Generalizations Of Ceva Theorem, By F.smarandache

GENERALIZATIONS OF CEVA’S THEOREM AND APPLICATIONS Florentin Smarandache University of New Mexico 200 College Road Gallup, NM 87301, USA E-mail: [email protected] In these paragraphs one presents three generalizations of the famous theorem of Ceva, which states: “If in a triangle ABC one draws the concurrent straight lines A B BC C A AA1 , BB1 , CC1 , then 1 ⋅ 1 ⋅ 1 = −1 “. A1C B1 A C1 B Theorem 1: Let us have the polygon A1 A2 ... An , a point M in its plane, and a circular permutation ⎛ 1 2 ... n − 1 n ⎞ p=⎜ ⎟ . One notes M ij the intersections of the line Ai M with the lines ⎜⎝ 2 3 ... n 1 ⎟⎠ Ai + s Ai + s +1 ,..., Ai + s + t −1 Ai + s + t (for all i and j , j ∈{i + s,..., i + s + t − 1}). If M ij ≠ An for all respective indices, and if 2s + t = n , one has: n ,i + s + t −1

∏

i , j =1,i + s

M ij Aj M ij Ap ( j )

= (−1) n ( s and t are natural non-zero numbers).

Analytical proof: Let M be a point in the plain of the triangle ABC , such that it satisfies the conditions of the theorem. One chooses a Cartesian system of axes, such that the two parallels with the axes which pass through M do not pass by any point Ai (this is possible). One considers M (a, b ) , where a and b are real variables, and Ai ( Xi , Yi ) where Xi and Yi are known, i ∈{1, 2,..., n}. The former choices ensure us the following relations: Xi − a ≠ 0 and Yi − b ≠ 0 for all i ∈{1, 2,..., n}. The equation of the line Ai M (1 ≤ i ≤ n) is: x−a y−b − = 0. One notes that d ( x, y; Xi , Yi ) = 0 . Xi − a Yi − b One has M ij Aj δ ( Aj , Ai M ) d ( X j , Y j ; X i , Yi ) D( j, i ) = = = M ij Ap ( j ) δ ( Ap ( j ) , Ai M ) d ( X p ( j ) , Yp ( j ) ; X i , Yi ) D( p ( j ), i )

1

where δ ( A, ST ) is the distance from A to the line ST , and where one notes with D(a, b ) for d ( Xa , Ya ; Xb , Yb ) . Let’s calculate the product, where we will use the following convention: a + b will mean p( p(...p (a)...)) , and a − b will mean p −1 ( p −1 (...p −1 (a)...)) 12 4 4 3 14 4244 3 b times

b times

i + s +t −1

M ij A j

j =i+ s

M ij A j +1

∏

=

i + s +t −1

∏

j =i + s

D( j,i) = D( j + 1,i)

=

D(i + s, i) D(i + s + 1, i) D(i + s + t − 1, i ) ⋅ ⋅⋅⋅ = D(i + s + 1, i ) D(i + s + 2, i ) D(i + s + t , i )

=

D(i + s,i) D(i + s,i) = D(i + s + t,i) D(i − s,i)

The initial product is equal to: n D(i + s,i) D(1 + s,1) D(2 + s, 2) D(2s, s) D( 2 s + 1, s + 1) = ⋅ ⋅⋅⋅ ⋅ ∏ D(n, s) D(1 − s,1) D(2 − s, 2) D(1, s + 1) i =1 D(i − s,i)

⋅

D(2s + 2, s + 2) D(2s + t, s + t) D(2s + t + 1, s + t + 1) ⋅⋅⋅ ⋅ ⋅ D(t, s + t) D(t + 1, s + t + 1) D(2, s + 2)

⋅

D(2s + t + 2, s + t + 2) D(2s + t + s, s + t + s) ⋅⋅⋅ = D(t + s, s + t + s) D(t + 2, s + t + 2)

=

D(1 + s,1) D(2 + s, 2) D(2s + t, s + t) D(s, n) ⋅ ⋅⋅⋅ ⋅⋅⋅ = D(1,1 + s) D(2, 2 + s) D(s + t, 2s + t) D(n, s)

