Annales Universitatis Paedagogicae Cracoviensis Studia Mathematica XV (2016)

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1 OPEN Ann. Univ. Paedagog. Crac. Stud. Math. 15 (2016), DOI: /aupcsm FOLIA 182 Annales Universitatis Paedagogicae Cracoviensis Studia Mathematica XV (2016) Naga Vijay Krishna Dasari, Jakub Kabat Several observations about Maneeals - a peculiar system of lines Abstract. For an arbitrary triangle ABC and an integer n we define points D n, E n, F n on the sides BC, CA, AB respectively, in such a manner that AC n AB n CDn BD n, AB n BC n AEn CE n, BC n AC n BFn AF n. Cevians AD n, BE n, CF n are said to be the Maneeals of order n. In this paper we discuss some properties of the Maneeals and related objects. 1. Introduction Given an arbitrary triangle ABC we consider certain cevians [1] defined (for given integral n) in the following way. Definition 1.1 Let ABC be a triangle and let n be an integer. There are points D n, E n, F n on the sides BC, CA, AB respectively, satisfying AC n AB n CD n BD n, AB n BC n AE n CE n, BC n AC n BF n AF n. We call the cevians AD n, BE n, CF n the order n Maneeals of the triangle ABC. It is easy to check, that medians, bisectors and symmedians of a triangle are examples of the Maneeals with integer n 0, n 1, n 2 respectively. Furthermore, for the given triangle ABC, the points D n, E n, F n (uniquely determined by the integer n), are vertices of a new triangle, which is said to be the Maneeal s AMS (2010) Subject Classification: 51M04, 51M15, 51A20. Keywords and phrases: Maneeals, Maneeal s Points, Maneeals triangle of order n, Maneeal s Pedal triangle of order n, Cauchy-Schwarz inequality, Lemoine s Pedal Triangle Theorem.

2 [52] Naga Vijay Krishna Dasari, Jakub Kabat triangle of order n. For Maneeals AD n, BE n, CF n we can use the Ceva s theorem in order to prove that they are concurrent. Their point of intersection M n is said to be the Maneeal s point of order n. For given M n we can choose points P n, Q n, R n on the sides BC, CA, AB respectively, in such a manner that line segments M n P n, M n Q n, M n R n are perpendicular to the corresponding sides of the triangle ABC. Points P n, Q n, R n are the vertices of a next triangle [11], which is said to be the Maneeal s pedal triangle of order n. In the present note we investigate properties of Maneeal s points and pedal triangles. In this paper, for a given triangle ABC and an integer n the points D n, E n, F n, P n, Q n, R n should be always understood in accordance with the definitions given above. Furthermore, we will use the following notation: AB c, BC a, CA b, a n + b n + c n, XY Z - the area of triangle XY Z, ABC, n DnE nf n, n PnQ nr n, R - the radius of the circumcircle of triangle ABC, S - circumcenter of triangle ABC. 2. First properties Theorem 2.1 (Ceva s Theorem [4, 5]) For a given triangle with vertices A, B, and C and non-collinear points D, E, and F on lines BC, AC, AB respectively, a necessary and sufficient condition for the cevians AD, BE, and CF to be concurrent (intersect in a single point) is that BD CE AF DC EA F B. C E D A F B Theorem 2.2 (Maneeal s points) Maneeals of order n are always concurrent. Proof. Since the points D n, E n, F n lie always between the vertices A, B, C, they cannot be collinear. The equality BD n CD n CE n AE n AF n BF n bn c n cn a n an c n 1 together with Theorem (2.1) proves the assertion.

