EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 4135-4146 ISSN 1307-5543 – ejpam.com Published by New York Business Global Generalizing the Equal Incircles Theorem: Insights from Sangaku Problems Perawit Boonsomchua 1Engineer Science Classroom (ESC), Learning Institute, King Mongkut’s University of Technology Thonburi, Bangkok, Thailand Abstract. Sangaku problems are traditional Japanese geometrical puzzles, often displayed in reli- gious temples, that have intrigued mathematicians for centuries. This study aims to generalize the Equal Incircles Theorem, extending Angela Drei’s proof to N -circles, by applying the trigonomet- ric method alongside foundational mathematical tools, including mathematical induction, Heron’s formula, and the telescoping product. A generalized equation for N circles based on the Equal Incircles Theorem is derived through explicit mathematical formulation and characterization. The findings deepen our understanding of geometric relationships, highlight the historical significance of Sangaku problems, and offer potential advancements for future engineering applications, math- ematics education, and research in mathematical history. 2020 Mathematics Subject Classifications: 51M04, 01A27 Key Words and Phrases: Sangaku problems, Equal Incircle Theorem, Angela Drei’s proof 1. Introduction Sangaku are wooden tablets inscribed with various geometrical problems and devoted to Shinto shrines and Buddhist temples, as shown in Figure 1, during the Japanese Edo period (1603-1868 CE) this is a book: [2]. The historical significance of this tradition was largely unrecognized by scholars until it was brought to light through the publication Japanese Temple Problems: Sangaku. This source discovered by an article in a collection: [5] notes that Japanese temple geometry often predated the discovery of several well-known Western geometric theorems, such as the Katayamahiko Temple Problem, Meiserinji Tem- ple Problem, and the Equal Incircles Theorem. Subsequently, the geometrical principles underlying Japanese temple architecture were documented in the book Sacred Mathemat- ics: Japanese Temple Geometry this is a book: [4]. Following this, an investigation led by R. J. Hosking [see also [7]] as a technical report, addressed a traditional mathematical approach to a Sangaku problem from Okayama prefecture, as shown in Figure 2. However, these Sangaku problems still lack modern alternative mathematical methods to gain deeper insights. Recent investigations have reported that these ancient problems have contributed to modern mathematical methods, particularly in the field of discrete geometry. Notable examples include the geometric inversion method [see also [10]] a journal article, Euclidean geometry [see also [6]] a journal article, and so on. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5300 Email address: perawit.boon@mail.kmutt.ac.th (P. Boonsomchua) https://www.ejpam.com 4135 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4136 Figure 1: A Sangaku board hanging under the roof of a temple in Japan. Figure 2: Sangaku Problems at Katayamahiko Shrine. Equal Incircles Theorem Let C be a point. Assume points Mi, for i = 1, 2, . . . , N (N > 3), lie on a line not passing through C. Assume further that the incircles of triangles M1CM2, M2CM3, . . . , MN−1CMN all have equal radii. Then the same is true for the triangles M1CM3, M2CM4, . . . , MN−2CMN , and also for the triangles M1CM4, M2CM5, . . . , MN−3CMN , and so on. Figure 3: The image depicts the diagram associated with the Equal Incircles Theorem. Several Sangaku problems from Japan remain unsolved or lack characterization in mathematical theorems that used to explicit mathematical formulations, including the “Equal Incircle Theorem” [see also [3]] as technical reports, which presents a configuration of multiple equal circles inscribed within the same number of sub-triangles, with larger equal circles inscribed within the larger triangles, as shown in Figure 3. Subsequently, Angela Drei’s proof provides initial mathematical formulations for two Sangaku problems with equal incircles, employed to characterize the geometrical relationships expressed in terms of trigonometric