D(i + s,i) n ⎛ P(i + s) ⎞ =∏ =∏ ⎜ − = (−1)n ⎟ P(i) ⎠ i =1 D(i,i + s) i =1 ⎝ n

because:

X r − a Yr − b − D ( r , p ) X p − a Yp − b P(r ) ( X − a )(Yr − b) , = =− r =− X a Y b − − D ( p, r ) P( p) ( X p − a )(Yp − b) p p − X r − a Yr − b the last equality resulting from what one notes: (Xt − a)(Yt − b) = P(t) . From (1) it results that P(t) ≠ 0 for all t from {1, 2,..., n} . The proof is completed.

Comments regarding Theorem 1:

2

t represents the number of lines of a polygon which are intersected by a line Ai0 M ; if one notes the sides Ai Ai +1 of the polygon, by ai , then s + 1 represents the order of the first line intersected by the line A1 M (that is as +1 the first line intersected by A1 M ).

Example: If s = 5 and t = 3 , the theorem says that : - the line A1 M intersects the sides A6 A7 , A7 A8 , A8 A9 . - the line A2 M intersects the sides A7 A8 , A8 A9 , A9 A10 . - the line A3 M intersects the sides A8 A9 , A9 A10 , A10 A11 , etc. Observation: The restrictive condition of the theorem is necessary for the M ij Aj . existence of the ratios M ij Ap ( j ) Consequence 1.1: Let’s have a polygon A1 A2 ...A2 k +1 and a point M in its plan. For all i from {1, 2,..., 2 k + 1}, one notes M i the intersection of the line Ai Ap (i ) with

the line which passes through M and by the vertex which is opposed to this line. If n M i Ai M i ∉ Ai , Ap(i ) then one has: ∏ = −1 . i =1 M i Ap ( i )

{

}

The demonstration results immediately from the theorem, since one has s = k and t = 1 , that is n = 2k + 1 . The reciprocal of this consequence is not true. From where it results immediately that the reciprocal of the theorem is not true either. Counterexample: Let us consider a polygon of 5 sides. One plottes the lines A1 M 3 , A2 M 4 and A3 M 5 which intersect in M .

M 3 A3 M 4 A4 M 5 A5 ⋅ ⋅ M 3 A4 M 4 A5 M 5 A1 Then one plots the line A4 M 1 such that it does not pass through M and such that it forms the ratio: M 1 A1 = 1 / K or 2 / K . (One chooses one of these values, for which (2) M 1 A2 A4 M 1 does not pass through M ). Let us have K =

At the end one traces A5 M 2 which forms the ratio

1 M 2 A2 = −1 or − in 2 M 2 A3

function of (2). Therefore the product: 5 M i Ai = -1 without having the respective lines concurrent. ∏ i =1 M i Ap ( i )

3

Consequence 1.2: Under the conditions of the theorem, if for all i and j , j ∉ i, p −1 (i ) , one notes M ij = Ai M ∩ A j Ap ( j ) and M ij ∉ A j , Ap( j ) then one has:

{

{

}

n

M ij A j

∏M

i, j =1

ij

= (−1)n .

Ap( j )

{

}

}

j ∉ i, p −1 (i)

Effectively one has s = 1 , t = n − 2 , and therefore 2s + t = n . Consequence 1.3: For n = 3 , it comes s = 1 and t = 1 , therefore one obtains (as a particular case ) the theorem of Ceva. An Application of the Generalizations of Ceva’s Theorem is presented below. Theorem 2: Let us consider a polygon A1 A2 ...An inserted in a circle. Let s and t be two non zero natural numbers such that 2s + t = n . By each vertex Ai passes a line di which intersects the lines Ai + s Ai + s +1 ,...., Ai + s +t −1 Ai + s +t at the points M i,i + s ,..., M i + s + t −1

respectively and the circle at the point M i' . Then one has: n i + s + t −1

∏∏ i =1

j =i + s

M ij Aj M ij Aj +1

n

M i' Ai + s

i =1

M i' Ai + s +t

=∏

.

