A plane extension of the symmetry relation

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1 A plane extension of the symmetry relation Alexandru Blaga National College Mihai Eminescu Satu Mare, Romania The idea of this article comes as a response to a problem of symmetry. The defining of this new symmetry leads in particular cases at the symmetry to a line or to a point, in the plane. More exactly, considering the symmetric of a point P 0 from the sides of a triangle, to a line which includes one of the triangle s vertices,we obtain extensions of a classic problem. Let P 0 be a point on the side [BC] and a line d, C d and d AB. The symmetric of P 0 is defined lie this: a) S(P 0 ) = P 0, P 0 [AC]; b) [P 0P 0 ] d = {C 1 }; c) [P 0 C 1 ] [C 1 P 0] (figure 1). This ind of construction is possible in two ways: 1. If S is the midpoint of the segment [AP 0 ] and Z the midpoint of the segment [P 0 C], so [SZ] d = {C 1 } and SZ AC and the mid-segment in the triangle ABC. Hence, C 1 is the midpoint of the segment [P 0 P 0]. 1

2 . Let P 0 M AC, M d and MK BC, K AC. We have P 0 MKC a parallelogram, which means that the diagonals are halved, [P 0 K] [MC] = {C 1 }. So, from this construction we can deduce the terms in which this symmetric can be considered: in the first construction, d (SP 0 ) and in the second one M must be a point inside the triangle. In every situation that we are going to approach, the position of the points whose symmetric we will consider, they will meet one of this conditions. It can be proved that they are equivalent. If line d is the bisector of the angle, then the defined symmetric is that one from the classic acceptance. It is necessary that P 0 (AC). Based on the sense defined by this symmetry at a), b) and c), we will note the symmetric of P 0 with S(P 0 ). We are interested in the following problem: Considering a point P 0 [BC] and S(P 0 ) = M, M [AC]. For d 1 BC, A d 1 and S(M) = R, R [AD] and for d [AC], B d, S(R) = V, V [BC]. Following this described method in this way: in relation to d, in relation to d 1, in relation to d we obtain the points M, R, V on the order. So we performed this symmetry in the order d, d 1, d six times. The problem that interests us is in which conditions V = P 0, so the point P 0 comes bac in the initial position. In support of this property, we consider the article [1].

3 In the mentioned article, it is proved the fact that after six steps it reaches the starting point, if the lines which the symmetry are related to are angle bisectors. It will be also studied the case in which after three steps the point come bac in the initial position and also the concurrence of the lines BM, CR, AV. Theorem 1. Let AN, CK, BI be three cevians of the triangle ABC and AN CK BI, where N (BC), K (AB), I (AC). If P 0 (BC) and CP 0 CB = 1, BK KA = 1 p 1, CN NB = 1 p (i) p 1 > 1 > 1 ; (ii) p 1 (1 p ) < 1; (iii) (1 + p 1 p 1 p ) > 1 so that then after six consecutive symmetries, the point P 0 comes bac in P 0. Proof. We will show the steps by which we mae the construction. Step 1 (figure 3). We have [AS] [SP ], [CZ] [ZP 0 ], CK [SP 0 ] = {L}. Using the Menelaus s Theorem in ABP 0 with transversal C L K, we have 1 1 p 1 AL LP 0 = 1, so LP 0 = 1 AL p 1 and LP 0 AP 0 = 1 1+p 1. We deduce that SL = SP 0 LP 0 = AP 0 AP 0 1+p 1 = AP 0 p 1 +1 in SP 0 Z with the transversal C C 1 L, we obtain 1 LP 0 SC 1 C 1 Z p 1 1 = p 1 1 = AM AM and MC AC p 1 > 1, this condition also assures that L [SP 0 ]. = p 1 1 p 1 3. Using the same theorem, but SC 1 SL C 1 Z = 1, as (1) and this fraction exists because

4 {V 1 }. Step (figure 4). We have [S 1 B] [S 1 M], [AU] [MU], [AN] [BM] = We have AM CN BV 1 = 1, p 1 1 AC NB V 1 M p 1 1 p BV 1 = 1, V 1M V 1 M BV 1 p 1 1. So S +p 1 1 1V 1 = S 1 M V 1 M = BM With the same theorem in S 1 MU, we deduce that 1 V1M = p 1 1 and V 1M BM = BM p 1 1 = BM p 1p p 1 +1 V 1 S 1 S1A 1 A 1 U S 1 A 1 = BR = p 1p p 1 +1 A 1 U RA p 1 (), so that BR = p 1p p BA condition (ii).moreover, from the condition (ii), we have V 1 M < BM {R 1 }.. = 1 and. which exists from the = SM. Step 3 (figure 5). We have [CW ] [W R], [BJ] [RJ], [BI] [W R] = 4

