Logiciel libre de géométrie, d'analyse et de simulation multiplateforme par Yves Biton
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New theorem : Case of three cocyclic points. Image of the circumscribed circle of these points.

publication jeudi 14 août 2014.

IV) Case of three cocyclic points. Image through T of the cricumscribed circle to these points.

We shall note here T=TABC.

1. Let us first demonstrate that the image through T=TABC of plane E is included in the interior of a triangle

Le’s study the restriction of T to the circumscribed circle of triangle ABC.

$\Gamma$ stands for the circumscribed circle to triangle ABC.

We will use the usual notations in ABC : AB = c, BC = a, CA = b etc..

M being a some point of plane E, we consider the barycenter M’ of points (A,MA), (B,MB) et (C, MC), so M’ = T(M).

First we notice that if M=A, then M=A’ =TBC (A) where A’ is the barycenter of (B, AB) and (C, AC) (since MA = 0) and that, if M=B then M’=B’=TCA(B) where B’ is the barycenter of (C, BC) et (A, BA).

We will demonstrate that, if M belongs to the circle arc of $\Gamma$ with summits A and B which does’nt contain C, with M$\ne$A and M$\ne$B then M’ belongs to segment ]A’B’[.

A’ = bar (B, c) ; (C ; b). So, for each point N, $c\overrightarrow {NB} + n\overrightarrow {NC} = \left( b + c \right)\overrightarrow {NA’} $ .

In the same way :

$a\overrightarrow {NC} + c\overrightarrow {NA} = \left( c + a \right)\overrightarrow {NB’} $ (1).
and $b\overrightarrow {NA} + a\overrightarrow {NB} = \left( b + a \right)\overrightarrow {NC’} $ (2).

Replacing N with A’ in(1),we get :$c\overrightarrow {AB} + b\overrightarrow {AC} = \left( {b + c} \right)\overrightarrow {AA’} $.

Replacing N with A in (2) we get : $\overrightarrow {AC} = \frac{c + a}{a}\overrightarrow {AB’} $.

Replacing N with A in (3) we get : $\overrightarrow {AB} = \frac{b + c}{a}\overrightarrow {AC’} $.

Finally : $\frac{c\left( b + a \right)}{a}\overrightarrow {AC’} + \frac{b\left( c + a \right)}{a}\overrightarrow {AB’} = \left( b + c \right)\overrightarrow {AA’} $.

Or : $ - a\left( b + c \right)\overrightarrow {AA’} + b\left( c + a \right)\overrightarrow {AB’} + c\left( b + a \right)\overrightarrow {AC’} = \overrightarrow 0 $.

So $A = bar\left\{ {\left( {A’ ; - a\left( b + c \right)} \right) ;\left( {B’ ;b\left( c + a \right)} \right) ;\left( {C’ ;c\left( b + a \right)} \right)} \right\}$ (Sum of coefficients 2bc).

Or $A=bar\left\{ \left( A’ ;-a^2\left( b+c \right) \right) ;\left( B’ ;ab\left( c+a \right) \right) ;\left( C’ ;ac\left( b+a \right) \right) \right\}$(sum of coefficients 2abc).

In the same way :

$B=bar\left\{ \left( B’ ;-b^2\left( c+a \right) \right) ;\left( C’ ;bc\left( a+b \right) \right) ;\left( A’ ;ba\left( c+b \right) \right) \right\}$(Sum of coefficients 2abc).

et $C=bar\left\{ \left( C’ ;-c^2\left( a+b \right) ;\left( A’ ;ca\left( b+c \right) \right) \right) ;\left( B’ ;cb\left( a+c \right) \right) \right\}$(Sum of coefficients 2abc).

If M’ = T(M), according to homogeneity property of barycenters , M’ is the barycenter of (A ;2abcMA), (B,2abcMB) and (C,2abcMC) then,using the property of partial barycenters, we get :

M’ = bar {(A’, $a\left( {b + c} \right)\left( { - a{\text{MA}} + b{\text{MB}} + c{\text{MC}}} \right)$ ) ;(B’,$b\left( {c + a} \right)\left( - b{\text{MB}} + c{\text{MC}} + a{\text{MA}} \right)$) ;(C’,$c\left( {a + b} \right)\left( { - c{\text{MC}} + a{\text{MA}} + b{\text{MB}}} \right)$)}.

The Ptomemy’s theorem (proposition 1.2) shows that the three coeffcients are positive or null,which proves that M’ is inside triangle A’B’C ’ is on one of the edge segments of the triangle if, and only if, one of the three coefficients is null, which, still from Ptolemy’s theorem, is equivalent to say that M lies on one of the three circle arcs.

Fo instance, point M will belong to segment [B’C’]if, and only if :
$ - a{MA} + b{MB} + c{MC} = 0 \Leftrightarrow {AB} \times {MC} = {BC} \times {MA} + {AB} \times {MC}$ which, from Ptolemy’s theorem, is equivalent to say that M belongs to the circumscribed circle of ABC ant that the quadrangle ABMC is convex, which is equivalent to say that M lies on the circle arc of $\Gamma$ with summits B and C that does’nt contain A. The figure on the right illustrates this property :

Proposition 4.1 :The image through TABC of a some point M of plane E is a point M’ inside triangle A’B’C ’, M’ being on the edges of A’B’C ’ if, and only if, point M is on the circumscribed circle of ABC.