3 Several observations about Maneeals [53] In the next Lemma we collect a number of properties connected with Maneeals and related objects. Lemma 2.3 The lenghts of the segments of Maneeal s triangle of order n are given by the following formulas: E n F n 2 b2n c 2 (a n + c n ) 2 + b 2 c 2n (a n + b n ) 2 b n c n (a n + b n )(a n + c n )(b 2 + c 2 a 2 ) (a n + b n ) 2 (a n + c n ) 2, E n D n 2 a2n b 2 (c n + b n ) 2 + a 2 b 2n (c n + a n ) 2 a n b n (c n + a n )(c n + b n )(a 2 + b 2 c 2 ) (c n + b n ) 2 (c n + a n ) 2, F n D n 2 a2n c 2 (b n + c n ) 2 + a 2 c 2n (b n + a n ) 2 a n c n (b n + a n )(b n + c n )(a 2 + c 2 b 2 ) (b n + c n ) 2 (b n + a n ) 2. Lemma 2.4 The lenghts of the segments in which the Manneals of order n divide the sides of the triangle are given by the following formulas: BD n CE n AF n cn a b n + c n, CD n bn a b n + c n, an b c n + a n, AE n cn b c n + a n, bn c b n + a n, BF n an c b n + a n. Lemma 2.5 The Manneal s point M n of order n divides the Manneal s segments of order n in the following ratios: AM n M n D n cn + b n a n, BM n M n E n cn + a n b n, CM n M n F n an + b n c n. Lemma 2.6 It is also convenient to keep, for further reference, record of the following identities: AM n AD n cn + b n, BM n BE n cn + a n, CM n CF n an + b n, M n D n AD n M n E n BE n M n F n CF n an, bn, cn, (1)

4 [54] Naga Vijay Krishna Dasari, Jakub Kabat and AD n 2 BE n 2 CF n 2 b 2 c 2 (b n + c n ) 2 [(bn + c n )(b n 2 + c n 2 ) a 2 b n 2 c n 2 ], a 2 c 2 (a n + c n ) 2 [(an + c n )(a n 2 + c n 2 ) b 2 a n 2 c n 2 ], b 2 a 2 (b n + a n ) 2 [(bn + a n )(b n 2 + a n 2 ) c 2 b n 2 a n 2 ]. (2) Lemma 2.7 The Maneeal s point of order n and every pair of vertices of the given triangle ABC determine new triangles. Areas of these triangles are related to the areas of the triangle ABC in the following way: BMnC an, CMnA bn, BMnA cn. (3) We can also consider the triangles determined by pairs of points D n, E n, F n and one of the vertices of the triangle ABC. Their areas are given by the following formulas: a n c n BDnF n (b n + c n )(b n + a n ), a n b n CDnE n (c n + b n )(c n + a n ), b n c n AEnF n (a n + c n )(a n + b n ). Lemma 2.8 It is also worth to note some properties connected with the lengths of line segments between the vertices of pedal Maneeals triangles of order n, the Maneeals points of order n, and vertices of the triangle ABC. M n P n 2 an 1, M n Q n 2 bn 1, M n R n 2 cn 1. (5) (4) BP n 1 a 2 c 2 (a n + c n )(a n 2 + c n 2 ) b 2 a n c n 4a 2n 2 2, BR n 1 a 2 c 2 (a n + c n )(a n 2 + c n 2 ) b 2 a n c n 4c 2n 2 2, CP n 1 a 2 b 2 (a n + b n )(a n 2 + b n 2 ) c 2 a n b n 4a 2n 2 2, CQ n 1 a 2 b 2 (a n + b n )(a n 2 + b n 2 ) c 2 a n b n 4b 2n 2 2, AQ n 1 c 2 b 2 (c n + b n )(c n 2 + b n 2 ) a 2 c n b n 4b 2n 2 2, AR n 1 c 2 b 2 (c n + b n )(c n 2 + b n 2 ) a 2 c n b n 4c 2n 2 2. (6)

5 Several observations about Maneeals [55] Corollary 2.9 By definition we have BP n + CP n a, CQ n + AQ n b, BR n + AR n c. Hence, by adding all equations in (6), we get (a + b + c) (a n + b n + c n ) a 2 c 2 (a n + c n )(a n 2 + c n 2 ) b 2 a n c n 4a 2n a 2 c 2 (a n + c n )(a n 2 + c n 2 ) b 2 a n c n 4c 2n a 2 b 2 (a n + b n )(a n 2 + b n 2 ) c 2 a n b n 4a 2n a 2 b 2 (a n + b n )(a n 2 + b n 2 ) c 2 a n b n 4b 2n c 2 b 2 (c n + b n )(c n 2 + b n 2 ) a 2 c n b n 4b 2n c 2 b 2 (c n + b n )(c n 2 + b n 2 ) a 2 c n b n 4c 2n 2 2. Now we are in a position to prove the following new identity. Theorem 2.10 n 2 n n. a n b n c n (a n + b n )(b n + c n )(c n + a n ), Proof. We will start with simple consequences of the definition of Maneeals: A F n En M n B D n C From Lemma 2.4 we obtain AE n cn a n E nc, AE n AC BF n an b n F na. cn c n + a n, BF n an BA a n + b n.

