Some Metric Properties of Semi-Regular Hexagons and a Constructive Task

Home , Equilateral triangle, Hexagon, Right triangle, Simple polygon

Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.9, (2020), Issue 2, pp.96-106

SOME METRIC PROPERTIES OF SEMI-REGULAR HEXAGONS AND A CONSTRUCTIVE TASK

NENAD STOJANOVIC´

ABSTRACT. A simple polygon A A ...A that has equal either all sides or all inte- A6 ≡ 1 2 6 rior angles is called a semi-regular hexagon. In terms of this deﬁnition, we can distinguish between two types of semi-regular hexagons: equilateral (they have equal all sides and different interior angles) and equiangular (they have equal interior angles and different sides). Unlike regular polygons, one characteristic element is not enough to analyze the metric properties of semi-regular polygons, so the additional one is needed. To select this additional characteristic element, note that regular triangles A A A , A A A △ 1 3 5 △ 2 4 6 can be inscribed to each semi-regular equilateral hexagon by joining the vertices. Let us ∠ use the mark δ = (a, b1) to mark the angle between side a of the semi-regular hexagon and side b of inscribed regular triangle A A A This parameter is present in every 1 △ 1 3 5 semi-regular hexagon, therefore, in interpreting the metric properties of the semi-regular ∠ hexagon in this paper, in addition to side a, we also use angle δ = (a, b1). The manner in which convexities, the possibility of construction, the calculation of the surface area and the problem of the inscribed circle and other metric properties depend on side a and an- ∠ gle δ = (a, b1) is elaborated in this paper. A formula has been designed to calculate the surface area of the semi-regular equilateral hexagon depending on parameters a and δ, as well as side a and radius r of the inscribed circle. The problem of the relation between the sides of inscribed triangles i ; i = 1, 2 is also pondered upon. P3

1. 1 INTRODUCTION

A simple polygon 6 A1 A2 A3 A4 A5 A6 that has equal all sides and different interior angles is called a semi-regularA ≡ hexagon [1]. In terms of this deﬁnition, we can distinguish between two types of semi-regular hexagons: equilateral hexagons (they have equal all sides and different interior angles) and equiangular hexagons (they have equal interior angles and different sides). We consider vertex Ai, i = 1,...,6 of a semi-regular equilateral hexagon to be in an even, or in an odd position, if index i is an even, or an odd number, respectively,(Figure1),[1],[2]. In this paper, we dealt with the metric properties of convex equilateral semi-regular hexagons. Unlike with regular polygons, the length of one of the sides is not sufﬁcient for analyzing the metric properties of equilateral semi-regular polygons, so another characteristic element is needed,[3-5].

2010 Mathematics Subject Classiﬁcation. Primary 51MO5; Secondary 11DO4. Key words and phrases. semi-regular polygons,semi-regular hexagon, an inscribed and circumscribed circle, a ratio of surface areas of semi-regular polygons.