5 BR AI CR 1 BA IC R 1 R leads to AI CN BK IC NB KA p 1 +1 p 1 p CR 1 CR p 1p +p p 1 +1 W B 1 W J = CV = +p 1 1 CB = 1 and because the concurrence AN CK BI AI = 1, then = p IC 1p. We can write the relation = 1 and we deduce that R 1R = p 1p p 1 +1 ; W R R 1 R CR + p = From the Menelaus s Theorem in W RJ, we obtain (3), which exists from the condition (iii). Moreover, from the condition (iii), we have R 1 R < CR = W R. We get from the succesive application of the first three steps, the points M, R, U starting from the point P 0 [BC]. In the following calculations, we put +p 1 1 = p 3. Point V can tae the place of P 0, because CV CB = p 3 < 1: + p 1 1 < leads to p 1 (1 p ) < 1, so (ii). Step 4 (figure 6). We have [S A] [S V ], [CZ ] [Z V ], [CK] [S V ] = {L }. From the relations CV BK AL CB AK L V that L V = p 3 AL p 1, L V = AV and finally AM AC = p 1 p 3 p 1 > +p 1 1 CZ = 1 and V L S C 1 CV L S C 1 Z = 1, we conclude p 3 1+p 3, S L = AV p1 p 3 S p 1 +p 3, C 1 = AM = p 1 p 3 C 1 Z M C p 3 p 1 (4). The inequality p 1 > p 3 is equivalent with or + 1 >, which is provided from (i) and (ii), because + 1 > p 1 > (p 1 1) > 0. Step 5 (figure 7). [S 1M ] = {V 1}. We have [BS 1] [S 1M ], [AU ] [M U ], [AN] 5

6 From the relations AM CN 1 = 1 and AV AC V 1 M that V 1 M = p 1 p 3 BV p 1 p, V 1 M = p 1 p 3 BM p 1 p 3 +p 1 p, S 1V S 1 A 1 A 1 U 1 NB BV M V AM 1 V 1 S 1 1 = BM = BR = p 3 p 1 +p 1 p R A p 1 p > 1, which is provided from the condition (i). S 1 A 1 A 1 U = 1, we conclude V 1M = BM p3 p 1 +p 1 p p 1 p 3 +p 1 p, (5). Condition p 3 p 1 + p 1 p > 0 is equivalent with Step 6 (figure 8). We have [CW ] [W R ], [BJ ] [R J ], [W R ] [BI] = {R 1}. From the relations BR AI CR 1 = 1 and BJ R R BA IC R 1 R BR 1 W W R 1 1 B 1 B 1 J that R 1 R CR 1 = 1, we conclude = p 3 p 1 +p 1 p 1, because from the concurrence of the lines BI, AN 6

7 and CK we have AI = p R IC 1p. On the other hand, R 1 = p 3 p 1 +p 1 p CR 1+p 3 p 1 +p 1 p, W R 1 = CR 1 p 3+p 1 p 1 p 1+p 3 p 1 +p 1 p, so we have W B 1 = CV = 1 p 3+p 1 p 1 p B 1 J V B p 3 p 1 +p 1 p, then CV = 1 p 3+p 1 p 1 p (6). CB 1 The conclusion V = P 0 goes to 1 p 3 + p 1 p 1 p = 1 or 1 + p 1 p 1 p ( ) 1 + p1 p 1 p 1 = 1, so the theorem is proved. Theorem. In the conditions from the first theorem, the concurrence of the lines BM, CR, AV is equivalent with the condition P 0 = V. Proof. This tas actually mean the condition after three succesive symmetries, the point P 0 comes bac in the initial position. From the third step we have CV = +p 1 1 and from the first three steps, we obtain the relations (3), () and (1) CV = +p 1 1 CB V B +1 p 1, BR = p 1p p 1 +1 RA p 1, AM = p 1 1.The 1 MC 1 concurrence of the lines BM, CR, AV is equivalent from the Ceva s Theorem, with AM CV BR = 1, so +p MC V B RV 1 1 = 1, or (1+p 1 p 1 p ) =, so 1 + p 1 p 1 p = (7). The condition of returning is CV = CP 0, so +p 1 1 = 1, or 1 + p CB CB 1 p 1 p =, so relation (7). In the given conditions from the Theorem 1, the point P 0 is not unique, if the concurrence of cevians are fixed. We are going to analyze two particular cases, when the cevians AN, BI, CK are medians and angle bisectors, cases were we determine the points which returns in the initial position after three consecutive symmetries. (A) Let AN, BI, CK be medians. In this case, p 1 = p = 1 and the condition of returning is = 1, or =, P 0 = N, so P 0 is the midpoint of [BC]. Because [P 0 C 1 ] [C 1 M] and [BK] [AK], we deduce that [P 0 M] is a mid-segment, namely M = I. So the symmetry leads in this case at the medians feet, so P 0 = N. (B) Let AN, BI, CK be angle bisectors, then we have the usual symmetry and as a result of [1], P 0 is the tangent point of the inscribed circle with the side [BC]. It s true the fact that in this case, p 1 = b a, p = c b, p 1p = c a and 1+p 1 p 1 p = a+b c is equivalent with = CP 0 = CP 0, so CP a CB a 0 = a+b c. This is the segment determined by point C and the tangent point of the inscribed circle with the side [BC], where a = BC, b = AC and c = AB. 7

