(S(x)) = x+ x x x x r r − r · a − 2 a − r · −ra − −a (cid:18) (cid:19) 1 (cid:18) 1 (cid:19) (cid:26) 2 (cid:18) 1 (cid:19)(cid:27) 1 (a ra )2 = 1 − 2 x2. ra − 2a2a r2 1 1 2 We denote this area by V . 4 If 1 < x < 0, then S(x) is a pentagon as in the figure 11, which is − the same shape as that in the case where ra /(a ra ) < x < 0 on 2 1 2 − − (1-1-1). Then Area(S(x)) = V . 3 Korff F-signature of Hirzebruch surfaces 9 Therefore s(X,D) = Vol(P ) Σ,D = −a1−a1ra2 1 (a ra )x+(a +ra ) 2dx Z−aa12−+rraa22 2a21a2r2 { 1 − 2 1 2 } − ra2 −1 0 a1−ra2 + V dx+ V dx+ V dx 1 4 3 Z−a1−a1ra2 Z−a1r−ar2a2 Z−1 a 2 = , a (a a r) 1 1 2 − (see Section 4 (%o2)). (1-2) We assume that K < E < L . That is, (r + 1)/a < 1/a < t t t 1 2 (r + 2)/a . This is equivalent to (r + 1)a < a < (r + 2)a . Then 1 2 1 2 2a < a . We have I G . Since (r +1)a < a and a ra < a , 2 1 t t 2 1 2 2 1 ≤ ≤ we have (r +2)a < 2a . Therefore L < F . The point T denotes the 2 1 t t same in the case (1-1). (See the figure 12.) If C = L and D = K, then ( x/a + 1/a ,x/r + 1/r) = ((r + 1 1 − 2)/a , 1). Hence x = (r +1). If B = E , then (1/a )x+1/a = 1 t t 1 1 − − − 1/a . Hence x = (a a )/a . By the same argument in the case 2 1 2 2 − − (1-1), we have the following four values of x. If D = T, then x = a /(a ra ). If B = T, then x = ra /(a ra ). If C = I and 1 1 2 2 1 2 − − − − D = H, then x = 1. If A = O and B = H, then x = 0. − Since (r + 1)a < a , we have r(r + 1)a + a < (r + 1)a . Hence 2 1 2 1 1 a < (r + 1)(a ra ). Therefore (r + 1) < a /(a ra ). If 1 1 2 1 1 2 − − − − a /(a ra ) > (a a )/a , then a2 (r+2)a a +ra2 > 0. We have 1 1− 2 1− 2 2 1− 2 1 2 (r +2) √r2 +4 (r+2)+√r2 +4 − a < a < a . 2 1 2 2 2 By the assumption that (r +1)a < a < (r +2)a , we have a /(a 2 1 2 1 1 − ra ) > (a a )/a in case (1-2). Therefore a /(a ra ) < (a 2 1 2 2 1 1 2 1 − − − − − a )/a . Since 2a < a , we have a < a a . Hence (a a )/a < 2 2 2 1 2 1 2 1 2 2 − − − 1. The assumption that (r + 1)a < a gives the inequalities ra < 2 1 2 − a a and a < a ra . Those imply ra /(a ra ) < (a 1 2 2 1 2 2 1 2 1 − − − − a )/a . Therefore (a a )/a < ra /(a ra ). Finally, we com- 2 2 1 2 2 2 1 2 − − − − pare the inequality between 1 and ra /(a ra ). Suppose that 2 1 2 − − − 1 < ra /(a ra ). Then a > 2ra . With the assumption that 2 1 2 1 2 − − − a < (r+2)a , we have 2ra < (r+2)a . Hence r < 2. That is, r = 1. 1 2 2 2 Therefore if r = 1, then a a a ra 1 1 2 2 (r +1) < < − < 1 < < 0. − −a ra − a − −a ra 1 2 2 1 2 − − 10 Daisuke Hirose and Tadakazu Sawada By the same argument, if ra /(a ra ) < 1, then 1 < r. Therefore 2 1 2 − − − if r 2, then ≥ a a a ra 1 1 2 2 (r +1) < < − < < 1 < 0. − −a ra − a −a ra − 1 2 2 1 2 − − We remark that ra /(a ra ) = 1, because r is integer. 2 1 2 − − 6 − Forvariousvaluesofx, weconsider shapesofS(x)andareasofthem. If (r+1) < x < a /(a ra ), then S(x) is a trapezoid as in the 1 1 2 − − − figure 13. Hence 1 1 1 1 1 1 1 1 Area(S(x)) = x+ x + x+ 2 −a a − −ra − ra −a a − a (cid:26)(cid:18) 1 1 (cid:18) 2 2(cid:19)(cid:19) (cid:18) 1 1 2(cid:19)(cid:27) 1 1 x+ ( 1) × r r − − (cid:18) (cid:19) 1 = (a 2a r)x+2a r+a a r (x+r +1). 2a a r2 { 1 − 2 2 1 − 1 } 1 2 If a /(a ra ) < x < (a a )/a , then S(x) is a pentagon as 1 1 2 1 2 2 − − − − in the figure 14. Hence 1 1 1 1 1 1 a 2 Area(S(x)) = x+ ( 1) x x ( 1) r r − − · a − 2 a − −a · a − − (cid:26) (cid:27) 1 (cid:26) 2 (cid:18) 1 (cid:19)(cid:27) (cid:26) 1 (cid:27) 1 1 = (x+r +1) (a x+a )2. a r − 2a2a 2 1 1 1 2 We denote this area by V . 5 (1-2-1) Suppose that r = 1. If (a a )/a < x < 1, then S(x) is a 1 2 2 − − − trapezoidasinthefigure15, which isthesameshapeasthatinthecase where a /(a ra ) < x < 1 on (1-1-1). Hence Area(S(x)) = V . 1 1 2 1 − − − If 1 < x < ra /(a ra ), then S(x) is a trapezoid as in the 2 1 2 − − − figure 16, which is the same shape as that in the case where 1 < x < − ra /(a ra ) on (1-1-1). Hence Area(S(x)) = V . 2 1 2 2 − − If ra /(a ra ) < x < 0, then S(x) is a pentagon as in the figure 2 1 2 − − 17, which isthesameshape asthat inthecasewhere ra /(a ra ) < 2 1 2 − − x < 0 on (1-1-1). Hence Area(S(x)) = V . 3