96 Some metric properties of semi-regular hexagons and constructive task

For the selection of this characteristic element of equilateral semi-regular hexagons, note that triangles A1 A3 A5, A2 A4 A6 can be inscribed to every semi-regular equilateral hexagon by joining△ odd, or△ even vertices. Equilateral triangles 1 A A A , 1 A A A are the inscribed triangles to P3 ≡ △ 1 3 5 P3 ≡ △ 2 4 6 semi-regular hexagon 6. Let us mark the sides of these triangles with b1, and b2, respec- A π ∠ tively, and their interior angles with γ; γ = 3 . Let us use the mark δ = (a, bi); i = 1, 2 to mark the angle between side a of semi-regular hexagon and the side of the inscribed A6 regular triangle. For triangle 1 A A A the following applies δ = ∠(a, b ). Isosce- P3 ≡△ 1 3 5 1 les triangles A1 A2 A3, A3 A4 A5 and A5 A6 A1 are triangles that border semi-regular hexagon (Figure△ 1), [6].△ △ A6 In this way, besides side Ai Ai+1 = a; Ai+1 = A1, i = 1,2,...,6 of the semi-regular ∠ hexagon, we have also included the other characteristic element, i.e. angle δ = (a, b), b1 = b which is required in analyzing the metric properties. In this paper, the following marks are used: - the side of semi-regular polygon 6 is marked with a, - the number of sides of the ”inscribed”P regular polygon is marked with n - the side of regular triangle i ; i = 1, 2 ”inscribed” to semi-regular hexagon = P3 A6 A1 A2...A6, and constructed by joining its even vertices, or odd vertices is marked with bi, i N, i = 1, 2. ∈ - the interior angles of the semi-regular hexagon at the vertices of ”inscribed” regular triangle 1 A A A are marked with α, and at the other vertices with β. P3 ≡△ 1 3 5 For the interior angles the following applies π π π π π α = + 2δ, β = π 2δ = + 2( δ)= + 2ϕ, ϕ = δ. (1.1) 3 − 3 3 − 3 3 − The parameters which are used in the analysis of the metric properties of the semi- ∠ regular hexagon are the length of its side a and the value of angle δ = (a, b1) where b1 indicates the side of the equilateral triangle inscribed to it and formed by joining the a,δ odd vertices. Mark 6 indicates that the semi-regular equilateral convex hexagon is determined by side aAand angle δ,[5].

2. MY RESULT

2.1. The convexity of a semi-regular hexagon. The following theorem shows the dependence of the convexity of a semi-regular equilateral hexagon on angle δ = ∠(a, b), where b = b is the side of inscribed regular triangle 1 A A A . 1 P3 ≡△ 1 3 5 Theorem 2.1. Semi-regular equilateral hexagon 6 with interior angles defined by relations (1.1) is: A π π 1) convex for all values δ 0; 3 , (for δ = 6 , 6 it is a regular hexagon), ∈h iπ π A 2) non-convex for all values δ [ 3 , 2 and >∈ π i 3) not defined for all values δ 2 . Proof: By definition, a simple polygon is said to be convex, if all its interior angles are smaller than π(Figure 1),[2]. Furthermore, using this definition, it is easy to determine

97 Nenad Stojanovic

Figure 1. Basic elements of a semi-regular equilateral hexagon

< < π the values of interior angles α, β for angle 0 ϕ 3 . π < < π π < < 2π 1. For all δ 0; 6 from inequality 0 δ 6 we get that 3 α 3 , from which it follows that∈ h all valuesi of the interior angles are equal to angle α < π and are in the π 2π interval 3 ; 3 . By a similarh procedure,i it is easy to show that the interior angles equal to angle β are < < π also smaller than π. Indeed, from inequality 0 δ 6 by a series of elementary 2π < < transformations we find that 3 β π, that is, all values of the interior angles equal 2π to angle β are in interval 3 ; π . Since all the interior anglesh of thei semi-regular equilateral hexagon are smaller than π, π for all angle values δ 0; 6 it is convex for the values of angle δ. π ∈h i For δ = 6 the interior angles of the hexagon are equal, so for this value of angle δ,itisa convex regular hexagon. π π π < < π 2. If δ 6 ; 3 ] then from inequality 6 δ 3 we find that the values of the interior ∈ h 2π angles of a semi-regular hexagon equal to angle α are in interval 3 ; π . By a similar procedure, it is shown that the values of the interior angles equal to angleh βi are in interval π 2π 3 ; 3 . Based on this, it follows that the semi-regular equilateral hexagon is also convex h i π π for all angle values δ 6 ; 3 . ∈hπ π i π < π 3. Let it be that δ [ 3 ; 2 . Starting from inequality 3 δ 2 we find that for the ∈ i ≤4π interior angles equal to angle α the interval of values is [π; 3 . Therefore, all the interior angles equal to angle α are greater than π, then the semi-regulari equilateral hexagon is π π non-convex for all the values of angle δ [ 3 ; 2 . π ∈ i 4. If δ 2 , all the interior angles that are equal to angle α are greater than or equal to angle π≥, while all the interior angles are equal to angle β 0. So, for these values of angle δ, the semi-regular hexagon does not exist. ≤

