EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 1, 2020, 180-184 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Some applications of Ptolemy’s theorem in secondary school mathematics Samed J.Aliyev1,∗, Shalala A.Hamidova1, Goncha Z.Abdullayeva1 1 Department of Methods of Mathematics and its Teaching, Faculty of Mechanics and Mathematics, Baku State University, Baku, Z.Khalilov str.23,, AZ1148, Azerbaijan Abstract. We consider some applications of Ptolemy’s theorem. In particular, we find a criterion for constructing an inscribed hexagon. 2020 Mathematics Subject Classifications: 97G10, 97G40 Key Words and Phrases: Ptolemy’s theorem, inscribed rectangle, regular heptagon, inscribed hexagon. Today, to improve the quality of teaching mathematics is one of biggest challenges faced in the field of education. To do so, you have to diversify the process of teaching, to improve the methods of teaching, to start using new approaches for some problems. Since ancient times, the researchers have been studying the properties of inscribed polygons. The question of ”Under what conditions is it possible to draw a polygon inside a circle? has always been of interest, and some criteria were found for inscribed polygons. The criterion for an inscribed rectangle provided in modern textbooks of secondary school geometry involves internal angles of rectangle. It says: for a rectangle to be inscribed in a circle, it is necessary and sufficient that the sum of its opposite angles be equal to 180◦. These textbooks do not mention the relationships between the sides and diagonals of inscribed rectangle. Though, this is exactly what Ptolemy’s famous theorem is about. Not included in modern school textbooks, Ptolemy’s theorem is a good criterion for constructing an inscribed rectangle which involves the sides and diagonals of a rectangle. It says: for a rectangle to be inscribed in a circle, it is necessary and sufficient that the sum of the products of its opposite sides be equal to the product of its diagonals. In this work, we consider some problems which have been earlier solved by traditional methods, and we easily solve these problems by using new approaches, namely, Ptolemy’s theorem. The results obtained reveal the effectiveness of this new approach. It incites students attention to the subjects studied during geometry classes and encourages them ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i1.3605 Email addresses: samed59@bk.ru (S. J.Aliyev), shalalahamidova@hotmail.com (S. A.Hamidova), a.q.z.41@mail.ru (G. Z.Abdullayeva) http://www.ejpam.com 180 c© 2020 EJPAM All rights reserved. S. J.Aliyev, S. A.Hamidova, G. Z.Abdullayeva / Eur. J. Pure Appl. Math, 13 (1) (2020), 180-184 181 to learn more about these subjects. As a result, students get more interested in geometry and start loving it. Inscribed polygons have been studied in [1], [2], [5]. Also, the criteria for constructing some inscribed polygons have been found in [3], [4]. In this work, we consider some applications of Ptolemy’s theorem. Namely, using Ptolemy’s theorem, 1) we prove some property of a point lying on a circle circumscribed about some regular triangle; 2) we prove some property of a regular heptagon; 3) we find a criterion for constructing an inscribed hexagon. We first prove that if the point M lies on a circle circumscribed about the regular triangle ABC, then one of the intervals MA, MB, MC is equal to the sum of two other intervals. Figure 1: Circle circumscribed about the regular triangle ABC. Let the point M lie on the arc AB (Figure 1). Let’s prove that MA + MB = MC. (1) We will use traditional methods. As ∠BMC = ∠CMA, due to angle bisector property of a triangle we have MB MA = BK KA or BK ·MA = MB ·KA, (2) Regarding similar triangles BMC and KMA , we have BC KA = MC MA or BC ·MA = MC ·KA, As BC = BK + KA , from the last equality, using (2), we get the validity of (1): (BK + KA) ·MA = MC ·KA S. J.Aliyev, S. A.Hamidova, G. Z.Abdullayeva / Eur. J. Pure Appl. Math, 13 (1) (2020), 180-184 182 BK ·MA + MA ·KA = MC ·KA, MB ·KA + MA ·KA = MC ·KA, or MB + MA = MC. But, this relation can be obtained quite easily without using bisector property and similar triangles, just by means of Ptolemy’s theorem. As the rectangle MABC is inscribed, by Ptolemy’s theorem we have MA ·BC + MB ·AC = MC ·AB. Considering the conditions BC = AC = AB , we get the validity of (1). Now let’s prove that if the points A,B,C,D are the consecutive vertices of a regular heptagon (Figure 2), then the following relation holds: 1 AB = 1 AC + 1 AD . (3) Figure 2: Regular heptaqon ABCDEFK. Let the point E be a vertex of a regular heptagon next to D. The triangles ABC and ACD are equal to the triangles CDE and AFE, respectively. Hence AC = CE and AD = AE. Taking into account the equality of these intervals and the conditions DE = CD = AB , by Ptolemy’s theorem applied to the rectangle ACDE we get the validity of (3): AC ·DE + AE · CD = CE ·AD, AC ·AB + AD ·AB = AC ·AD, 1 AB = 1 AC + 1 AD . Finally, let’s find a criterion for constructing an inscribed hexagon, which involves its sides and large diagonals, using Ptolemy’s theorem. In other words, let’s prove that the product of large diagonals of an inscribed hexagon is equal to the sum of products of its S. J.Aliyev, S. A.Hamidova, G. Z.Abdullayeva / Eur. J. Pure Appl. Math, 13 (1) (2020), 180-184 183 consecutive non-adjacent sides and products of its opposite sides with a large diagonal which does not intersect any of these sides. Let the hexagon ABCA1B1C1 be inscribed in a circle (Figure 3). Figure 3: The hexagon ABCA1B1C1 be inscribed in a circle. We must prove that the following relation is true in the hexagon ABCA1B1C1 : AA1 ·BB1 · CC1 = AB · CA1 ·B1C1 + BC ·A1B1 ·AC1+ +AB ·A1B1 · CC1 + BC ·B1C1 ·AA1 + CA1 ·AC1 ·BB1. (4) Applying Ptolemy’s theorem to the rectangles ACA1C1 ,A1B1C1B ,ABCC1 ,ABCA1, respectively, we obtain AA1 · CC1 = AC ·A1C1 + AC1 · CA1, (5) A1C1 ·BB1 = BA1 ·B1C1 + A1B1 ·BC1, (6) AC ·BC1 = AB · CC1 + BC ·AC1, (7) AC ·BA1 = AB · CA1 + BC ·AA1. (8) Then, multiplying (5), (6), (7), (8) by BB1, AC, A1B1, B1C1, respectively, and adding up the resulting equalities we obtain the relation (4). The equality (4) is a relationship between the sides and diagonals of an inscribed hexagon. That’s why it can well be called an analogue of Ptolemy’s theorem for an inscribed hexagon. REFERENCES 184 References [1] V.A. Dalinger; Graphics teaches how to think , Matematika v shkole, (1990), No4, p.32-36. 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