EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 1110-1117 ISSN 1307-5543 – ejpam.com Published by New York Business Global Solving geometry problems by alternative methods in mathematics education Samed J.Aliyev1,∗, Maftun N.Heydarova2, Shahin M.Aghazade1 1 Department of Methods of Mathematics and its Teaching, Faculty of Mechanics and Mathematics, Baku State University, Baku, Z.Khalilov str. 23, AZ 1148, Azerbaijan 2 Department of Methods of Mathematics and its Teaching, Faculty of Mathematics, Sumgait State University, Sumgait, district 43, AZ 5008, Azerbaijan Abstract. Solving geometry problems is both difficult and interesting. Difficult because there is no general algorithm to solve more or less non-trivial problems as every single problem requires individual and creative approach. At the same time, this is a very interesting activity, because for almost every problem there are plenty of ways to solve it. In this work, we present the method of auxiliary circle divided into equal parts. This method allows finding solution algorithm for some geometry problems which are hard to solve by the method of additional constructions. 2020 Mathematics Subject Classifications: 97G10, 97G30, 97G40 Key Words and Phrases: Geometry problem, Ceva’s theorem, regular polygon, solution algo- rithm. 1. Introduction In geometry, there is a class of problems for which the traditional methods (such as the method of equal triangles, the method of geometric transformations, the vector method, etc.) are either inapplicable or provide complicated and cumber-some solutions. In many cases, it becomes possible to solve such kind of problems by adding some lines (so called additional constructions) to the drawing. Sometimes these constructions suggest themselves, but in other cases situation may depend on your experience, expertise and geometric intuition. The method of solving problems by adding some constructions to the drawing is called the method of additional constructions. The essence of the method of additional con- structions lies in the fact that you add some new (auxiliary) elements to the drawing of ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4721 Email addresses: samed59@bk.ru (S,J.Aliyev), meftun.heydarova.82@mail.ru (M.N.Heydarova),shahinaghazade@bsu.edu.az (Sh.M.Aghazade) https://www.ejpam.com 1110 © 2023 EJPAM All rights reserved. S. J.Aliyev, M. N.Heydarova, S. M.Aghazade / Eur. J. Pure Appl. Math, 16 (2) (2023), 1110-1117 1111 Figure 1: (Triangle ABC with the given angles) your problem. As a result, the relationships between the problem data and the unknown quantities, which have been hard to see before, become more tangible, or even obvious. There are problems where additional constructions are the only way of solution. Solv- ing geometry problems by the method of additional constructions is a very hard task, because there is no general algorithm to solve your problem by this method. Even the schoolchildren with a good knowledge of geometry are not able to solve such kind of problems. In this work, we overcome the above mentioned difficulties by using the method of auxiliary circle divided into equal parts. Namely, we present a solution algorithm for geometry problems which are hard to solve by the method of additional constructions. In conclusion, let’s mention the works [1–3, 5, 7] and [6], which deal with the method of additional constructions. We solve some of problems considered in these works by using our new method. 