Published: 1953 Received: -Available on J-STAGE: August 29, 2006Accepted: -
Advance online publication: -
Revised: -
Correction information
Date of correction: August 29, 2006Reason for correction: -Correction: AUTHORDetails: Wrong : Tadahiko KUBOTA, Denzaburo HEMAH Right : Tadahiko KUBOTA, Denzaburo HEMMI
Date of correction: August 29, 2006Reason for correction: -Correction: CITATIONDetails: Right : (1) J. Pál: Ein Minimumproblem für Ovale. Math. Ann. 83 (1921). Cf. throughout this note as a reference the excellent report by T. Bonnesen u. W. Fenchel, Theorie der konvexen Körper. (Ergebn, d. Math. III 1.) Berlin (1934). (2) T. Kubota: Einige Ungleichheitsbeziehungen über Eilinien und Eiflächen. Sci. Rep. Tôhoku Univ. 12 (1923). (3) M. Yamanouchi: Notes on closed convex figures. Proc. Phys. -Math. Soc. Japan 14 (1932). (4) Lebesgue: Sur le probléme des isopérimètres et sur les domaines de largeur constante. Bull. Soc. Math. France C. R. (1914). (5) H. Lebesgue: Sur quelques questions de minimum, relatives aux courbes orbiformes, et sur leurs rapports avec le calcul des variations. J. Math. Pures Appl. 4 (1921). (6) W. Blaschke: Konvexe Bereiche gegebener konstanter Breite und kleinsten Inhalts. Math. Ann. 76 (1915). (7) T. Kubota: Eine Ungleichheit für Eilinien. Math. Z. 20 (1924). (8) In 1917 (?) Mr. K. Yanagihara proved the following theorem at the ordinary meeting of Mathematical Institute in Tohoku University: Let E0, E1, E2 be three congruent ovals which are situated in a homogetic positions and touch with each other. If we construct a ring of congruent ovals E3, E4, … around E0, so that Ei, Ei-1, E0 are located in homogetic positions and touch with each other, then E6 touches E1, that is, E7 is the very oval E1. If we denote the internal common tangent of E0, Ei by ti, then, by placing E1 in a suitable position, we can make the three pairs of opposite sides of the convex hexagon formed by t1, t2, …, t6 have respectively equal lengths. Cf. “Suni Zasso” (Miscellaneous notes in mathematics) 2, Tokyo-Butsuri-Gakko-Zasshi (Journal of Physics School) 26 (1917). The hexagon A1A2 … A6 in Fig. 3 is applicable to the hexagon in the above mentioned Yanagihara's theorem, and the parallelogram in Fig. 4 is a special case of it. (9) The determination of the arc to which Lemma 2 is applied, owes to Prof. Hombu. Our previous proof was as follows: If every side of P1P2 … P6 is not of length D, then by applying the method of Lemma 2, or by repetitions of it to every side of base hexagon, if necessary, we can make two sides of the base hexagon be of length D. But in doing so the area of the asymmetric oval becomes smaller. (10) We have considered the case when D and Δ or L and Δ are given, and not the case when D and L are so given that 3D≤L≤πD. For this case, by a property of the convex polygon in Favard's paper (Ann. Ecole Norm. 46, 1929), we get 2F≥D2{[π/2θ]sin2θ+sin(2θ[π/2θ])-√3} (≥DLcosθ-√3D2), where [ ] is Gauss's notation and θ is the root of [π/2θ]sinθ+cos(θ[π/2θ])=L/(2D) in the interval (0, π/6). The equality occurs when and only when the oval is an asymmetric curve whose base hexagon is a regular hexagon of sides D and whose breadth curve is a regular 6n-polygon inscribed in a circle with radius D. This inequality differs in some respects from Theorem 2 or 3, for the equality does not occur unless L=6nD sin (π/6n), n=1,2,3, …. (*) Read at the annual meeting of the Math. Soc. of Japan held in June 2, 1951. Added in proof by D. Hemmi. After we wrote this paper I received D. Ohmann's paper (Math. Z. 55, 1952. 347-352) and M. Sholander's (Trans. Amer. Math. Soc. 73, 1952. 139-173). The former does not touch the case 3D<L<πD and the latter gives the partial results and conjectures a property of the solution. The proofs of Sholander's conjecture and the inequality in Notes (10) may be seen in Bull. Yamagata Univ. (Natural science) 2 (1953) 157-170 and 3 (1953) 1-11.
Date of correction: August 29, 2006Reason for correction: -Correction: PDF FILEDetails: -