98 Some metric properties of semi-regular hexagons and constructive task

Figure 2. Semi-rectangular hexagon of side a and angle δ

2.2. Surface area of a semi-regular hexagon. The dependence of the surface of a convex equilateral semi-regular hexagon on the length of its side a and angle δ = ∠(a, b) 1 closed with sides a and b of inscribed equilateral triangle 3 A1 A3 A5 is shown in a Theorem P ≡△

Theorem 2.2. The surface area (P6), of a semi-regular equilateral convex hexagon of a side a and angle δ = ∠(a, b) is determinedA in a relation π (P )= 2√3 a2 cos δ cos( δ). (2.1) A 6 · · 3 − Proof: Observe that the surface area of the semi-angular equilateral hexagon is equal to the sum of the surface areas of the isosceles triangles constructed above each side of regular triangle 1 A A A which is inscribed to it (Figure 2). That is P3 ≡△ 1 3 5 (P )= 3 (P )+ (P ) (2.2) A 6 ·A 1 A 2 where (P1) is the surface area of the isosceles bordering triangle, and (P2) is the surface areaA of the inscribed equilateral triangle. A As the surface area of inscribed regular triangle A1 A3 A5, where side b is b = 2a cos δ 2 △ · is (P )= b √3 , and with the surface area of isosceles triangle A A A , A 2 4 △ 1 2 3 (P )= a2 sin δ cos δ, we get that A 1

(P ) = 3 (P ) + (P ) A 6 ·A 1 A 2 = 3a2 sin δ cos δ + a2√3cos2 δ = √3a2 cos δ(√3sin δ + cos δ). √ π π If we notice that 3sin δ + cos δ = tan 3 sin δ + cos δ = 2cos( 3 δ) we ﬁnd that the surface area of the semi-regular equilateral hexagon is − π (P )= 2a2√3cos δ cos( δ). A 6 3 −

99 Nenad Stojanovic

π Corollary 2.1. For angle δ = 6 the hexagon is regular, and its surface is π π π 3√3 (P )= 2a2√3cos cos( )= a2 A 6 6 3 − 6 2 · which is consistent with the formula for the surface area of the regular hexagon, which is obtained 1 2 α π 2π if in (Pn) = 4 na cot 2 we add that n = 6 and α = 3 , wherein α = n is the central angle of the regularA n-side polygon. The surface area of the semi-regular hexagon can be calculated also if the length of side b of the inscribed regular triangle and angle δ are known, i.e. it applies that Corollary 2.2. The surface area of the semi-regular hexagon, with side b of the inscribed regular triangle and angle δ is calculated by the following formula √3 cos π δ (P ) = b2 3 − δ (2.3) A 6 2 · cos where b is the side of the inscribed regular triangle and angle δ=∠(a, b). Proof. From enequality b = 2a cos δ we express side a and use the formula for the surface area of the semi-regular hexagon

b π (P ) = 2( )2√3cos δ cos( δ) A 6 2cos δ 3 − √3 cos( π δ) = b2 3 − . 2 · cos δ 2.3. The radius of the inscribed circle. A circle can be inscribed, but not circumscribed to an equilateral semi-regular hexagon. Let us analyze the ratio of the radius of the inscribed circle (apothem), side a and angle δ, as well as the calculation of the surface area depending on the radius of the semi-regular equilateral hexagon. The following theorem shows the relation between the radius of the inscribed circle and parameters a and δ of the semi-regular hexagon. Theorem 2.3. The radius of the circle inscribed to the semi-regular equilateral hexagon with side a and angle δ 0; π , δ = π is expressed by the following ∈h 3 i 6 6 2a π r = cos δcos δ (2.4) √3 · 3 −

Proof: Note triangle A1 A2O (Figure 3). The following applies for the angles of this ∠ α π ∠ β π triangle OA1 A2 = 2 = 6 + δ and A1 A2O = 2 = 2 δ. From the special right α r − triangle OA1K it follows that tan 2 = x , and from there we have the following: α r = x tan (2.5) · 2 β and x = d(A1, K). Similarly, from special right triangle OKA2, it follows that tan 2 = r a x ,and from that we get − β r =(a x) tan , (2.6) − 2 and d(K, A )= a x. 2 −

100 Some metric properties of semi-regular hexagons and constructive task

Figure 3. Radius of the inscribed circle.

