How To Solve An Old Japanese Geometry Problem

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1 116 Taget circles i the ratio 2 : 1 Hiroshi Okumura ad Masayuki Wataabe I this article we cosider the followig old Japaese geometry problem (see Figure 1), whose statemet i [1, p. 39] is missig the coditio that two of the vertices are the opposite eds of a diameter. (The authors implicitly correct the omissio i the proof they provide o page 118.) We deote by O(r) the circle with cetre O, radius r. Problem [1, Example 3.2]. The squares 0 0 ad 0 have acommo vertex, ad the vertices ad 0, 0 ad lie o the circle O(R) whose diameter is 0 0, lyig withi the circle. The circle O 1 (r 1 ) touches ad ad also iterally touches O(R), ad O 2 (r 2 ) is the icircle of triagle. Show that r 1 = 2r 2. (1) Figure 1. Figure 2. We shall see that 0 0, beig a diameter, is merely a suciet coditio. Figure 2 shows that some additioal property ivolvig 0 ad 0 is required for deducig (1). I Theorem 1 we give a simple coditio that implies (1). Theorem 1. For ay triagle, if 0 is the reectio i of a poit, the the radius of oe of the circles iterally touchig the circumcircle of 0 ad also touchig ad is twice the size of the radius of the icircle of. opyright c 2001 aadia Mathematical Society

2 117 Proof. Let 0 itersect at P ad the circumcircle (say) of 0 agai at, ad let Q be the foot of the perpedicular from to 0 (see Figure 3). The \ 0 = \Q sice the right triagles 0 P ad Q share acommo agle \ 0. Moreover \ 0 = \. Hece, \Q = \. This implies that H ad are symmetric i, where H is the orthocetre of 0. Thus, is the orthocetre of, ad therefore, ad share a commo ie-poit circle (say), ad touches the icircle (say) of iterally by Feuerbach's Theorem. Therefore, the dilatatio of magicatio 2 with cetre carries ito the circumcircle of, which is, ad ito oe of the circles touchig, ad iterally. This implies that the last circle is twice the size of, ad the proof is complete. We metioed \oe of the circles" i the theorem, but if we itroduce orietatios of lies ad circles, the circle is determied uiquely. Let us assume that ad have couter-clockwise orietatios. The the circle of twice the size of (illustrated byadotted lie i Figure 3) is the oe which touches, ;! ad ;! so that the orietatios at the poitsoftagecy are the same (see the arrows i Figure 3), ad the poit of tagecy of the circle with is the reectio of i the poit of tagecy of ad. 0 H Q P Figure 3. The followig two properties follow immediately (see Figures 4 ad 5). They ca be foud without a proof i [1, p. 29].

3 118 orollary 1. If is a right triagle with right agle at, the the circle touchig the circumcircle of iterally ad also touchig ad is twice the size of the icircle of. orollary 2. If 0 is a isosceles triagle with 0 =, ad is a poit lyig o the lie 0, the oe of the circles touchig the circumcircle of 0 iterally ad also touchig ad is twice the size of the icircle of. 0 Figure 4. Figure 5. The last corollary holds sice the reectio i of lies o the circumcircle of 0. Though Figure 5 illustrates oly the case where lies o the segmet 0, ad this is the case stated i [1], the corollary does ot eed this coditio (see Figure 6b). oversely, for a triagle,let 0 be the reectio i of,ad let 0 be the itersectio of the lie ad the circumcircle of 0. The = 0 holds. Hece, with the two sides ad ad their itersectios with the circumcircle, we ca costruct two similar isosceles triagles (see Figures 6a ad 6b, where the ratio of the two smaller circles is 2:1). lies withi the circumcircle if ad oly if \ > 90, ad Figure 1ca be obtaied by lettig \ = Figure 6a. 0 Figure 6b. 0

4 119 The followig property with a proof usig trigoometric fuctios ca be foud i [2, p. 75] (see Figure7). orollary 3. If 0 E ad 0 0 E are tworegular petagos sharig the side E, the the circle iterally touchig the circumscribed circle of 0 0 ad the sides ad is twice the size of the icircle of E E 0 Figure " 2+1;[=2] Figure8. The corollary ca be geeralized yet further (see Figure 8). Theorem 2. If 1, 2,, 2+1 are vertices of a regular (2 +1){go lyig i this order, ad i is the reectio of i i the lie 1 2+1,ad is the circle passig through, +1, +1,, the oe of the circles touchig the lies 1 2, 1 2 ad iterally is twice the size of the icircle of the triagle made by the lies 2 2+1, ad 2+1;[=2] 2+1;[=2], where [x] is the largest iteger which does ot exceed x. Proof. 2+1 is the cetre of, ad the reectio of 2+1;[=2] i the lie through the cetres of the two regular polygos is 1+[=2]. Let us produce to P, where P lies o ad 1 lies o the segmet 2+1 P,ad let Q be the itersectio of 1 P ad 1+[=2] 1+[=2]. y simple calculatio we have 2+1 P = 2r cos 2(2 +1), = 2r si 2 +1, where r is the circumradius of Now let us suppose that is odd the [=2] = ( ; 1)=2 adwehave 1 Q = r si = r si 2 2 ; ; si Q = Q = r si si , ; ,

5 120 1 Q Q = 1 2r si 2 +1 = 2r si 2 ; 1 2(2 +1) = 1 2r cos 2(2 +1). Therefore, we get 1 Q Q = 2+1 P, ad this implies that Q is the mid-poit of 1 P. Similarly, we ca prove the same fact i the case of beig eve. Thus, the ratio of the two similar isosceles triagles made by the lies 1 2, 1 2, 1+[=2] 1+[=2], ad 1 2, 1 2, the taget of at P,is1:2ad the theorem is proved. P 1+(;1)=2 Q 1+(;1)= Figure 9 (whe is odd). The authors would like to thak the referee for suggestig helpful commets. Refereces 1. H. Fukagawa ad. Pedoe, Japaese Temple Geometry Problems, harles abbage Research etre aada, H. Fukagawa ad. Sokolowsky, Japaese mathematics - how may problems ca you solve? Volume 2, Morikita Shuppa, Tokyo, 1994 (i Japaese). Hiroshi Okumura Masayuki Wataabe Maebashi Istitute oftechology Kamisadori Maebashi Guma , Japa okumura@maebashi-it.ac.jp wataabe@maebashi-it.ac.jp

Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is

Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is 0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values