forms. So, this paper aims to develop a mathematical framework that generalizes the mul- tiple incircles problem, enabling non-equal radii, by extending from the Equal Incircles Theorem. The study also explores three setups: (1) two tangent circles within a triangle in Section 1; (2) an analysis of the inclusion of incircles in a sectorial triangular configura- tion in Section 2; (3) the analysis of the inclusion of incircles and ex-circles in a sectorial triangular configuration in Section 3. P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4137 2. Main results This section has formulated an equation that describes generalized adjacent triangles with inscribed circles in the initial part and ex-circles in the remaining part, based on An- gela Drei’s proof. This formulation is extended to other scenarios, including perpendicular and angle-bisector cases. Section 1 Two Tangent Circles in a Triangle Theorem 1 Let a triangle △AB1C1 with vertices A, B1, and C1. Inside this triangle, there are two circles. Let circles O1 (green) and O2 (blue) as the incircles of the triangles ∆AB1B2 and ∆AC1B2 with radii r1 and r2 respectively. Circle O1 is tangent to the sides AB1, AB2 and B1C1 at points F1, E1, and D1 respectively, and circle O2 is tangent to the sides AC1, AB2, and B2C2 at points E2, F2 and D2 respectively by specifying that, (i) |B1C1| = a, |AB1| = c, |C1A| = b. (ii) |AF1| = x2, |F1B1| = x1, |AE2| = y1, |D2B2| = x3, |E2C1| = y2 (iii) ∠AB1C1 = B, ∠AC1B1 = C, and ∠B1AC1 = A (iv) semi-perimeter of AB1B2 triangle : s1 = x1 + x3 + y1 (v) semi-perimeter of AC1B2 triangle : s2 = x3 + y1 + y2 (vi) semi-perimeter of ABC triangle : s = a+b+c 2 = x1 + y3 + y2 + x3 And then, r1 r2 = tan B 2 · (s1 − (x3 + y1)) tan C 2 · (s2 − (x3 + y1)) Figure 4: The image depicted the Sangaku Two Circles Problem. Proof. Let us begin to consider the radius of two incircles (green and blue), using the Incircle of a Triangle formulation. We get, r1 = [AB1B2] s1 (1) r2 = [AB2C1] s2 (2) P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4138 Divided the Equation (1) by Equation (2). Then, r1 r2 = [AB1B2] [AC1B2] · s2 s1 (3) From the area-based ratio, we can formulate the area proportion of two sub-triangles ∆AB1B2 and ∆AC1B2 depending on its bases. [AB1B2] [AC1B2] = (s1 − x2) (s2 − y1) (4) Substituting Equation (3) into Equation (4) and we arrive at r1 r2 = s2 s1 · (s1 − x2) (s2 − y1) (5) This equation is straightforward; to derive in terms of bisector-angle trigonometry. tan( B 2 ) = r1 s1 − (y1 + x3) , tan( C 2 ) = r2 s2 − (y1 + x3) Hence, tan B 2 tan C 2 = r1 r2 · (s2 − (y1 + x3)) (s1 − (y1 + x3)) ⇔ r1 r2 = tan B 2 · (s1 − (x3 + y1)) tan C 2 · (s2 − (x3 + y1)) Suppose the biggest right triangle, ∆AC1B1 , obtained a perpendicular line from vertex A. In that case, this scenario can extend to the scope of mathematical formulation that describes two tangent circles in a triangle with a perpendicular line. Corollary 1 (Perpendicular triangle scenario) Consider a triangle △AB1C1 with vertices A, B1, and C1. Inside this triangle, there are three inscribed circles. Let O1 (green), O2 (blue), and O3 (light gray) be the incircles of triangles △AB1B2, △AC1B2, and △AB1C1 with radii r1, r2, and r respectively. Circle O1 is tangent to sides AB1, AB2 and B1B2 at points F1 and E1, and D1 respectively, and circle O2 is tangent to sides AC1, AB2 and C1B2 at points E2, F2 and D2 respectively. Circle O3 is tangent to sides AB1, AC1, and B1C1 at points T1, T2, and T3, respectively such that the following conditions hold: (i) B1A ⊥ C1A. (ii) AB2 ⊥ B1C1. And then, AB2 = √ (s1 − x2)(s2 − y1) Proof. We know that the incenter of circle O lies on the line AB2, with points B2 and T1 coinciding in this case. Therefore, AT1IT2 forms a square, as all its sides are equal to the radius of the circle O. It follows that b+ c = 2r + a Consider the incircles ∆AB1B2 and ∆AC1B2 can express the connection between incircles radius and side AB1, influencing useful mathematical formulation after combined for each other. P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4139 Figure 5: The image depicts the Sangaku Two Circles Problem, illustrating the perpendicular line AB2. Figure 6: The image depicts