A Generalization of A Leibniz Geometrical Theorem
1 A GENERALIZATION OF A LEIBNIZ GEOMETRICAL THEOREM
Mihály Bencze, Florin Popovici,
Department of Mathematics, Áprily Lajos College, Bra ş ov, Romania Florentin Smarandache, Chair of Department of Math & Sciences, University of New Mexico, 200 College Road, Gallup, NM 87301, USA, E-mail: [email protected]
Abstract : In this article we present a generalization of a Leibniz’s theorem in geometry and an application of this.
Leibniz’s theorem.
Let M be an arbitrary point in the plane of the triangle ABC , then MA + MB + MC =
13 ( a + b + c ) + MG , where G is the centroid of the triangle. We generalize this theorem: Theorem.
Let’s consider A , A ,..., A n arbitrary points in space and G the centroid of this points system; then for an arbitrary point M of the space is valid the following equation: MA i i = n ∑ = n A i ≤ i < j ≤ n ∑ A j + n ⋅ MG . Proof.
First, we interpret the centroid of the n points system in a recurrent way. If n = then is the midpoint of the segment. If n = , then it is the centroid of the triangle. Suppose that we found the centroid of the n − points created system. Now we join each of the n points with the centroid of the n − points created system; and we obtain n bisectors of the sides. It is easy to show that these n medians are concurrent segments. In this manner we obtain the centroid of the n points created system. We’ll denote i G the centroid of the A k ,
1, 2,..., 1, 1,..., k i i n = − + points created system. It can be shown that ( 1) i i n A G GG − = . Now by induction we prove the theorem. If n = the MA + MA = A A + MG or MG =
14 2 MA + MA ( ) ( ) , where G is the midpoint of the segment A A . The above formula is the side bisector’s formula in the triangle MA A . The proof can be done by Stewart’s theorem, cosines theorem, generalized theorem of Pythagoras, or can be done vectorial. Suppose that the assertion of the theorem is true for n = k . If A , A ,..., A k are arbitrary points in space, G is the centroid of this points system, then we have the following relation: MA i i = k ∑ = k A i ≤ i < j ≤ k ∑ A j + k ⋅ MG k . Now we prove for n = k + . Let A k + ∉ A , A ,..., A k , G { } be an arbitrary point in the space and let G be the centroid of the A , A ,..., A k , A k + points system. Taking into account that G is on the segment A k + G and k ⋅ A k + G = GG , we apply Stewart’s theorem to the points M , G , G , A k + , from where: MA k + ⋅ GG + MG ⋅ GA k + − MG ⋅ A k + G = GG ⋅ GA k + ⋅ A k + G . According to the previous observation A k + G = kk + A k + G and GG = kk + A k + G . Using these, the above relation becomes: MA k + + k ⋅ MG = kk + A k + G + ( k + MG . From here k ⋅ MG = MA i i = k ∑ − k A i ≤ i < j ≤ k ∑ A j . From the supposition of the induction, with M ≡ A k + as substitution, we obtain A i A j i = k ∑ = k A i ≤ i < j ≤ k ∑ A j + k ⋅ A k + G and thus kk i k i ji i j k k A G A A A Ak k k k + += ≤ < ≤ = −+ + + ∑ ∑ . Substituting this in the above relation we obtain that k k i i j i ki i j k i MA A A A A k MGk k k k + += ≤ < ≤ = ⎛ ⎞= − + + + =⎜ ⎟+ +⎝ ⎠ ∑ ∑ ∑ ( ) i j i j k A A k MGk ≤ < ≤ + = + ++ ∑ . With this we proved that our assertion is true for n = k + . According to the induction, it is true for every n ≥ natural numbers. Application 1.
If the points A , A ,..., A n are on the sphere with the center O and radius R , then using in the theorem the substitution M ≡ O we obtain the identity: i ji j n OG R A An ≤ < ≤ = − ∑ . In case of a triangle: OG = R − a + b + c ( ) . In case of a tetrahedron: OG = R − a + b + c + d + e + f ( ) . Application 2.
If the points A , A ,..., A n are on the sphere with the center O and radius R , then i ji j n A A n R ≤ < ≤ ≤ ∑ . The equality holds if and only if G ≡ O . In case of a triangle: a + b + c ≤ R , in case of a tetrahedron: a + b + c + d + e + f ≤ R . Application 3.
