Open source cross-platform software of geometry, analysis and simulation - Yves Biton

# New theorem : A few properties of triangle A’B’C’ in the case of three points.

publication Thursday 14 August 2014.

### VI. A few properties of triangle A’B’C’ in the case of three points.

1. First property

 A’ is the barycenter of (B, c) et (C, b). B’ is the barycenter of (C, a) et (A, c). C ’ is the barycenter de (A; b) et (B; a). Let us demonstrate that lines (AA’), (BB’) et (CC’) are convergent. We name H the barycenter of the three balanced points (A, bc), (B, ca) et (C, ab). The partial barycenter of (B, ca) and (C, ab) is also the barycenter of (B, c) et (C, b). Therefore it is A’. According to the partial barycenter theorem, H is therefore the barycenter of points (A, bc) et (A’, ca+ab). In particular, H belongs to segment [AA’]. In the same way we have H belonging to segments [BB’] and [CC’].

H is therefore the convergent point of segments [AA’], [BB’] et [CC’].

Property 6.1 : In a triangle ABC, we consider A’, barycenter of (B,AB) and (C,AC), B’, barycenter of (C, BC) et (A, BA) and C’, barycenter of (A; CA) et (B; CB). The lines (AA’), (BB’) et (CC’) are convergent.

To be noticed: The lines (AA’), (BB’) et (CC ’) are Ceva’s lines of triangle ABC.

2. Second property.

Property 6.2 : With the previous notations, the area of triangle A’B’C’ is always inferior or equal to the quarter of the aera of triangle ABC.

We name SA, SB, SC the respective areas of triangles AB’C ’, BC’A’ et CA’B’.

$S_A = \frac{1}{2}AB’AC’\sin \hat A$

B’ is the barycenter of (A; c) et (C; a), $\overrightarrow {AB’} = \frac{a}{a + c}\overrightarrow {AC}$.

C ’ is the barycenter of (B; a) et (A; b), $\overrightarrow {AC’} = \frac{a}{a + b}\overrightarrow {AB}$.
So we get $AB’ = \frac{ab}{a + c}$ et $AC’ = \frac{ac}{a + b}$.
And $S_A = \frac{1}{2}\frac{a^2bc}{(a + b)(a + c)}\sin \hat A$.

In the same way, $S_B = \frac{1}{2}\frac{b^2ca}{(b + c)(b + a)}\sin \hat B$ et $S_C = \frac{1}{2}\frac{c^2ab}{(c + a)(c + b)}\sin \hat C$.

Let us name S the area of ABC et S ’ the area of A’B’C ’.

We have $S = \frac{1}{2}bc\sin \hat A = \frac{1}{2}ca\sin \hat B = \frac{1}{2}ab\sin \hat C$

Therefore $S_A = \frac{a^2S}{(a + b)(a + c)}$, $S_B = \frac{b^2S}{(b + c)(b + a)}$ et $S_B = \frac{c^2S}{(c + a)(c + b)}$.

We know that A’, B’ et C ’ are on segments [BC], [CA] and [AB], so we get : $S’=S- S_A-S_B-S_C$.

Then $S’=S\left[ 1-\frac{a^2}{\left( a+b \right)\left( a+c \right)}-\frac{b^2}{\left( b+c \right)\left( b+a \right)}-\frac{c^2}{\left( c+a \right)\left( c+b \right)} \right]$.

$S’=\frac{S}{\left( a+b \right)\left( b+c \right)\left( c+a \right)}\left[ \left( a+b \right)\left( b+c \right)\left( c+a \right)-a^2\left( b+c \right)-b^2\left( c+a \right)-c^2\left( a+b \right) \right]$.

$S’=S\times \frac{2abc}{\left( a+b \right)\left( b+c \right)\left( c+a \right)}$.

We want to prove that $\frac{S’}{S} \le \frac{1}{4}$.

Let us note p = a + b + c.

Then $\left( a+b \right)\left( b+c \right)\left( c+a \right)=\left( p-a \right)\left( p-b \right)\left( p-c \right)=p^3-p^2\left( a+b+c \right)+p \left( ab+bc+ca \right) -abc$.

So $\left( a+b \right)\left( b+c \right)\left( c+a \right)=p^3-p^2\times p+p\left( ab+bc+ca \right)-abc=p\left( ab+bc+ca \right)-abc$.

Therefore $\frac{\left( a+b \right)\left( b+c \right)\left( c+a \right)}{abc}=p\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)-1=\left( a+b+c \right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)-1=2+\left( \frac{a}{b}+\frac{b}{a} \right)+\left( \frac{b}{c}+\frac{c}{b} \right)+\left( \frac{c}{a}+\frac{a}{b} \right)$

So we get $\frac{\left( a+b \right)\left( b+c \right)\left( c+a \right)}{abc}=8+\frac{\left( a-b \right)^2}{ab}+\frac{\left( b-c \right)^2}{bc}+\frac{\left( c-a \right)^2}{ca}$.

So we always have $\frac{\left( a+b \right)\left( b+c \right)\left( c+a \right)}{abc}\ge 8$ et $\frac{2abc}{\left( a+b \right)\left( b+c \right)\left( c+a \right)}\le \frac{1}{4}$, the equality occuring only when triangle ABC is équilateral.

Therefore $S’ \le \frac{1}{4}S$,the equality occuring only when triangle ABC is équilateral. In this case, A’, B ’ et C ’ are the midpoints of triangle’s ABC sides.

Continuation of the article