Proof: Let i be fixed. 1) The case where the point M i,i + s is inside the circle.

There are triangles Ai M i,i + s Ai + s and M i' M i,i + s Ai + s +1 which are similar, since the angles M i,i + s Ai Ai + s and M i,i + s Ai + s +1 M i' on one side, and Ai M i,i + s Ai + s and Ai + s +1 M i,i + s M i' are equal. It results from it that: M i,i + s Ai AA (1) = 'i i + s M i,i + s Ai + s +1 M i Ai + s +1

4

Ai+s+1

Ai

MMM Mi,i+s

M i' Ai+s

In a similar manner, one shows that the triangles M i,i + s Ai Ai + s +1 and M i,i + s Ai + s M i' are similar, from which: M i,i + s Ai AA (2) = i ' i + s +1 . Dividing (1) by (2) we obtain: M i,i + s Ai + s M i Ai + s (3)

M i,i + s Ai + s M i,i + s Ai + s +1

=

M i' Ai + s ' i

M Ai + s +1

⋅

Ai Ai + s . Ai Ai + s +1

2) The case where M i,i + s is exterior to the circle is similar to the first, because the triangles (notations as in 1) are similar also in this new case. There are the same interpretations and the same ratios; therefore one has also the relation (3). Ai+s+1

Ai+s

Ai

M i'

Let us calculate the product: 5

Mi,i+s

i + s + t −1

M ij A j

j =i+ s

M ij A j +1

∏

=

⋅

M i' Ai + s

⋅

=

i + s +t −1

∏

j =i+ s

M i' Ai + s +1

M i' Ai + s +1 M i' Ai + s + 2

⎛ M i' A j Ai A j ⎞ ⋅ ⎜ ' ⎟= ⎜⎝ M A ⎟⎠ A A i j +1 i j +1 ⋅⋅⋅

M i' Ai + s +t −1 M i' Ai + s +t

⋅

Ai Ai + s Ai Ai + s +1 AA M 'A AA ⋅ ⋅ ⋅ ⋅ i i + s + t −1 = ' i i + s ⋅ i i + s Ai Ai + s +1 Ai Ai + s + 2 Ai Ai + s +t M i Ai + s +t Ai Ai + s +t

Therefore the initial product is equal to: n ⎛ M i' Ai + s Ai Ai + s ⎞ M i' Ai + s ⋅ = ∏ ⎜ ' ⎟ ∏ Ai Ai + s +t ⎠ i =1 M i' Ai + s +t i =1 ⎝ M i Ai + s +t since: n

n

Ai Ai + s

∏AA i =1

⋅

i

=

i + s+t

A1 A1+ s A2 A2 + s AA ⋅ ⋅ ⋅ ⋅ s 2s ⋅ A1 A1+ s + t A2 A2 + s + t As +1 A1

As + 2 A2 s + 2 A A A A A A AA ⋅ ⋅ ⋅ s + t n ⋅ s + t +1 1 ⋅ s + t + 2 2 ⋅ ⋅ ⋅ n s = 1 As + 2 A2 As + t At As + t +1 At +1 As + t + 2 At + 2 An As + t

(by taking into account the fact that 2 s + t = n ). Consequence 2.1: If there is a polygon A1 A2 ,...., A2 s −1 inscribed in a circle, and from each vertex Ai one traces a line di which intersects the opposite side Ai + s −1 Ai + s in M i and the circle in M i' then: n M i' Ai + s −1 M i Ai + s −1 =∏ ∏ ' i = 1 M i Ai + s i = 1 M i Ai + s n

n +1 . 2 If one makes s = 1 in this consequence, one finds the mathematical note from [1], pages 35-37.

In fact for t = 1 , one has n odd and s =

Application: If in the theorem, the lines di are concurrent, one obtains: n

M i' Ai + s

∏M A i =1

' i

= (−1)n .

i + s+t

6

Reference:

[1] Dan Barbilian - Ion Barbu – “Pagini inedite”, Editura Albatros, Bucharest, 1981 (Ediţie îngrijită de Gerda Barbilian, V. Protopopescu, Viorel Gh. Vodă).

7