JAPANESE THEOREM: A LITTLE KNOWN THEOREM WITH MANY PROOFS – PART I

Mangho Ahuja, Wataru Uegaki, and Kayo Matsushita

[Note: (A), (U), (M) refer to the authors’ names. The pronoun I always means (A).]

\underline{Japanese Theorem}. Triangulate a cyclic polygon by lines drawn from any vertex. The sum of the radii of the incircles of the triangles is independent of the vertex chosen.

1. Background. I (A) found this theorem in a 1993 article [5], where the author Nick Mackinnon wrote “ … I have used the above theorem as a starter for course work, not expecting a proof of the theorem (I can’t even prove it myself) … ” It was the author’s remark in parenthesis that intrigued me. In fall 1995 I assigned this theorem as a problem to Cathy, a Masters Degree student. By spring 1996 both Cathy and I had a proof of the theorem [3]. But one question remained: Why is it called the Japanese Theorem? My long search for the answer ended when I received a 15–page fax in English, French, and Japanese from Professor Yoshida of Kyoto University.

The theorem (Quadrilateral case) had originated in China [6]. So, when it came to Japan around 1900, it was known as the “Chinese Theorem” [4]. Later, when Y. Mikami generalized it from a quadrilateral to a polygon, the name remained the “Chinese Theorem”. So, who coined the term “The Japanese Theorem”? This theorem, without a name, appeared in a 1906 article entitled, “Japanese Mathematics” [2]. We believe this led the later authors to call it the “Japanese Theorem”.

This theorem is displayed on a wooden tablet in a Shinto shrine. The hanging of such tablets showing mathematical theorems was a common custom in Edo era in Japan [1]. These tablets, called Sangaku in Japanese, can be seen all over Japan. The Sangaku of our theorem was once hanging in the Tsuruoka–San’nosha shrine in the Ushu area (at present Yamagata and Akita prefectures of Japan), but it has since disappeared.

To learn more about the history of the theorem, in summer 1999 I traveled to Japan, where (M) and I searched the libraries of Kyoto University for references. At the same time Professor (U) of Mie University was examining and analyzing every available document relevant to the theorem. His findings are published in a paper (in Japanese) in the Journal of Mie University [7]. The authors plan to write a detailed history of the theorem in another paper, leaving the present paper to focus on the geometry only.

2. Plan. In part I, we state a few elementary results, E1, E2, …, E5, and then derive results G1, G2, …, G8 in Geometry, of which a few are well–known. We end part I with the oldest proof of the theorem. In part II we show a variety of proofs – five different proofs by Japanese mathematicians. We end our paper with a discussion of the generalized (polygonal) case of the theorem and the conclusion.

\underline{Elementary Results}. In any triangle ABC, the following are true:

(E1) sin A + sin B + sin C = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2} .

(E2) sin A + sin B − sin C = 4 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2} .

(E3) cos A + cos B + cos C = 1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} .

(E4) If chords AB and CD of a circle intersect at P, then P A · P B = P C · P D.

(E5) \underline{Ptolemy’s Theorem}. If a quadrilateral ABCD is inscribed in a circle, then AB · CD + BC · AD = AC · BD.

3. Results from Plane Geometry. We will prove results G1 to G8 (some of these are known theorems) needed for the proofs that follow. For a triangle ABC, let O, R, and I, r denote the center and radius of the circumcircle and the incircle, respectively.

(G1) r = 4R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} .

\underline{Proof}. Now

a = BC = r cot \frac{B}{2} + r cot \frac{C}{2} = r[\frac{cos \frac{B}{2} sin \frac{C}{2} + cos \frac{C}{2} sin \frac{B}{2}}{sin \frac{B}{2} sin \frac{C}{2}} ]

= r[\frac{sin (\frac{B}{2} + \frac{C}{2} )}{sin \frac{B}{2} sin \frac{C}{2}}] = r[\begin{matrix}cos & \frac{A}{2} \\ sin & \frac{B}{2} & sin & \frac{C}{2} \end{matrix} ].

Thus,

r = a[\begin{matrix}sin & \frac{B}{2} & sin & \frac{C}{2} \\ cos & \frac{A}{2} \end{matrix} ].

But

a = 2R sin A = 4R sin \frac{A}{2} cos \frac{A}{2},

hence,

r = 4R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} .

(G2) If AI produced meets the circumcircle in F, then F B, F C, F I are equal. In other words the points B, C, and I lie on a circle with center F.

\underline{Proof}. Clearly, ^{}B̸IF = ^{}I̸BA+^{}B̸AI = B/2+A/2. Also, ^{}I̸BF = ^{}I̸BC + C̸BF = ^{}I̸BC + ^{}C̸AF = B/2 + A/2. Thus, ^{}B̸IF = ^{}I̸BF and F I = F B. Similarly, F I = F C. So the points B, I, and C lie on a circle with center F.

(G3) R^{2} − 2Rr = OI^{2} (This result is also known as Chapple’s Theorem.)