2. The Method Let’s illustrate our new method on the problem solved by using the method of addi- tional constructions in [4]. Problem 1. The point M is given inside the triangle ABC so that ∠ABM = 40◦, ∠CBM = 10◦, ∠ACM = 20◦, ∠BCM = 10◦ (Fig.1). Prove that ∠CAM = 30◦. Before proceeding to solve this problem by our method, let’s recall how it was solved by the method of additional constructions in [4]. Let the circle centered at M with a radius BM = CM intersect the continuation of AC at the point D . Then, as DM = CM , the triangle CMD is equilateral and ∠MDC = ∠MCA = 20◦ (Fig.2). If we join the points D and B, then the triangle BDM becomes equilateral, because ∠BMC = 160◦ and ∠CMD = 140◦. In the triangle ABD, we find ∠BAD = 180◦−(20◦+ 80◦) = 80◦ . Hence, BD = AD. On the other hand, as BD = BM , we have AB = BM , i.e. the triangle ABM is isosceles. Since the angle at the vertex of this triangle is equal to 40◦, the adjacent angles at the base are equal to 70◦. Consequently, ∠BAM = 70◦, and therefore, ∠CAM = 30◦. Now let’s solve this problem by our method. S. J.Aliyev, M. N.Heydarova, S. M.Aghazade / Eur. J. Pure Appl. Math, 16 (2) (2023), 1110-1117 1112 Figure 2: (Triangle BDM is equilateral, while triangles ABD and ABM are isosceles ) Divide the circle into 18 equal parts and name the division points. Degree measure of the arc between two consecutive points is equal to 360◦ 18 = 20◦. If we place the triangle ABC inside the circle as shown on Fig.3 (A ≡ C18, B ≡ C15, C ≡ C5 ), then the conditions of the problem are satisfied: ∠C18C14C4 = 40◦, ∠C5C15C4 = 10◦, ∠C18C5C16 = 20◦, ∠C15C5C16 = 10◦. From Fig.3 we can see that the sought angle C5C18C8 is equal to 30◦ . Consequently, the considered problem is equivalent to the following statement: In the regular octadecagon, the diagonals C4C15, C5C16 and C8C18 intersect each other at one point. To verify the intersection of these three diagonals at one point, it is convenient to use the following trigonometric (angular) theorem of Giovanni Ceva [6]. Theorem. Let the points A1, B1, C1 be given on the sides BC,AC,AB of the triangle ABC, respectively. Then the segments AA1, BB1 and CC1 intersect each other at one point if and only if (Fig.4) sin∠BAA1 sin∠CAA1 · sin∠CBB1 sin∠ABB1 · sin∠ACC1 sin∠BCC1 = 1. So, the verification of whether three diagonals C4C15, C5C16 and C8C18 on Fig.3 intersect each other at one point is reduced to the verification of the following identity: sin10◦ sin40◦ · sin20 ◦ sin10◦ · sin70 ◦ sin30◦ = 1 And the validity of this identity follows from the equalities sin70◦ = cos20◦, sin40◦ = 2sin20◦cos20◦. . If we compare the above two methods used to solve Problem 1, then we can easily see that the second one is simpler and more effective. Besides, the first method requires additional constructions, while the second one only needs the accurate drawing. S. J.Aliyev, M. N.Heydarova, S. M.Aghazade / Eur. J. Pure Appl. Math, 16 (2) (2023), 1110-1117 1113 Figure 3: (Triangle C18C15C5, equal to triangle ABC, placed inside the circle) This example shows that many problems, even very difficult ones, can be solved using one property of regular polygon diagonals: property of intersection at one point. When solving Problem 1 by our method, we divided the circle into 18 equal parts. This is because the triangle in this problem has the angles which are the multiples of 10◦. Depending on the sizes of the angles, the problem can be solved by dividing the circle into less than 18 equal parts. For example, to solve the problem below, we will divide the circle into 12 equal parts. Problem 2. The point M is given inside the triangle ABC so that ∠ABM = ∠CBM = ∠BCM = 15◦ and ∠ACM = 30◦ (Fig.5). Prove that ∠CAM = 75◦. As the angles given in this problem are the multiples of 15◦, we divide the circle into 12, not 18 equal parts. Degree measure of the arc between two consecutive points is 360◦ 12 = 30◦. If we place the triangle ABC inside the circle as shown on Fig.6 (A ≡ P1, B ≡ P10, C ≡ P3), then the conditions of the problem are