Using the arithmetic and harmonic mean inequality, from the previous application, it results the following inequality: ( )
22 21
11 4 i j n i j nA A R ≤ < ≤ −≥ ∑ . In the case of a triangle: 1 a + b + c ≥ R , in case of a tetrahedron: 1 a + b + c + d + e + f ≥ R . Application 4.
Considering the Cauchy-Buniakowski-Schwarz inequality from the Application 2, we obtain the following inequality: ( 1) 2 i ji j n n nA A nR ≤ < ≤ −≤ ∑ . In case of a triangle: a + b + c ≤ R , in case of a tetrahedron: a + b + c + d + e + f ≤ R . Application 5.
Using the arithmetic and harmonic mean inequality, from the previous application we obtain the following inequality ( 1) ( 1)1 2 2 i j n i j n n nA A R ≤ < ≤ − −≥ ∑ . In case of a triangle: a + b + c ≥ R , in case of a tetrahedron: 1 a + b + c + d + e + f ≥ R
32 .
Application 6.
Considering application 3, we obtain the following inequality: ( 1) 14 ki j ki j n i j n i j n n A A A A ≤ < ≤ ≤ < ≤ ⎛ ⎞⎛ ⎞− ≤ ≤⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ∑ ∑ ( ) ( 1) ( 1) if is even,16 2( ) ( 1) 4( ) ( 1) if is odd16 2 M m n n n nM mM m n n M m n nM m ⎧ + − −⎪⎪ ⋅≤ ⎨ + − − − −⎪⎪ ⋅⎩ where m = min A i A jk { } and M = max A i A jk { } . In case of a triangle: 9 ≤ a k + b k + c k ( ) a − k + b − k + c − k ( ) ≤ M + M ⋅ m + m M ⋅ m , in case of a tetrahedron: ( ) ( ) ( ) k k k k k k k k k k k k M ma b c d e f a b c d e f M m − − − − − − +≤ + + + + + + + + + + ≤ ⋅ . Application 7.
Let A , A ,..., A n be the vertexes of the polygon inscribed in the sphere with the center O and radius R . First we interpret the orthocenter of the inscribable polygon A A ... A n . For three arbitrary vertexes, corresponds one orthocenter. Now we take four vertexes. In the obtained four orthocenters of the triangles we construct the circles with radius R , which have one common point. This will be the orthocenter of the inscribable quadrilateral. We continue in the same way. The circles with radius R that we construct in the orthocenters of the n −
1 sides inscribable polygons have one common point. This will be the orthocenter of the n sides, inscribable polygon. It can be shown that O , H , G are collinear and n ⋅ OG = OH . From the first application i ji j n OH n R A A ≤ < ≤ = − ∑ and ( )
222 2 21
11 1 i ji j n
GH n R A An ≤ < ≤ ⎛ ⎞= − − −⎜ ⎟⎝ ⎠ ∑ . In case of a triangle OH = R − a + b + c ( ) and GH = R − a + b + c ( ) . Application 8.
In the case of an A A ... A n inscribable polygon i ji j n A A n R ≤ < ≤ = ∑ if and only if O ≡ H ≡ G . In case of a triangle this is equivalent with an equilateral triangle. Application 9.
Now we compute the length of the midpoints created by the A , A ,..., A n space points system. Let S =
1, 2,..., i − i + n { } and G be the centroid of the A k , k ∈ S , points system. By substituting M ≡ A i in the theorem, for the length of the midpoints we obtain the following relation: A i G = n − A i A k k ∈ S ∑ − n − ( ) A u A v u , v ∈ S : u ≠ v ∑ . 5 Application 10.
In case of a triangle m a = b + c ( ) − a
4 and its permutations. From here: ( ) a b c m m m a b c + + = + + , m a + m b + m c ≤ R , m a + m b + m c ≤ R . Application 11.
In case of a tetrahedron m a =
19 3 a + b + c ( ) − d + e + f ( ) ( ) and its permutations. From here: m a = ∑ a ∑ ( ) , m a ≤ ∑ R , m a ≤ ∑ R . Application 12.