the Sangaku Two Circles Problem featuring the inscribed circles O. 2(r1 + r2) +AB1 +AC1 = B1B2 +AB2 + C1B2 +AB2 Further simplifying, we get: AB1 = r1 + r2 + r (6) We derived the right-hand side of the main formulation but needed to show that the left-hand side of the equation is equal to the right-hand side, demonstrating the equality between both expressions. According to the similar triangle theorem, and consider the triangle ∆AB1B2 ∼ ∆AB2C1, which implies the following. B1B2 AB1 = AB1 B2C1 =⇒ AB2 1 = B1B2 ·B2C1 (7) Substituting Equation (6) into Equation (7) and we arrive at AB2 2 = (B1B2) · (C1B2) = (x1 + x3 + y1 − x2) · (x3 + y2) = (s1 − x2)(s2 − y1) Therefore, AB2 = √ (s1 − x2)(s2 − y1) Consider a triangle ∆AC1B1, where a bisector line is drawn from vertex A to the opposite side. Within this configuration, we explored this condition that used to describe two circles that are tangent to each other and tangent to two sides of the triangle with the bisector line. Corollary 2 (Triangle angle bisector scenario) Consider a triangle ∆AB1C1 with vertices A, B1, and C1. Inside this triangle, there are two circles. Let the circles O1 (green) and O2 (blue) as the incircles of the triangles ∆AB1B2 and ∆AC1B2 with radii r1 and r2 respectively. Circle O1 is tangent to the sides AB1, AB2 and B1C1 at points F1, E1, and D1 respectively, and circle O2 is tangent to the sides AC1, AB2, and B2C2 at points E2, F2 and D2 respectively such that the following conditions hold: (i) ∠B1AB2 = ∠C1AB2 = θ (ii) The cevian AB2 internal bisects the angle ∠B1AC1 P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4140 And then, s2 − y1 s2 − x3 = s1 − x2 s1 − (x3 + y1 − x2) Figure 7: The image depicts the Sangaku Two Circles Problem, illustrating the angle-bisecting scenario. Proof. According to the angle-bisector theorem in the triangle ∆AB1C1 so that x1 + x2 y1 + y2 = x1 + x3 + y1 − x2 x3 + y2 Therefore, s2 − y1 s2 − x3 = s1 − x2 s1 − (x3 + y1 − x2) The next section is inspired by the American Invitational Mathematics Examination 2018 Problem 13 [see also [1]] as a technical report. This problem is the best sample to point out the extended scope of this study. However, it obtained a new mathematical formulation that can describe the minimum area of a triangle, which contains two incir- cles centered in the base of the triangle. The relation can be expressed in the form of trigonometry. Corollary 3 (Area minimum on bisecting triangle angle scenario) Consider a triangle △AB1C1 with vertices A, B1, and C1. Inside this triangle, there are two circles. Denote the circles O1 (green) and O2 (blue) as the incircles of the triangles ∆AB1B2 and ∆AC1B2 with radii r1 and r2 respectively. Circle O1 is tangent to the sides AB1, AB2 and B1C1 at points F1, E1, and D1 respectively, and circle O2 is tangent to the sides AC1, AB2, and B2C2 at points E2, F2 and D2 respectively such that the following conditions hold: (i) The cevian AB2 internal bisects the angle ∠B1AC1 and the angle ∠I1AI2. And then, the minimum area of the triangle AI1I2 is: (s− a)(s− b)(s− c) a Proof. Assume that the angle ∠AB2B1 = δ, and that we have drawn BI1 and CI2, which bisect the angles ∠AB1C1 and ∠AC1B1 respectively. ∡AB1I1 = ∡I1B1C1 = B 2 , ∡AC1I2 = ∡I2C1B1 = C 2 , ∡AI1B1 = 90◦ + δ 2 , P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4141 Figure 8: This image presents the line AB2 internally bisecting the angles ∠B1AC1 and ∠I1AI2. Figure 9: This image illustrates multiple lines bisecting angles within this configuration. ∡AB2C1 = 180◦ − δ, ∡AI2C1 = 180◦ − δ 2 , ∡I1AI2 = α+ β Next, using the sine law is important for deriving relationships between the sides of the triangle and the relevant angles in this configuration: AI1 = sin ( B 2 ) sin ( 90◦ + δ 2 ) · c (5) AI2 = sin ( C 2 ) sin ( 180◦ − δ 2 ) · b (6) Substituting the Equation (5) and (6) into the Area of Triangle Trigonometry formu- lation. [AI1I2] = 1 2 ·AI1 ·AI2 · sin ( A 2 ) = 1 2 · b · c · sin ( A 2 ) · sin ( B 2 ) · sin ( C 2 ) sin ( 90◦ + δ 2 ) · sin ( 180◦ − δ 2 ) And then, [AI1I2] = b · c · sin(A2 ) · sin( B 2 ) · sin( C 2 ) sin δ noting that 1 ≤ 1 sin δ and sin(A2 ) = ± √ 1−cosA 2 . We deduce the minimum area of triangle AI1I2 from the following inequality establishes a lower bound. [AI1I2] ≥ b · c · (√ (1− cosA)(1− cosB)(1− cosC) 8 ) (7) We applied the Law of Cosines to the concyclic angles. This is an important point be- cause it reduces the trigonometric calculations involving cosines to a relationship between the sides of the triangle. 