Logiciel libre de géométrie, d'analyse et de simulation multiplateforme par Yves Biton

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Nouveau theorem : The case of two points

modification lundi 14 août 2014.

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III. The case of two points

In this section, we consider deux points A and B distincts of plane E and we study l’The transformation T defined by T(M)= bar (A,MA) ; (B,MB).

1. Image of T.

MA et MB being positives, T takes it’s values in segment [AB]. T is obviously surjective because, in restriction to segment [AB], T is the symmetry through I, midpoint of [AB].

2. Geometrical construction of the image through T of a point M non aligned with A and B.

We consider $\Gamma$ the circumscribed circle of triangle MAB.

Let’s name P the intersection point of the perpendicular bisector of segment [AB] with the circle arc of summits A and B which does’nt contain M, and N the intersection point of line (MP) with segment [AB].

Then we know that N is the foot of the internal bisector of M in the triangle AMB.

Proposition 1.1 implies that $\frac{MA}{MB} = \frac{NA}{NB}$ . N being the barycentre of (A, NB) and (B, NA), we deduct that N is the barycenter of (A, MB) and (B, MA). The symmetric point M ’ of M through I (midpoint of [AB]) is therefore the barycenter of (A, MA)and (B, MB).

In other words : T (M) = M’.

The inverse construction prooves that if $\Gamma$ is a circle going through A and B and M ’ a point of segment ]AB[, There is an unique point M of the circle arc of $\Gamma$ which does’nt contain P such as M ’=T(M).

In other words :

Proposition 3.1 : If $\Gamma$ is a circle and A and B two distincts points of this circle, the transformation T sets une bijection of each of the two circle arcs of $\Gamma$ with summits A et B in the segment [AB].

3. Fibers of T.
We will call fibers of T the sets $T^{ - 1} \left( {\left\{ C \right\}} \right)$ for C $\in$ [AB].

When C = A (resp. C = B) ,The fiber of C is reduced in {B} (resp. to {A}).

If C $\in$ [AB], C $\ne$ A and C $\ne$ B, then C is barycenter of (A, $\lambda$) and (B, $1-\lambda$) with 0 < $\lambda$ < 1.

If M $\in$ E - {A, B}, T (M) = C is equivalent to say that the barycenter of points (A, MA) et (B, MB) and the barycenter of (A, $\lambda$) and (B, $1-\lambda$) are the same, which is true if, and only if, $\frac{MA}{MB} = \frac{\lambda }{{1 - \lambda }}$ .

So the fiber of C is the set $C_{A,B,\frac{\lambda }{1 - \lambda }}$.

We have therefore a criterion for T to be injective when applied to a set $\Gamma$ : the ecessary and sufficient condition is that $\Gamma$ don’t intercept two times the circles CA,B,k with k>0. It is the case when we have an arc of circle joining A and B, but also in many other occasions : an arc of ellipsa, a broken line [AC] $\cup$ [CB] where C is a point of the perpendicular bisector of [AB], ot, as shown on the figure above, a well chosen Bezier curve .

Continuation of the demonstration

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