Generalisations of a Four-Square Theorem. Hiroshi Okumura and John F. Rigby. results on a wooden board and dedicated it to a shrine or a temple.

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1 20 enealisations of a ousquae Theoem Hioshi Okumua and John. Rigby In the 17 th {19 th centuies, Japanese people often wote thei mathematical esults on a wooden boad and dedicated it to a shine o a temple. Then the boad was hung unde the oof thee. Such a boad iscalled a sangaku. In this pape, we shall discuss genealisations of a sangaku poblem involving fou squaes, which we state as the following theoem (see igue 1.). [Ed. see RUX poblem 1496 [1989: 298; 1998: 56].] Theoem 1. [1, p.47]let E 0, 0, D, 0 D be squaes as in igue 1. Then D, E, D 0 ae collinea if and only if E 0 is half the size of igue 1.,!,! The angle, denoted by \, is,! the angle,! between and ; it is positive if the diection of otation fom to is anticlockwise, and negative othewise. The following lemma is the key of ou genealizations (see igue 2). Lemma 1. If the vetices of two tiangles and ED lie anticlockwise in this ode, and if they satisfy, then 2\DE =, \. = E ; \ + \ED = ; (1) D Poof. Let f be a dilative otation with cente mapping into D. If we denote the atio of magnication by, then = D = E, Reseach suppoted in pat by NS gant DMS opyight c 1999 anadian Mathematical Society

2 21 D E P igue 2. by the st pat of (1). Let P = f(). Then lies on the segment PE since \DP + \ED = by the second pat of (1), while PE = P+E = + E = = E, = = DP : Hence EP D is an isosceles tiangle with summit P. 2\DE =, \P D =, \. E, + E Theefoe, we get Reecting igue 2 in a line, we get a simila lemma with opposite oientations: If the vetices of two tiangles and ED lie clockwise in this ode, and if they satisfy, = E D ; \ + \ED =,; then 2\DE =,, \. Lemma 2. If \E = =2 in Lemma 1, then the angle bisecto of \ is paallel to DE. Poof. In this situation, the tiangle DP is obtained fom the tiangle by a dilative otation with cente and angle of otation =2. In the following examples, we shall agee that 0 E, 0, D and D 0 have anticlockwise oientations. omplete poofs will be given late. Example 1. (igue 3) The atio of the sides of the two squaes 0 E and 0 is n : n+1 (n is an intege). The two ectangles consist of n squaes as in the gue. Let E = na; then = (n +1)a. Hence (, )= = a= =(na)=(n) = E=D. nd we can show that D, E, D 0 ae collinea and that this line is paallel to the angle bisectos of \ and \ 0 0. Example 2. (igue 4) ou simila ectangles whee the atio of the sides of E 0 and 0 is 1:2. Then D, E, D 0 ae collinea, and this line is paallel to the angle bisectos of\ and \ 0 0 (this is a special case of Example 4 below).

3 igue 3. : = n :(n+1) (n =2). igue 4. : =1:2 Example 3. (igue 5) ou simila ectangles whee : E = s : a and : = s 2 : s We get =(s 2 +1)=s 2 =(s 2 +1) 0 =s, and, as well as 0, = = (s2 +1)=s 2, D=s s, = = =s D = E D ; (s2 +1)0, 0 = s0 = E 0 : sd 0 0 D 0 0 D 0 0 nd we can show that D, E, D 0 ae collinea, and that this line is paallel to the angle bisectos of\ and \ 0 0. Example 4. (igue 6) ou simila cyclic quadilateals. The cyclic quadilateals E 0, 0, D, 0 0 D 0 0 ae simila, with \E = =2 and =2. In the gue, we get (, )= = = = E=D. uthe, we can show that D, E, D 0 ae collinea, and this line is paallel to the angle bisectos of\ and \ igue 5. igue 6. Example 5. (igue 7) ou simila hamonic cyclic quadilateals: E 0 is acyclic quadilateal with the popety E 0 = E 0. The quadilateals 0 D, 0 0 D 0 0 ae simila to E 0, and

4 23 =2. In the gue,, = = E 0 E 0 = E D (=E) (= D) = D D = E ( 0 =E 0 ) D (= D) E D ; and we can show that D, E, D 0 ae collinea. In example 4, the quadilateals 0, D, 0 0 D 0 0 ae diectly simila, and E 0 is oppositely simila to the othe thee, wheeas, in example 5, it is 0 that is oppositely simila to the othe thee quadilateals. In all the above examples, the tiangles and ED satisfy the hypotheses of Lemma 1 (see igue 8). lso, the tiangles 0 0 and 0 ED 0 satisfy the hypotheses of the eected lemma. Thus we get \DE = =2, \=2, and \D 0 E 0 =,=2, \ 0 0 =2. Hence we get \DED 0 = \DE + \E 0 + \ 0 ED 0 = 1 2 (, \)+(,\0 )+ 1 2 (+\0 0 ) = 2, 1 2 (\ +2\0 + \ 0 0 ): ut since \ 0 = \ 0, we have 1 2 (\ +2\0 + \ 0 0 ) = 1 2 (\ + \0 + \ \ 0 ) 0 (mod ) : Hence \DED 0 0 (mod ). This implies that D, E, D 0 ae collinea. lso, examples 1, 2, 3, and 4 satisfy the hypotheses of Lemma 2, and DD 0 is paallel to the angle bisecto of \ in all fou examples igue 7. igue 8.

5 24 Example 6. (igues 9 and 9) ou egula 2n{gons. The smallest one is half the size of the one above. The collineaity descibed in the gue is obtained fom example 4. igue 9(n=5). igue 9(n=5). Refeence [1.] H. ukagawa and Pedoe, D., Japanese Temple eomety Poblems, hales abbage Reseach ente, Winnipeg, anada, Hioshi Okumua John. Rigby Maebashi Institute of Technology School of Mathematics 4601 Kamisadoi Maebashi Univesity of Wales adi unma 371, Japan adi 2 4, Wales, UK

On the Inradius of a Tangential Quadrilateral

On the Inradius of a Tangential Quadrilateral Foum Geometicoum Volume 10 (2010) 27 34. FORUM GEOM ISSN 1534-1178 On the Inadius of a Tangential Quadilateal Matin Josefsson bstact. We give a suvey of known fomulas fo the inadius of a tangential quadilateal,