1− cosA = (a− b+ c)(a+ b− c) 2bc , (8) 1− cosB = (b+ c− a)(b+ a− c) 2ca , (9) 1− cosC = (b+ c− a)(a+ c− b) 2ab . (10) P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4142 Substituting Equations (8), (9), and (10) into Equation (7). Therefore, the minimum area of the triangle AI1I2 is: (s− a)(s− b)(s− c) a Section 2 Generalization of Angela Drei’s Proof-Inspired Analysis of Inclusion In- in a Sectorial Triangular Configuration Theorem 2 Let a triangle △AB1Bn+1 with rays ABu from vertex A. For u ∈ {1, . . . , n}, suppose that Ou is the u-th inscribed circle with radii ru in the triangle ABuBu+1. It holds that Ou is touch to ABu, ABu+1, and BuBu+1 at Fu, Eu and Du respectively. Given the lengths AEu, and AFu be xu, and yu respectively, where su is the semi-parameter of triangle ABuBu+1. Then, r1 rn = sn s1 · n∏ i=2 ( si−1 − xi si − yi−1 ) Figure 10: The image illustrates the approach to the Generalized Sangaku Problem for N -circles. Proof. Consider the statement P (n) defined as: P (n) = rn−1 rn = sn sn−1 · ( sn−1 − xn sn − yn−1 ) (11) where it is known from Equation (11) that P (n) holds true for all positive integers n. Substituting the values from n = 2 to n = n into the equation, we obtain: n∏ i=2 P (i) = ( r1 r2 )( r2 r3 ) · · · ( rn−1 rn ) = ( s2 s1 )( s3 s2 ) · · · ( sn sn−1 ) · n∏ i=2 ( si−1 − xi si − yi−1 ) Further simplifying. Therefore, P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4143 r1 rn = sn s1 · n∏ i=2 ( si−1 − xi si − yi−1 ) Remark 1. This is a remark about the telescoping product from Equation (X), which presented the inscribed radii fractions is shown, generalized to the independent variables j and k for all j, k ∈ {2, . . . , n}. Therefore, rk rj+1 = sj+1 sk · j+1∏ o=k ( so − xo+1 so+1 − yo ) . Section 3 Generalization of Angera Drei’s Proof-Inspired Analysis of Inclusion In- and Ex-Circles in a Sectorial Triangular Configuration Theorem 3 Let a triangle △AB1Bn+1 with extended rays ABu from vertex A. For u ∈ {1, . . . , n}, suppose that Ou is the u-th inscribed circle with radii ru in the triangle ABuBu+1, and contains O′ u is the escribed circle with radii Ru that are tangent to the common base BnBn+1 by supposing the angle ∠ABiBi+1 is set to Bi−1. Then, n∏ i=1 ri Ri = tan ( B 2 ) · tan ( C 2 ) Figure 11: The image illustrates the approach to the Generalized Sangaku Problem for N -inscribe and escribed circles. Proof 1. By trigonometric formulations for bisecting angles, sin ( A 2 ) = √ (s− b)(s− c) bc , cos ( A 2 ) = √ s(s− a) bc , P. Boonsomchua / Eur. J. Pure Appl. Math, 17 (4) (2024), 4135-4146 4144 tan ( A 2 ) = √ (s− b)(s− c) s(s− a) . Applying Heron’s formula, the expression can be written as follows the relationship between any consecutive angles B is given by: tan ( B 2 ) · tan ( B1 2 ) = √ (s−B1B2)(s−AB1)(s−AB1)(s−AB2) s2(s−AB2)(s−B1B2) = r1 R1 (12) Similarly, this mathematical formulation allows us to repeat angle terms for each sub- stitution in Equation (12): r2 R2 = tan ( 180◦ −B1 2 ) · tan ( B2 2 ) , r3 R3 = tan ( 180◦ −B2 2 ) · tan ( B3 2 ) , ... rn Rn = tan ( 180◦ −Bn−1 2 ) · tan ( C 2 ) . Upon multiplying these previous n equations, the final product is derived as follows: n∏ i=1 ri Ri = tan ( B 2 ) · tan ( C 2 ) · n−1∏ i=1 tan ( 180◦ −Bi 2 ) and simplifying using the tangent identity, we deduce: n∏ i=1 ri Ri = tan ( B 2 ) · tan ( C 2 ) 3. Discussions and Conclusions By extending the Equal Incircles Theorem, this paper aims to develop a mathematical framework that generalizes the multiple incircles problem, allowing for non-equal radii. This study enhances the understanding of the Equal Incircles Theorem based on the generalized Sangaku problem. In Section 1, this paper revealed the relationship between the fraction of sub-circles in terms of trigonometry for both initial angles (B,C). Although the numerator and denominator are similar in form, they differ in the initial angles and the semi-perimeter