8 Annotation. Theorem 1 ensures enough conditions for the returning of the point P 0 after six steps in the initial position. We are going to prove that in more restrictive conditions, the condition of concurrence of the cevians AN, BI, CK is also required. We will note BI = α R. It is obvious the IC fact that if after three consecutive symmetries returns there. The case in which after three steps the point P 0 returns bac and this thin is possible if α p 1p p 1 +1 = 1 1, doesn t show interest. Following the previous process, we will exclude this case. CP 0 CB Theorem 3. Let P 0 [BC] and AN, BI, CK three cevians for which = 1, BK = 1 KA p 1, CN = 1 NB p such that: (i) after six consecutive symmetries, the point P 0 returns in the initial position; (ii) p 1 > 1 > 1 ; (iii) p 1 (1 p ) < 1; (iv) α < BA BR ; (v) α < BA BR (1 p 1 + p 1 p ). Then α = p 1 p, so that the cevians are concurrent. Proof. The ratio BI = α is considered just at the third step, so we are IC going to retae the reasoning from there. Step 3 p. We have the relations (figure 5): 1 p p 1 +1 α CR 1 = 1, R 1 R BJ BR R 1 R W R 1 W B 1 B 1 = 1, so that W B 1 J α( p 1 +1) = p 3 (7). Step 4. We have the relations (figure 6): CV 1 CB p 1 AL L V 1, p 3 p 1 AL = 1, S C L 1 = p 1 p V C 1 3 = AM AM = 1, = p 1 p Z p 3 M 3 C AC p 1 (8). S 1 A 1 B 1 = p 1p α( p 1 +1) J α( p 1 +1) Step 5. We have the relations (figure 7): AM AC = CV V B, CV CB = CZ = 1, L V S C CV L S 1 = C 1 Z CN BV 1 = AU NB V 1 M M V AM 1 V 1 S 1 = 1, from where we deduce that V A 1M = BM p 1 p 1 U 3 BR p 1 p +p 1 p 3, = BA p 1 p +p 3 p 1 p 1 p (9). Step 6. We have the relations (figure 8): BR α CR 1 = BJ R R RA R 1 R BR 1 W B W R 1 1 = B 1 J 1, from where we deduce that R 1 R CR α(p 1 p +p 3 p 1) p 1 p +α(p 1 p +p 3 p 1, CV CB = p 1p α(p 1 p +p 3 p 1) p 1 p (10). The condition of returning is CV CB = CP 0 CB, where from p 1p α(p 1 p +p 3 p 1) p 1 p = 8

9 1. Transforming the last relation after the replacing of p 3, we obtain p 1 p αp 1 p +αp 1 α(1 α+ α p α ) = p 1p, (p 1p α)+α(α p 1 p )+ α p (p 1 p α) + 1 (α p 1 p )(α + p 1 p ) = 0. After remaning the common factor, we ( ) have (α p 1 p ) α 1 α p + α+p 1p = 0. The case α 1 α p + α+p 1p = 0 leads to α ( p 1 + 1) = 1 1, 1 = CV = CP 0, so V = P CB CB 0. This was removed, so α = p 1 p, which proves the required concurrence. Comment. The condition (iv) ensures relation (7) and the condition (v) ensures relation (10). References [1] J. Berglund, R. Taylor, Bisections and reflections Math. Mag. 87(014), [] J. Berglund, Creating a full mathematical experience in the classroom, JOMA, 005. [3] Z. Usisin, J. Griffin, D. Witonsy, E. Wilmore, The classification of quadrilaters: a study in definition, Information Age Pub. Charlotte, NC, 008. Summary We reconsidered the notion of symmetry of a point conditioned by two lines. The symmetries defined lie this are applied succesively to three cevians. Considering a given point and performing six succesive symmetries, the point returns in the initial position if the three cevians are concurrent. Under certain conditions, the reverse is true. 9

Classical Theorems in Plane Geometry 1

BERKELEY MATH CIRCLE 1999 2000 Classical Theorems in Plane Geometry 1 Zvezdelina Stankova-Frenkel UC Berkeley and Mills College Note: All objects in this handout are planar - i.e. they lie in the usual