6 [56] Naga Vijay Krishna Dasari, Jakub Kabat On the other hand we can observe, that AE n AC ABE n and BF n BA BE nf n. BEnA Finally we can express the area of the triangle BE n F n by the formula BEnF n BF n BA BEnA an a n + b n BE na an a n + b n AE n AC a n c n (a n + b n )(c n + a n ). an a n + b n Strictly analogously, we can provide the following formulas: c n a n BEnD n (c n + b n )(a n + c n ), a n c n BFnD n (a n + b n )(c n + b n ). c n c n + a n To end the first part of the theorem, we only need to note, that Finally we get n BEnD n + BEnF n BFnD n. c n a n n (c n + b n )(a n + c n ) + a n c n (a n + b n )(c n + a n ) a n c n (a n + b n )(c n + b n ) a n c n (a n + b n )(b n + c n )(c n + a n ) [(an + b n ) + (b n + c n ) (c n + a n )]. a Above consideration implies that n 2 n b n c n (a n +b n )(b n +c n )(c n +a n ). The last part of the proof is quite formal. n 2 a n b n c n (a n + b n )(b n + c n )(c n + a n ) a n b n c n 2 (a n + b n )(b n + c n )(c n + a n ) a2n b 2n c 2n a 2n b 2n c 2n a n b n c n 2 (a n + b n )(b n + c n )(c n + a n ) n.

7 Several observations about Maneeals [57] Now we pass to the Maneeal s pedal triangle P n Q n R n. Lemma 2.11 The area of Maneeals pedal triangle of order n is given by the following formula n 22n 2 n+1 R n 2 (a n + b n + c n ) 2 [a2 n + b 2 n + c 2 n ]. Furthermore, sides of Maneeals pedal triangle have the following lengths: P n Q n 2 (a n 2 + b n 2 )(a n + b n ) a n 2 b n 2 c 2, Q n R n 2 (b n 2 + c n 2 )(b n + c n ) b n 2 c n 2 a 2, R n P n 2 (c n 2 + a n 2 )(c n + a n ) c n 2 a n 2 b 2. Theorem 2.12 For any point X in the plane we have the following formula M n X 2 an AX 2 + b n BX 2 + c n CX 2 a n + b n + c n a 2 b 2 c 2 (a n + b n + c n ) 2 (an 2 b n 2 + b n 2 c n 2 + c n 2 a n 2 ). X B D n C Proof. From the Stewart Theorem [3] for triangle XBC, we get XB 2 D n C + XC 2 D n B BC [ XD n 2 + D n C D n B ], or equivalently XB 2 D nc BC + XC 2 D nb BC XD n 2 + D n C D n B. If we use formulas from Lemma 2.4 and make a few simple transformations, then we will get XD n 2 cn c n + b n XC 2 + bn c n + b n XB 2 b n c n (c n + b n ) 2 BC 2.

8 [58] Naga Vijay Krishna Dasari, Jakub Kabat A X M n B D n C Furthermore, an analogous consideration for the triangle AXD n shows that XM n 2 AM n AD n D nx 2 + D nm n AD n AX 2 AM n D n M n AM n AD n D nx 2 + D nm n AD n AX 2 AM n AD n D nm n AD n 2 AD n ( c n + b n )( c n a n + b n + c n c n + b n XC 2 + bn c n + b n XB 2 bn c n a 2 ) (c n + b n ) 2 a n + a n + b n + c n AX 2 an (c n + b n ) (a n + b n + c n ) 2 AD n 2 Now we will use formula an AX 2 + b n BX 2 + c n CX 2 a n + b n + c n an (c n + b n ) (a n + b n + c n ) 2 AD n 2. AD n 2 a 2 b n c n (a n + b n + c n )(b n + c n ) b 2 c 2 (b n + c n ) 2 [(bn + c n )(b n 2 + c n 2 ) a 2 b n 2 c n 2 ] to simplify the following part of the main formula a 2 b n c n (a n + b n + c n )(b n + c n ) an (c n + b n ) (a n + b n + c n ) 2 AD n 2 a 2 b n c n (a n + b n + c n )(b n + c n ) an (c n + b n ) (a n + b n + c n ) 2 b 2 c 2 (b n + c n ) 2 [(bn + c n )(b n 2 + c n 2 ) a 2 b n 2 c n 2 ]