\underline{Proof}. Extend AI till it meets the circumcircle in F. We draw two diameters F OG and KIOL through O. Let IJ be the perpendicular from I to side AB, then IJ = r. Since ^{}B̸AF = ^{}B̸GF, the two right triangles AJI and GBF are similar. Hence, r/AI = BF/2R, or 2Rr = AI ·BF = AI ·IF, which by (E4) equals LI · IK = (R + OI)(R − OI) = R^{2} − OI^{2}. Hence, R^{2} − 2Rr = OI^{2}.

(G4) (Carnot’s Theorem) Let a_{1}, b_{1}, and c_{1} denote the lengths of perpendiculars from O to sides BC, CA, and AB, respectively. Then R + r = b_{1} + c_{1} + a_{1}, if O lies within the triangle ABC, and R + r = b_{1} + c_{1} − a_{1} if say OA intersects BC.

\underline{Proof}. Since ^{}B̸OC = 2^{}A̸, the length a_{1} is either R cos A, or R cos(π − A) = −R cos A, depending upon the position of O. Thus, b_{1} +c_{1} ±a_{1} = R[cos B +cos C + cos A] which equals R[1 + 4 sin A/2 sin B/2 sin C/2] by (E3), and equals R + r by (G1). Thus, R + r = b_{1} + c_{1} ± a_{1}.

(G5) Let ABCD be a quadrilateral inscribed in a circle. If E, F, G, and H denote the midpoints of the arcs AB, BC, CD, and DA, respectively, then the lines EG and F H are perpendicular.

\underline{Proof}. If we suppose that the arcs AB, BC, CD, and DA make angles 2α, 2β, 2γ, and 2δ at the center of the circle, then the sum 2α + 2β + 2γ + 2δ = 2π. Now the arc EF makes angle α + β at the center, so ^{}E̸HF = (1/2)(α + β). Similarly, arc GH makes angle γ + δ at the center, hence, ^{}G̸EH = (1/2)(γ + δ). If the lines EG and F H intersect in S, then the sum of the two angles SHE and HES of triangle ESH is (1/2)(α + β + γ + δ) = (1/2)π, making the third angle ^{}E̸SH a right angle.

(G6) Let ABCD be a rectangle whose diagonals intersect at S. Let P be any point in a plane. Then P A^{2} + P C^{2} = P B^{2} + P D^{2} = 2P S^{2} + 2AS^{2}.

\underline{Proof}. Since P S is the median for triangles P CA and P BD, we have P A^{2} + P C^{2} = 2AS^{2} + 2P S^{2} = P B^{2} + P D^{2}.

(G7) If I_{1}, I_{2}, I_{3}, I_{4} are the incenters of triangles ABC, BCD, CDA, and DAB, respectively, then the figure I_{1}I_{2}I_{3}I_{4} is a rectangle.

\underline{Proof}. Let E, F, G, H denote the midpoints of the arcs AB, BC, CD, and DA, respectively. Then E lies on the bisectors of angles BCA and BDA. Similarly, each of the points F, G, and H lies on two angle bisectors as shown in the diagram. This means that the bisectors AF and CE intersect at I_{1}, and the same for I_{2}, I_{3}, and I_{4}. The equal arcs AH and HD make equal angles at F, hence, F H is the bisector of angle AF D. For the same reason, F H is also the angle bisector of angle BHC. By using (G2) on triangles ABC and BDC, we get F I_{1} = F B = F C = F I_{2}. This makes I_{1}I_{2} perpendicular to the bisector of angle I_{1}F I_{2}, which is F H. Similarly, I_{3}I_{4} is perpendicular to F H, and this makes I_{1}I_{2} and I_{3}I_{4} parallel lines. Similarly, I_{1}I_{4} and I_{2}I_{3} are parallel lines, both being perpendicular to EG. But (G5) says that EG and F H are perpendicular. Hence, the figure I_{1}I_{2}I_{3}I_{4} is a rectangle.

(G8) Let IL, IM, IN be the perpendiculars from the incenter I to the sides BC, CA, and AB of triangle ABC. Then, AM = AN = (1/2)(b + c − a), BL = BN = (1/2)(c + a − b), and CL = CM = (1/2)(a + b − c).

\underline{Proof}. Note that AM = AN, BL = BN, CL = CM. Thus, 2AM +2BL+2CL equals a + b + c, or 2AM + 2a = a + b + c. Hence, AM = (1/2)(b + c − a).

4. Earliest Attempt. Japan was a closed society until 1854, when Commodore Perry forced open its doors, and Japan began to exchange goods and knowledge with the western countries. The native Japanese mathematics prior to its contact with the European world was known as Wasan, which means Japanese Mathematics. The earliest proof of the Japanese Theorem is found in a book on Wasan written by Tameyuki Yoshida [9]. We are not sure if Yoshida himself gave this proof. The interesting thing about this proof, besides being the earliest, is that it is based on just two properties of the circle – (1) from any point outside the circle, tangents drawn to the circle have equal length, and (2) an arc of a circle makes the same angles at any point on the circle. We now present this proof.