If we equate (2.5) and (2.6), and after completing the elemental transformations we ﬁnd that

a tan β x = · 2 . (2.7) α β tan 2 + tan 2 If we put this into equality (2.5) we get that a tan β tan α r = · 2 2 , (2.8) α β tan 2 + tan 2 i.e. a tan( π + δ) tan( π δ) r = · 6 2 − . (2.9) tan( π δ)+ tan( π + δ) 2 − 6 Since π sin( π δ) π cot δ + √3 tan( δ)= 2 − = cot δ, tan( + δ)= 2 − cos( π δ) 6 √3 cot δ 1 2 − − we ﬁnd that a cot δ(cot δ + √3) 2a π r = = ... = cosδcos( δ). √3(1 + cot2 δ) √3 3 − Proposition 2.1. The dependence of the radius of the circle inscribed on side b of the inscribed special right triangle and angle δ is given in the following formula:

1 π r = b cos( δ) (2.10) √3 · 3 − Proof. It is easy to show that the equality is valid using equality b = 2a cos δ and equality r = 2a cos δ cos( π δ) in which we ﬁnd that equality is valid. √3 · 3 −

101 Nenad Stojanovic

The manner of calculating the surface area of the semi-regular equilateral hexagon, if the length of the radius of inscribed circle r and angle δ are given, is seen in the following Theorem Theorem 2.4. The surface area of the semi-regular equilateral hexagon of the radius of inscribed circle r and angle δ 0, π and δ = π is calculated by the following formula ∈h 3 i 6 6 3√3 r2 (P )= . (2.11) A 6 2 cos δcos( π δ) 3 − Proof: If from equality r = 2a cos δ cos( π δ) we express side a using r and δ, we ﬁnd √3 · 3 − that √3 r a = · . (2.12) 2cos δ cos( π δ) 3 − If we substitute this equality in the formula for the surface area of semi-regular hexagon π (P )= 2a2√3cos δcos( δ) A 6 3 − after calculation we ﬁnd that 3√3 r2 (P )= . A 6 2 cos δcos( π δ) 3 − π Corollary 2.3. If δ = 6 the surface area of the semi-regular hexagon is equal to 3√3 r2 (P ) = A 6 2 cos δcos( π δ) 3 − 3√3 r2 = 2 cos π cos( π π ) 6 3 − 6 3√3 r2 = = 2√3r2 2 √3 2 ( 2 ) which coincides with the formula for the surface area of the regular hexagon if radius r of the inscribed circle is given.

π √3r Example 2.1. If we note that cos δcos( 3 δ)= 2a then from the formula for the surface area of the semi-regular hexagon we ﬁnd that the− surface area of the semi-regular hexagon, with given side a and radius r of the inscribed circle is calculated by the following formula

π √3r (P )= 2a2√3cos δcos( δ)= 2a2√3a2 = 3ar. (2.13) A 6 3 − 2a 2.4. A constructive task on semi-regular hexagons. The triangles inscribed to the equilateral convex hexagon by joining odd or even vertices generate a new equiangular convex hexagon, to which we can construct the inscribed equilateral triangles in the same way, and which then generate a new equilateral convex hexagon, etc. Between the semi- regular hexagons constructed in this way, a relation can be established, that is, the following theorem holds