9 Several observations about Maneeals [59] Finally we have a 2 b 2 c 2 (a n + b n + c n ) 2 [an 2 b n 2 + b n 2 c n 2 + c n 2 a n 2 ]. M n X 2 an AX 2 + b n BX 2 + c n CX 2 a n + b n + c n a 2 b 2 c 2 (a n + b n + c n ) 2 (an 2 b n 2 + b n 2 c n 2 + c n 2 a n 2 ). (7) Corollaries 2.13 From (7) we obtain for n 0, 1: M 0 X 2 AX 2 + BX 2 + CX 2 a2 + b 2 + c 2, 3 9 M 1 X 2 a AX 2 + b BX 2 + c CX 2 abc a + b + c (a + b + c). Note that M 0 is the centroid of the triangle ABC and M 1 is the incenter of ABC. From (7), (1), (2) we obtain the last corollary, which states, that the distance between two Maneeal s points, of order m and n, is given by the following formula where M m M n 2 1 (a m + b m + c m ) 2 (a n + b n + c n ) 2 V, V a 2{ [b m c n b n c m ] 2 [b m c n b n c m ][a m (b n c n ) a n (b m c m )] [a m b n a n b m ][a m c n a n c m ] } + b 2{ [c m a n c n a m ] 2 [c m a n c n a m ][b m (c n a n ) b n (c m a m )] [b m c n b n c m ][b m a n b n a m ] } + c 2{ [a m b n a n b m ] 2 [a m b n a n b m ][c m (a n b n ) c n (a m b m )] [c m a n c n a m ][c m b n c n b m ] }. In particular if we let m 1, n 0, we will get M 1 M 0 2 Corollary 2.14 From (2.12) for XS we have 1 (a + b + c) [a AM b BM c CM 0 2 abc]. M n S 2 an R 2 + b n R 2 + c n R 2 a n + b n + c n a 2 b 2 c 2 (a n + b n + c n ) 2 (an 2 b n 2 + b n 2 c n 2 + c n 2 a n 2 ) 0.

10 [60] Naga Vijay Krishna Dasari, Jakub Kabat Therefore, we have In particular, for n 1, since therefore R 2 a2 b n c n + b 2 a n c n + c 2 a n b n qn 2. R 2 abc (a + b + c) 2Rr R 2r. Other proofs of Euler s inequality you can find at [2, 13, 14, 6, 7, 8, 9, 10]. Now we can obtain a few relationships from the Cauchy-Schwarz inequality. They will be important part of proofs of several subsequent theorems. Lemma 2.15 Any non-zero real numbers a, b, c satisfy the following inequalities: ( a 2n a 2 + b2n b 2 + c2n c 2 ) (an + b n + c n ) 2 a 2 + b 2 + c 2, (8) (a 2n + b 2n + c 2n ) 1 3 (an + b n + c n ) 2, (9) ( a 2 ) a n + b2 b n + c2 (a + b + c)2 c n a n + b n + c n. (10) Proof. Indeed, these are special cases of the Cauchy-Schwarz inequality with the substitutions (x x x 2 3)(y y y 2 3) (x 1 y 1 + x 2 y 2 + x 3 y 3 ) 2 x 1 an a, x 2 bn b, x 3 cn c, y 1 a, y 2 b, y 3 c for (8) x 1 a n, x 2 b n, x 3 c n, y 1 y 2 y 3 1 for (9) and a 2 b x 1 a n, x 2 c 2 b n, x 2 3 c n, y 1 a n, y 2 b n, y 3 c n for (10). Lemma 2.16 (Inequality of arithmetic and geometric means) For any real numbers x 1,..., x n there is x x n n n x 1... x n. (11)

11 Several observations about Maneeals [61] The Symmedian point M 2 has a special feature, which we will describe by following Theorem 2.17 Let S(n) M n P n 2 + M n Q n 2 + M n R n 2. Then S(2) S(n) for all n Z. Proof. Using (5) we get [ S(n) 4 2 a 2n ] qn 2 a 2 + b2n b 2 + c2n c 2. A R n Q n M n B P n C Using (8) we get S(n) 4 2 q 2. Now, an easy computation shows, that This finishes the proof. S(2) M 2 P M 2 Q M 2 R q 2. Not only the Symmedian point, but also the Centroid M 0, has a special feature: Theorem 2.18 Let T (n) a 2 M n P n 2 + b 2 M n Q n 2 + c 2 M n R n 2. n Z. Then T (0) T (n) for all