\underline{Japanese Theorem (Quadrilateral Case)}. Let ABCD be a quadrilateral inscribed in a circle. Let r_{1}, r_{2}, r_{3}, r_{4} be the radii of the circles C_{1}, C_{2}, C_{3}, and C_{4} inscribed in triangles ABC, BCD, CDA, and DAB, respectively. Then r_{1} + r_{3} = r_{2} + r_{4}.

\underline{Proof}. Let P_{i}, Q_{i}, and H_{i} denote the points where the circle C_{i} touches the sides as shown in diagrams (A) and (B). The first step of the proof is to show that H_{1}H_{3} and H_{2}H_{4} are equal. To do so, let E denote the expression AD + BC − AB − CD. From diagram (A), E = (AQ_{3} + DQ_{3}) + (BQ_{1} + CQ_{1}) − (AP_{1} + BP_{1}) − (CP_{3} +DP_{3}). But AP_{1} = AH_{1}, AQ_{3} = AH_{3}, CQ_{1} = CH_{1}, CP_{3} = CH_{3}, and so on. Hence, E = AH_{3} + CH_{1} − AH_{1} − CH_{3} = (AH_{3} − AH_{1}) + (CH_{1} − CH_{3}) = 2H_{1}H_{3}. From diagram (B), E = (AP_{4}+DP_{4})+(BP_{2}+CP_{2})−(AQ_{4}+BQ_{4})−(CQ_{2}+DQ_{2}). But BP_{2} = BH_{2}, BQ_{4} = BH_{4}, and so on. Thus, E = DH_{4} +BH_{2} −BH_{4} −DH_{2} = (DH_{4} − DH_{2}) + (BH_{2} − BH_{4}) = 2H_{2}H_{4}. Hence, H_{1}H_{3} = H_{2}H_{4}.

Let arcs AB, BC, CD, DA make angles 2α, 2β, 2γ, and 2δ, respectively at points on the circle. This means ^{}A̸CB = ^{}A̸DB = 2α, ^{}B̸AC = ^{}B̸DC = 2β, C̸AD = ^{}C̸BD = 2γ, and ^{}D̸BA = ^{}D̸CA = 2δ. Then

tan α = \frac{r_{1}}{CH_{1}} = \frac{r_{4}}{DH_{4}} , tan β = \frac{r_{1}}{AH_{1}} = \frac{r_{2}}{DH_{2}} ,

tan γ = \frac{r_{3}}{AH_{3}} = \frac{r_{2}}{BH_{2}} , tan δ = \frac{r_{3}}{CH_{3}} = \frac{r_{4}}{BH_{4}} .

This gives us r_{1}DH_{4} − r_{4}CH_{1} = 0, r_{2}AH_{1} − r_{1}DH_{2} = 0, r_{3}BH_{2} − r_{2}AH_{3} = 0, and r_{4}CH_{3} −r_{3}BH_{4} = 0. On adding these, we get r_{1}(DH_{4} −DH_{2})+r_{3}(BH_{2} −BH_{4}) = r_{2}(AH_{3}−AH_{1})+r_{4}(CH_{1}−CH_{3}), or r_{1}(H_{2}H_{4})+r_{3}(H_{2}H_{4}) = r_{2}(H_{1}H_{3})+r_{4}(H_{1}H_{3}). But H_{2}H_{4} = H_{1}H_{3}, hence, r_{1} + r_{3} = r_{2} + r_{4}. This completes the proof.

\underline{References}

1. H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, Canada, 1989.

2. W. J. Greenstreet, “Japanese Mathematics,” The Mathematical Gazette, 3 (1906), 268–270.

3. C. Hawn, “A Study of the Japanese Theorem,” Masters Degree Thesis, Southeast Missouri State University, May 1996.

4. T. Hayashi, “Sur un Soi-Disant Theoreme Chinois,” Mathesis, III (1906), 257– 260.

5. N. Mackinnon, “Friends in Youth,” The Mathematical Gazette, 77 (1993), 20– 21.

6. Y. Mikami, “A Chinese Theorem in Geometry,” Archiv der Mathematik und Physik, 3 (1905), 308–310.

7. W. Uegaki, “On the Origin and History of the Japanese Theorem,” Journal of Mie University, March 2001.

8. H. Yoshida, Personal Letter, March 27, 1996.

9. T. Yoshida, Zoku Shinpeki Sanpo Huroku Kai (manuscript, date unknown).

Mangho Ahuja 2575 Fairlane Cape Girardeau, MO 63701 email: mangho@showme.net

Wataru Uegaki Department of Math Education Mie University 1515 Kamihama–cho Tsu–shi, MIE Japan 514–8507 email: uegaki@edu.mie-u.ac.jp

Kayo Matsushita Center for the Promotion of Excellence in Higher Education Kyoto University Kyoto 606-8501 Japan email: kmatsu@hedu.mbox.media.kyoto-u.ac.jp