102 Some metric properties of semi-regular hexagons and constructive task

Theorem 2.5. The intersection points of the sides of regular triangles 1 A A A , 2 P3 ≡ 1 3 5 P3 ≡ A2 A4 A6 inscribed to equilateral semi-regular hexagon 6 A1 A2 A3 A4 A5 A6 are the ver- A ≡ 1 2 tices of hexagon 6 B1B2B3B4B5B6, to which triangles 3 B1B3B5, 3 B2B4B6 are inscribed, andB their≡ intersection points of the sides are locatedT ≡ at the verticesT ≡ of hexagon C C C C C C , and so on.(Figure 4). The following needs to be proven: C6 ≡ 1 2 3 4 5 6 (1) that is an equiangular semi-regular hexagon, B6 (2) that 6 is an equilateral semi-regular hexagon, (3) thatC the ratio of the lengths of the smaller and the larger side of convex equiangular hexagon inscribed to equilateral convex semi-regular hexagon a,δ is in interval B6 A6 (0, √3 ) for all values of angle δ 0, π , 3 3 < R 3 ∈ π (4) that inequalities 1 r A 2 apply for all values of angle δ 0, 3 , with R being the B ≤ ∈ A length of the radius of the circle inscribed to hexagon 6, and r being the length of the radius of the circle circumscribed to hexagon . A B B6 Proof. 1.) Let A A A A A A (Figure 4.) be a semi-regular convex equilateral A6 ≡ 1 2 3 4 5 6 hexagon and 1 A A A ; 2 A A A be the regular triangles inscribed to it, the P3 ≡ 1 3 5 P3 ≡ 2 4 6 sides of which are marked with b1, b2 respectively. Let B1B2 = c1; B2B3 = c2 be the sides of this hexagon. To prove that 6 is an equiangular semi-regular hexagon, it needs to be proven that all interior anglesB are equal and the sides are different.

∠ Let us use mark δ = (a, b1) to mark the angle between side a of hexagon 6 and the 1 A side of equilateral triangle 3 A1 A3 A5. Note that for the interior angles at vertices P ≡∠ ∠ ∠ = π + A1, A3, A5 the following applies A1 ∼= A3 ∼= A5 3 2δ, and that for the interior ∠ ∠ ∠ = π + ( π ) angles at vertices A2, A4, A6 the following applies A2 ∼= A4 ∼= A6 3 2 3 δ . Since is an equilateral semi-regular hexagon, i.e. A A = A A = ... = A A =− a, it A6 1 2 2 3 6 1 follows that triangles A1B1 A2; A2B2 A3; A B A ; A B A ; △A B A ; △A B A are congruent, hence their interior angles △ 3 3 4 △ 4 4 5 △ 5 5 6, △ 6 6 1 2π ∠A B A ∠A B A ∠A B A ∠A B A ∠A B A ∠A B A = (2.14) 1 1 2 ≡ 2 2 3 ≡ 3 3 4 ≡ 4 4 5 ≡ 5 5 6 ≡ 6 6 1 3 are also congruent,(Figure 4). Note that the interior angles of hexagon 6 are transverse to the angles from the previous equality, and that the following is applicableB 2π ∠B = ∠B = ∠B = ∠B = ∠B = ∠B = ϕ = (2.15) 1 2 3 4 5 6 3 It follows that is the equiangular convex hexagon. Let us prove that the sides of B6 hexagon 6 are different from each other. Let us noteB triangles B A B ; B A B ; B A B . Let us observe now B A B . It △ 1 2 2 △ 3 4 4 △ 5 6 6 △ 1 2 2 has been shown that A1B1 A2 ∼= A2B2 A3 and from that it follows that line segments △ △ ∠ π B1 A2 = A2B2 are congruent. Since the interior angle is at vertex A2 = and point ∼ ∠ 3 S1 is on the bisector of base B1B2 it follows that S1 is a right angle and B1 A2S1 is ∠ π ∠ △π orthogonal to the angle at vertex B1 = 3 . Similarly, it turns out that B2 = 3 . On the basis of this, it follows that B1 A2B2 is an equilateral triangle and B1B2 = c1 is the side of hexagon (Figure 4). △ B6