12 [62] Naga Vijay Krishna Dasari, Jakub Kabat Proof. Using (5) we obtain By (9) we get Now an easy computation shows, that This finishes the proof. T (n) 4 2 q2n qn 2. T (n) a 2 M 0 P b 2 M 0 Q c 2 M 0 R Theorem 2.19 Let W (n) a M + b np n M + c nq n M nr n. Then W (1) S(n) for all n Z. Proof. Using (5) we get Using (10) we get W (n) q ( n a 2 ) 2 a n + b2 b n + c2 c n. W (n) (an + b n + c n ) ( a 2 ) 2 a n + b2 b n + c2 c n Now an easy computation shows, that This finishes the proof. a M 1 P 1 + b M 1 Q 1 + c M 1 R 1 (a + b + c)2. 2 (a + b + c)2. 2 Theorem 2.20 Let K(n) M n P n M n Q n M n R n. Then K(0) K(n) for all n Z. Proof. Using (5) we obtain By a special case of 2.16 we get K(n) 8an 1 b n 1 c n 1 (a n + b n + c n ) 3 3. (a n + b n + c n ) 3 3 a n b n c n. Equivalently, 1 (a n + b n + c n ) a n b n c n.

13 Several observations about Maneeals [63] Using this inequality, we get Furthermore, K(n) 8an 1 b n 1 c n 1 27a n b n c n abc 3. (a n + b n + c n ) 3 3 a n b n c n for n 0. Hence, M 0 P 0 M 0 Q 0 M 0 R 0 8a 1 b 1 c 1 This equality finishes the proof abc 3. Theorem 2.21 Let π be the circumcircle of the triangle ABC. For any n Z we choose points X n, Y n, Z n on π in such a manner, that chords AX n, BY n, CZ n contain the Maneeals AD n, BE n, CF n respectively. Let D n, E n, F n be intersection points of AX n, BY n, CZ n and Y n Z n, Z n X n, X n Y n respctively. Let finally m be an integer, X n D m, Y n E m, Z n F m be order m Maneeals of the triangle X n Y n Z n. Then the following conditions hold: D 2 D 2, E 2 E 2, F 2 F 2, if G m is an m-order Maneeal s point of triangle X n Y n Z n, then G 2 M 2. A Z n D n Y n F n E n M n E n F n B D n C X n

14 [64] Naga Vijay Krishna Dasari, Jakub Kabat Proof. Line segments BC and AX n are the chords of circle π, and D n is their point of intersection. Hence, Using 2.4 we get Strictly analogously we get E n Y n BD n D n C AD n D n X n. D n X n Now we can use (1) and (2) to obtain a 2 b n c n (b n + c n ) 2 AD n. b 2 a n c n (a n + c n ) 2 BE n, F c 2 a n b n nz n (a n + b n ) 2 CF n. M n X n M n D n + D n X n Strictly analogously we get Using (2) again, we get a2 b n c n + b 2 c n a n + c 2 a n b n (a n + b n + c n )(b n + c n ) AD n. a n AD n (a n + b n + c n ) + a 2 b n c n (b n + c n ) 2 AD n M n Y n a2 b n c n + b 2 c n a n + c 2 a n b n (a n + b n + c n )(c n + a n ) BE n, M n Z n a2 b n c n + b 2 c n a n + c 2 a n b n (a n + b n + c n )(a n + b n ) CF n. AX n AD n + D n X n AD n 2 a 2 b n c n + AD n (b n + c n ) 2 AD n c2 b n + b 2 c n (b n + c n ) AD n. Analogously we get BY n c2 a n + a 2 c n (a n + c n ) BE n, CZ n b2 a n + a 2 b n (a n + b n ) CF n. Now we observe, that triangles X n M n Z n and CM n A are similar. Indeed, we only need to note, that M n is the intersection point of chords AX n and CZ n and that respective angles are right. From the similarity, we derive that Since XnZn CA X n M n Z n CM n A X nm n 2 CM n 2 X nz n 2 CA 2 Z nm n 2 AM n 2. ZnMn AM n, therefore if we use (1), we get X n Z n AC Z nm n AM n b a 2 b n c n + b 2 c n a n + c 2 a n b n (a n + b n )(b n + c n ) AD n CF n.