of the triangle involving two sub-circles. When generalized to N sub-circles within the same triangle, separated by cevians AB2, the result revealed that the fraction between any two sub-inscribed circles radii forms a telescoping product, as described in Section 2. Also, this paper explores other generalized cases, illustrating the relationship between N sub-inscribed circles and sub-escribed circles, as shown in Figure 9. Our findings reveal that this formulation differs significantly from other sections, as the product of their relation provides meaningful interpretations. These relationships are directly tied to the radius of the largest circumcircle within the same triangle. REFERENCES 4145 On the other hand, the study considers the scenario of perpendicular triangles, as discussed in Corollary 1, presenting the sum of the radii of all circles in these configurations. This case guides us toward understanding the perpendicular scenario. At the same time, our findings also consider the angle-bisector case, a general concept frequently encountered in problem-solving across many sources. This finding demonstrates a novel approach with applications in many fields, such as mathematical education, particularly through the development of new geometry teaching materials [see also [9]] as a technical report. Additionally, it offers potential for the pro- gression of optimization methods [[8]] is a journal article and research in mathematical history. Acknowledgements The author conveys sincere appreciation to Mrs. Ratree Boonsomchua, an accom- plished mathematics teacher and advisor at Surasakmontri School, whose guidance was instrumental in preparing this paper. Her notable achievements include a gold award at the 2nd International Conference for Students in Science and Innovation 2022 (ISSI 2nd), organized by the Institute for the Promotion of Teaching Science and Technology (IPST). Further thanks to Mrs. Chalida Suwankosai, a mathematics teacher at Benjamarachanu- sorn School, for her insightful recommendations that have greatly expanded the study’s scope, with influences dating back to the author’s ninth-grade year. Profound gratitude is also extended to Mr. Taechasith Kangkhuntod, studying at RajsimaWittayalai School, and former Chief Executive Officer at CreativeLab, an independent organization. He orga- nized a crucial meeting on April 21, 2024, covering diverse fields like mathematics, physics, architecture, and history. This meeting significantly inspired the integration of modern mathematical methods with traditional Sangaku problems. References [1] 2018 AIME i problems, problem 13. American Mathematics Competitions (AMC), Mathematical Association of America, 2018. [2] E. A. Donovan, A. Hoots, and L. W. Wiglesworth. Japanese temple geometry. Teach- ing Mathematics Through Cross-Curricular Projects, 72:51–62, 2024. [3] A. Drei. Equal incircles theorem, angela drei’s proof. Cut-the-knot, 2011. [4] R. Y. A. N. Fameli. Math 400: Sangaku, Japanese Temple Geometry. https:// cklixx.people.wm.edu/teaching/math400/Fameli.pdf, 2020. Accessed: 2, 2023. [5] H. Fukagawa and T. Rothman. Japanese temple geometry problems. Scientific Amer- ican, 278(5):84–91, 1989. [6] N. Hartmann. Sangaku in multiple geometries: Examining japanese temple geometry beyond euclid, 2022. [7] R. Hosking. Solving sangaku: A traditional solution to a nineteenth-century japanese temple problem. Journal for History of Mathematics, 30:53–62, 2017. [8] D. Jargalsaikhan, D. Rentsen, and B. Darkhijav. Simulation on sangaku problem using optimization methods. Journal of Institute of Mathematics and Digital Tech- nology, 5(1):19–29, 2023. https://cklixx.people.wm.edu/teaching/math400/Fameli.pdf https://cklixx.people.wm.edu/teaching/math400/Fameli.pdf REFERENCES 4146 [9] H. Makishita. Solving problems from sangaku with technology—for good mathematics in education, 2010. [10] M. Tibi, G. Zanardo, E. Vendraminett, D. Pozzebon, L. Zaccaron, M. Breda, I. Emeliyanov, and L. Bonaldo. Two problems on touching circles and their con- nection to the stern-brocot tree. School: Liceo “M. Casagrande” – Pieve di Soligo, Treviso (Italy), 2021. Introduction Main results Discussions and Conclusions