103 Nenad Stojanovic

Similarly, triangles B3 A4B4; B5 A6B6 are proven to be equilateral, i.e. the following applies B A B △ B A B △ B A B . △ 1 2 2 ≡△ 3 4 4 ≡△ 5 6 6 Let us note now that B2 A3B3. From the already seen congruence A B B = A B B△ it follows that B A = B A . Since ∠A = π it follows that △ 2 2 1 ∼ △ 3 3 4 2 3 ∼ 3 3 3 3 B2 A3B3 is an equilateral triangle and that side B1B3 = c2 is the side of hexagon 6. In the△ same way it is shown that B A B ; B A B are mutually congruent equilateralB △ 4 5 5 △ 6 1 1 triangles, with bases B2B3 = B4B5 = B6B1 = c2. Equilateral triangles B1 A2B2 and B A B are not congruent, i.e. the following applies B B = c = B△B = c , and △ 2 3 3 1 2 1 6 1 3 2 therefore, based on this we can conclude that equiangular hexagon 6 does not have equal sides, i.e. it is an equiangular semi-regular hexagon. B 2) Let us show that 6 C1C2C3C4C5C6 is an equilateral semi-regular hexagon. Note, C ≡ ∠ ∠ for example, that, from the congruence of interior angles B1 ∼= B2 and sides B6B1 ∼= B2B3 = c2 of equilangular hexagon 6 and B1B2 ∼= B2B1 = c1 the congruence of triangles B B B = B B B follows. FromB the congruence of these triangles, follows the △ 6 1 2 ∼ △ 1 2 3 congruence of sides B2B4 ∼= B4B6 ∼= B6B2 = b. Similarly, the congruence of triangles B3B4B5 ∼= B4B5B6 ∼= B5B6B1 ∼= B6B1B2 is shown. From the congruence of these△ triangles, the△ congruence△ of equilateral△ triangles B1B3B5; B2B4B6 follows. From the congruence of these triangles, the congruence of △ ∠ △ ∠ = π angles C6B1C2 ∼= C6B2C2 3 follows. Furthermore, from the congruence of special right triangles B C T = B C T the △ 1 1 1 ∼ △ 2 1 1 congruence of sides B1C1 ∼= B2C2 follows, and from the congruence of transverse angles ∠C C B = ∠C C B the congruence of triangles B C C = B C C follows. The 6 1 1 ∼ 2 1 2 △ 1 1 6 ∼ △ 2 1 2 congruence of these triangles implies the equality of sides C6C1 ∼= C1C2 = a1 of hexagon 6. A similar procedure shows that all sides of hexagon 6 are equal, that is, it is an equilateralC hexagon. The difference of angles can be easily pCroven.

Figure 4. Relationship between the inscribed semi-regular hexagons

104 Some metric properties of semi-regular hexagons and constructive task

3). To determine the ratio of the sides of inscribed semi-regular hexagon , let us B6 express the lengths of the sides relative to the length of side a of hexagon 6 and angle ∠ A δ = (a, b1) (Figure 4). Note equilateral triangle B1 A2B2. From special right triangle A1 A2T1 we find that side c of hexagon is given△ by the following equality △ 1 B6 2√3 c = a sin δ. (2.16) 1 3 And from isosceles triangle A6 A1 A2 with base A2 A6 = A2B1 + B1B6 + B6 A6 = 2c1 + c2 △ ∠ ∠ π and with the value of angles being A2 A6 = 3 δ we find that side c2 of hexagon is given by the following equality ≡ − B6 2√3 π c = a sin( δ). (2.17) 2 3 3 − From equalities 2.16 and 2.17 we find that c sinδ 1 = (2.18) c sin( π δ) 2 3 − ∠ Let us consider the dependence of the equality from the value of angle δ = (a, b1). π Note that for δ = 6 , c1 = c2 and in this case, hexagons 6 and 6 are regular hexagons. πB A π Since semi-regular hexagon 6 is convex for δ (0, 3 ) and δ = 6 we distinguish two cases: A ∈ 6 (1) If δ (0, π ) then π δ > δ and sin( π δ) > sin δ i.e. ∈ 6 3 − 3 − π 2√3 2√3 π sin δ < sin δ a sin δ < sin( δ) c < c . 3 − ⇔ 3 3 3 − ⇒ 1 2 Since for 0 < δ < π it is 0 < sin δ < 1 and 1 < sin( π δ) < √3 we find 6 2 2 3 − 2 that the ratio of the smaller side,c1, and the greater side, c2, of the semi-regular √3 hexagon is in interval (0, 3 ) i.e. the following inequalities are valid √ c 2 3 a sin δ sin δ √3 0 < 1 = 3 = < . 2√3 π π c2 a sin( δ) sin( 3 δ) 3 3 3 − − (2) In a similar procedure, we see that if δ ( π , π ) then π δ < δ and sin( π δ) < ∈ 6 3 3 − 3 − sin δ and c1 > c2. So the ratio of the length of smaller side c2 and larger side c1 √3 of semi-regular equiangular hexagon 6 is in interval (0, 3 ) i.e. the following inequalities apply B √ c 2 3 a sin( π δ) √3 0 < 1 = 3 3 − < . √ c2 2 3 sin δ 3 3 a In both cases, the ratio of the length of the smaller and larger sides of isosceles hexagon which is inscribed to equilateral hexagon a,δ is in interval (0, √3 ). B6 A6 3 4) Using the marks as in Figure 4, and the previously obtained results on the radius of the circle which is inscribed to the equilateral hexagon, we find that R = 2√3 a A 3 ·