15 Several observations about Maneeals [65] Strictly analogously we get X n Y n c Y n Z n a a 2 b n c n + b 2 c n a n + c 2 a n b n (a n + c n )(b n + c n ) AD n BE n, a 2 b n c n + b 2 c n a n + c 2 a n b n (a n + c n )(a n + b n ) CF n BE n. By (3) and similarity of respective triangles we have X n M n Z n X nz n 2 CA 2 By the same taken we get Y n M n Z n X n M n Y n CM n A b n (a 2 b n c n + b 2 c n a n + c 2 a n b n ) 2 (a n + b n + c n )(a n + b n ) 2 (b n + c n ) 2 AD n 2 CF n 2. a n (a 2 b n c n + b 2 c n a n + c 2 a n b n ) 2 (a n + b n + c n )(a n + b n ) 2 (a n + c n ) 2 BE n 2 CF n 2, c n (a 2 b n c n + b 2 c n a n + c 2 a n b n ) 2 (a n + b n + c n )(c n + b n ) 2 (c n + a n ) 2 AD n 2 BE n 2. No we can determine the area of the triangle Z n BY n. Since therefore ZnM ny n ZnBY n Y nm n BY n, ZnBY n BY n Y n M n ZnM ny n an (a 2 b n c n + b 2 c n a n + c 2 a n b n )(c 2 a n + a 2 c n ) (a n + b n ) 2 (a n + c n ) 2 BE n 2 CF n 2. Strictly analogously we get XnBY n cn (a 2 b n c n + b 2 c n a n + c 2 a n b n )(a n c 2 + c n a 2 ) (c n + b n ) 2 (c n + a n ) 2 BE n 2 AD n 2. Furthermore we have the following relation On the other hand since we can conclude, that Z n E n X n E n Z nby n XnBY n an (b n + c n ) 2 AD n 2 c n (b n + a n ) 2 CF n 2. Y n Z n Y n X n a(bn + c n ) AD n c(b n + a n ) CF n, Y n Z n 2 Y n X n 2 Z ne n X n E n if and only if n 2.

16 [66] Naga Vijay Krishna Dasari, Jakub Kabat On the other hand, by definition, the cevian Y n E m, is an order m Maneeal of triangle X n Y n Z n if, and only if, the following condition holds Z n E m X n E m Y nz n m Y n X n m. Now it is easy to see, that the above condition is satisfied for n m 2. Hence, E 2 E 2. We can provide strictly analogously, that D 2 D 2, F 2 F 2. In particular, cevians X 2 D 2, Y 2 E 2, Z 2 F 2 are Symmedians of the triangle X 2 Y 2 Z 2. The fact, that G 2 M 2 we conclude by the definition (construction) of points X 2, Y 2, Z 2. Theorem 2.22 (Lemoine s Pedal Triangle Theorem [15]) Let D n E n F n be the order n Maneeal s triangle and P n Q n R n be the order n pedal Maneeal s triangle of the given triangle ABC. We can choose points T n, U n, W n, on sides R n P n, Q n P n, Q n R n, respectively, in such a manner that cevians R n U n, P n W n, Q n T n contain the line segments R n M n, P n M n, Q n M n respectively. Then the following conditions hold: if R n R m, Q n Q m, P n P m are order m Maneeals of the triangle P n Q n R n, then U 2 R 0, W 2 P 0, T 2 Q 0, Symmedian point of triangle ABC is the centroid of its pedals triangle P n Q n R n, AF1 F 1B P1U1 U 1Q 1, BD1 D 1C Q1W1 W 1R 1, CE1 E 1A R1T1 T 1P 1. A Rn F n W n Q n M n E n T n U n B D n P n C Proof. We use (5) to make simple computations P n U n U n Q n R nm np n RnM nq n 1 2 R nm n M n P n sin ( R n M n P n ) 1 2 R nm n M n Q n sin ( R n M n Q n )