satisfied: ∠P1P10P2 = ∠P3P10P2 = ∠P10P3P11 = 15◦,∠P1P3P11 = 30◦. From Fig.6 we can see that the sought angle P8P1P3 is equal to 75◦. Consequently, the considered problem is equivalent to the following statement: In the regular dodecagon, the diagonals P1P8, P2P10 and P3P11 intersect each other at one point. Let’s verify this by the above Ceva’s theorem. Then it is reduced to the verification of the identity sin15◦ sin15◦ · sin30 ◦ sin15◦ · sin30 ◦ sin75◦ = 1 And the validity of this identity follows from the equalities sin75◦ = cos15◦, sin30◦ = 2sin15◦cos15◦. S. J.Aliyev, M. N.Heydarova, S. M.Aghazade / Eur. J. Pure Appl. Math, 16 (2) (2023), 1110-1117 1114 Figure 4: (Segments AA1, BB1 and CC1 in the triangle ABC intersect each other at one point ) Figure 5: (Triangle ABC with the given angles) Note that the presented method can be used to solve different kinds of problems, not only the ones mentioned above. Let’s consider one of such problems. Problem 3. The points D and E are given on the sides AB and AC of the triangle ABC, respectively, so that ∠ABE = ∠CBE = 20◦,∠ACD = 30◦,∠BCD = 70◦ (Fig.7). Prove that ∠BED = 60◦. As the angles given in this problem are the multiples of 10◦, we divide the circle into 18 equal parts. If we place the triangle ABC inside the circle as shown on Fig.8 (B ≡ C11, C ≡ C5, E ≡ C3), then the conditions of the problem are satisfied: ∠AC11C3 = ∠C3C11C5 = 20◦, vAC5C18 = 30◦,∠C18C5C11 = 70◦. From Fig.8 we can see that the sought angle C17C3C11 is equal to 60◦. As for Problem 3, it is equivalent to the following statement: In the regular oc- tadecagon, the diagonals C1C11, C3C17 and C5C18 intersect each other at one point. Let’s S. J.Aliyev, M. N.Heydarova, S. M.Aghazade / Eur. J. Pure Appl. Math, 16 (2) (2023), 1110-1117 1115 Figure 6: (Triangle P1P10P3, equal to triangle ABC, placed inside the circle) prove this statement as follows. It suffices to show that ∠C17DC18 + ∠C18DC1 + ∠C1DC3 = 180◦. Indeed, ∠C17DC18 = 1 2 (∪C17C18 + ∪C3C5) = 10◦ + 20◦ = 30◦, ∠C18DC1 = 1 2 (∪C1C18 + ∪C5C11) = 10◦ + 60◦ = 70◦, ∠C1DC3 = 1 2 (∪C1C3 + ∪C11C17) = 20◦ + 60◦ = 80◦, Figure 7: (Triangle ABC with the given angles) REFERENCES 1116 Figure 8: (Triangle AC11C3, equal to triangle ABC, placed inside the circle) 3. Concluding Remarks This work deals with the solution of integer angled triangle problems in geometry. Usu- ally, these problems are solved by the method of additional constructions, which requires a lot of experience and geometric intuition as there is no general algorithm for solving problems by this method. In this work, we present the method of auxiliary circle divided into equal parts, justification of which is based on one property of diagonals of regular polygons: three diagonals in regular polygons intersect each other at one point. From the problems solved above it can be seen that our method is simple and easy. References [1] S. Aliyev, Sh. Aghazade, and G. Abdullayeva. Using area method in secondary school geometry. European Journal of Pure and Applied Mathematics, 14(2):601–607, 2021. [2] S. Aliyev, Sh. Hamidova, and G. Abdullayeva. Some applications of ptolemy’s theorem in secondary school mathematics. European Journal of Pure and Applied Mathematics, 13(1):180–184, 2020. [3] S. Krackov. Multivariate visual-graphic representation of mathematical problems. Mathematika v şkole, (1), 2013. (in Russian). REFERENCES 1117 [4] Y.P. Ponarin. Elementary geometry. MTsNMO, Moscow, 2008. (in Russian). [5] Michael Serra. Discovering Geometry An Investigative Approach. Key Curriculum Press, 2003. [6] I.F. Sharigin. Geometry. Drofa, M., 1999. (in Russian). [7] Hung-Hsi Wu. Teaching Geometry in Grade 8 and High School According to the Com- mon Core Standards. 2013. October 16, 2021, 202 p.