105 Nenad Stojanovic cos δ cos π δ . Furthermore, from special right triangle B S O there is 3 − △ 1 1 2 √3 2 √3 2 a√3 rB =( 3 a cos δ) +( 3 a sin δ) , and after calculation we get rB = 3 , with a being the length of the side of semi-regular hexagon a,δ. Further on A6 R π π 1 A = 2cosδcos δ = cos 2δ + r 3 − − 3 2 . B π < < π To estimate the value of the ratio, note that for δ (0, 6 ) from inequality 0 δ 6 it π < π < 1∈< π < follows that 3 2δ 3 0. From this we get 2 cos(2δ 3 ) 1 i.e. the following − −< R π 1 < 3 − inequalities are true 1 r A = cos 2δ 3 + 2 2 . B π π − π < < π By similar reasoning, for δ ( 6 , 3 ) from inequality 6 δ 3 we get the following inequalities 0 < 2δ π < π ∈, and from this we get thatinequalty − 3 3 R π 1 3 1 < A = cos 2δ + < r − 3 2 2 B is valid. π a,δ R 3 Note that for δ = 6 semi-regular hexagon 6 is regular, and then it is r A = 2 . A B 2.5. Conclusion. By using characteristic elements, i.e. side a and angle between side a of the semi-regular hexagon and side b1 of the regular triangle inscribed to the semi-regular hexagon, one can express in a very simple way many of its metric properties, such as: convexity, possibility of construction, surface area calculation, radius of the inscribed circle, and some other properties. Only one problem in the ratio between the sides of i inscribed triangles 3, i = 1, 2 has been considered here, however, other relations may also be considered. P

REFERENCES [1] Hincin, Ja, A., Aleksandrov, S.P., Markusevic, I. Poligoni i Poliedri, Enclikopedija elementarne matem- atike, Knjiga cetvrta, Moskva, 1963. [2] Radojcic,M., Elementarna Geometrija, Naucna knjiga, Beograd, 1961. [3] Stojanovic, N., Neka metricka svojstva uopstenih polupravilnih poligona, Disertacija, Sarajevo 2015. [4] Stojanovic, N., Inscribed circle of general semi-regular polygon and some of its features, International Journal of Geometry, Vol.2., 2013, N0.1, pp. 5 - 22 . [5] N.Stojanovic,Some metric properties of general semi-regular polygons, Global Journal of Advanced Research on Classical and Modern Geometries,Vol.1, Issue 2,(2012) pp.39-56. [6] N.Stojanovic,V. Govedarica, Jedan pristup analizi konveksnosti i racunanju povrsine jednakostranih polupravillnih poligona, II MKRS, Zbornik radova,Trebinje,(2013),pp.87-105.

UNIVERSITY OF BANJA LUKA FACULTY OF AGRICULTURE BOSNIA AND HERZEGOVINA. Email address: [email protected]

106