17 Several observations about Maneeals [67] and get 2c n 1 2a n 1 (a n +b n +c n ) (a n +b n +c n ) 2c n 1 2b n 1 (a n +b n +c n ) an 2 b n 2 sin ( 180 B) (a n +b n +c n ) sin ( 180 A) an 1 sin B b n 1 sin A P n U n U n Q n an 2. (12) bn 2 In particular, if we choose n 2, we obtain P n U n U n Q n. Therefore, by definition of order m Maneeals of the triangle P n Q n R n the cevian R 2 U 2 is a median of this triangle. Finally R 2 U 2 R 2 R 0. Strictly analogously we can show, that P 2 W 2 P 2 P 0, Q 2 T 2 Q 2 Q 0. On the other hand, if we put n 1 into (12), we obtain And analogously P 1 U 1 U 1 Q 1 b a AF 1 F 1 B. Q 1 W 1 W 1 R 1 c b BD 1 D 1 C,, R 1T 1 T 1 P 1 a c CE 1 E 1 A. For further properties see [12]. Final remarks. In this article we introduced Maneeal s points, which to the best of our knowledge, have non been studied in the literature before. We shared a number of properties of these points, related lines and triangles. There is surely much more to discover. We hope, this note will sparkle some interest in the construction and will lead further research in this area of very classical triangle geometry. Acknowledgement. The authors would like to thank Prof. Tomasz Szemberg for his help while preparing this manuscript. Additional thanks go to the referee for valuable comments and suggestions. References [1] Altshiller-Court, Nathan. "College geometry. An introduction to the modern geometry of the triangle and the circle." Reprint of the second edition. Mineola, NY: Dover Publications Inc.: Cited on 51. [2] Andreescu, Titu, and Dorin Andrica. "Proving some geometric inequalities by using complex numbers." Educaţia Matematică 1, no. 2 (2005): Cited on 60. [3] Chau, William. "Applications of Stewart s theorem in geometric proofs." Accessed April 7, pdf. Cited on 57.

18 [68] Naga Vijay Krishna Dasari, Jakub Kabat [4] Coxeter, Harold S.M. Introduction to Geometry. New York-London: John Wiley & Sons, Inc., Cited on 52. [5] Coxeter, Harold S.M., and S.L. Greitzer. "Geometry revisited." Vol. 19 of New Mathematical Library. New York: Random House Inc., Cited on 52. [6] Dasari, Naga Vijay Krishna. "The new proof of Euler s Inequality using Extouch Triangle." Journal of Mathematical Sciences & Mathematics Education 11, no. 1 (2016): Cited on 60. [7] Dasari, Naga Vijay Krishna. "The distance between the circumcenter and any point in the plane of the triangle." GeoGebra International Journal of Romania 5, no. 2 (2016): Cited on 60. [8] Dasari, Naga Vijay Krishna. "A few minutes with a new triangle centre coined as Vivya s point." Int. Jr. of Mathematical Sciences and Applications 5, no. 2 (2015): Cited on 60. [9] Dasari, Naga Vijay Krishna. "The new proof of Euler s Inequality using Spieker Center." International Journal of Mathematics And its Applications 3, no. 4-E (2015): Cited on 60. [10] Dasari, Naga Vijay Krishna. "Weitzenbock Inequality 2 Proofs in a more geometrical way using the idea of Lemoine point and Fermat point." GeoGebra International Journal of Romania 4, no. 1 (2015): Cited on 60. [11] Gallatly, William. Modern geometry of a triangle. London: Francis Hodgson, Cited on 52. [12] Honsberger, Ross. "Episodes in nineteenth and twentieth century Euclidean geometry." Vol 37 of New Mathematical Library. Washington, DC: Mathematical Association of America, Cited on 67. [13] Nelsen, Roger B. "Euler s Triangle Inequality via proofs without words." Mathematics Magazine 81, no. 1 (2008): Cited on 60. [14] Pohoaţă, Cosmin. "A new proof of Euler s Inradius Circumradius Inequality." Gazeta Matematica Seria B no. 3 (2010): Cited on 60. [15] Pohoaţă, Cosmin. "A short proof of Lemoine s Theorem." Forum Geometricorum 8 (2008): Cited on 66. Naga Vijay Krishna Dasari Department of Mathematics Kesava Reddy Educational Instutions Machilipatnam, Kurnool, Andhra Pradesh India vijay @gmail.com Jakub Kabat Institute of Mathematics Pedagogical University of Cracow Podchorazych Kraków Poland xxkabat@gmail.com Received: August 31, 2015; final version: September 5, 2016; available online: October 13, 2016.

A NEW PROOF OF PTOLEMY S THEOREM

A NEW PROOF OF PTOLEMY S THEOREM DASARI NAGA VIJAY KRISHNA Abstract In this article we give a new proof of well-known Ptolemy s Theorem of a Cyclic Quadrilaterals 1 Introduction In the Euclidean geometry,