Let us notice that A’ = TBC(A) may be constructed in a geometrical way as we already explained, as well as B’=TCA (B). Their construction is omitted on the following figure for more legibility.

Let us construct P = TAB(M) with the method exposed before with two points.

According to the partiel barycenter theorem, M’ is also barycenter of (P ; MA+MB) and (C ; MC).So M’ belongs to line (CP). Therefore it is the intersection point between line (CP) with line (A’B’).

Bijectivity of T restricted to the arc of summits A et B which does’nt contain C.

The study made for the two points case proves that in two distincts points M of the arc with summits A and B not containing C correspond two distinct points P=TAB(M) and therefore two distinct points M’ distincts on segment [A’B’]. So T is injective on this circle arc.

M’ is a point of segment [A’B’], distinct from A’ and B’. Let us call P the intersection point of line (CM’) with segment [AB]. Let us construct P’, symmetric point of P through I, midpoint of [AB]. Let us call H the intersection point of the perpendicular bisector of [AB] with the circle arc containing C. Then we construct point M, intersection point of line (HP’) with segment [AB].
Then we know that P=TAB(M) and obviously M’=TABC(M).
Conclusion :

Théorème 4.2 : ABC being a triangle inscribed in a circle $\Gamma$, the transformation TABC sets a bijection from $\Gamma$ to the triangle A’B’C’ where A’=TBC (A), B’=TCA(B) and C ’=TAB(C), TABC sending respectively the arcs BC, CA et AB not containing A, B and C on segments [B’C’], [C’A’] et [A’B’].

2. Construction of the image of a some M through T.

We only have to create the respective images P, Q, R of M through TBC, TCA, TAB via the method exposed for two points.
M’ = T (M) is therefore at the intersection of lines (AP), (BQ) and (CR) as shown on the figure beside ( we need only two of these lines of course)

3. Let us now search the points which have the same image as a given point M.

Lemma 4.3 : A, B et C are three points non aligned of the plane, $\Gamma$ is their circumscribed circle and R it’s radius.

We write T = TABC.

Let us consider M$\in$E. If M is not on $\Gamma$ and if M$\ne$ O, there is eaxctly a point N verifying T(M) = T(N) : it is the image i(M) through inversion i of center O and ratio R2 ,R being the radius of $\Gamma$. On $\Gamma$, application T is injective.

First we notice T(O) = G where G is the gravity center of triangle ABC and O is the unique point whose image is G.
We suppose that M $\in \Gamma$ is fixed and that M $\ne$ O. In proposition 2.1 we saw that T(M)=T[i (M)].
If we put N=i(M), then we get T(N)=T(M).

To demonstrate that N is the unique other point whose image through T is the same as image of M, we only have to get that que, given a point M, there are at most two points N such as T(N)=T(M).

Let us suppose that T(M)=T(N). (A, B, C) being an affine frame of the plane, it implies that the barycentric coordinates of M and N sur (A, B, C) are equal so :

$\frac{MA}{MA + MB + MC} = \frac{NA}{NA + NB + NC}$, $\frac{MB}{MA + MB + MC} = \frac{NB}{NA + NB + NC}$ et $\frac{MC}{MA + MB + MC} = \frac{NC}{NA + NB + NC}$.

Using the inverse numbers and solving, we get : $\frac{MA}{MB} = \frac{NA}{NB}$, $\frac{MA}{MC} = \frac{NA}{NC}$ and $\frac{MB}{MC} = \frac{NB}{NC}$.

So N is at the intersection of circles $C_{A,B,\frac{MA}{MB}}$ and $C_{A,B,\frac{NA}{NB}}$ (and also $C_{B,C,\frac{MB}{MC}}$) ), which proves indeed that there are at most two points N solutions (et therefore exactly two when M $\notin \Gamma$ and M $\ne$ O since then we have i(M) $\ne$ M).

4. Let us show that T is surjective on the interior of triangle A’B’C ’.

Lemma 4.4 : $\Gamma$ being a circle of center O, radius R going through points A and B. If k > 0, the circles CA, B, k are circles orthogonal to $\Gamma$. In particular, they are invariant through the inversion i of center O and ratio R2.

Demonstration : We suppose that k > 0, k $\ne$ 1 and that C is the circle CA, B, k. We consider a point M, intersection point of C and $\Gamma$. Naming I et J the feet of the bisectors from M in AMB, we know that [IJ] is a diameter of C and, according to proposition 1.5, the tangent to C in M is going through the midpoint of segment [IJ]. The circles C et $\Gamma$ are therefore orthogonal and proposition 1.6 proves that the circle is invariant through the inversion i of center O and ratio OA2.

Note 4.5 : The lemma 4.4 applies to the case k = 1 considering that the perpendicular bisector of [AB] is a circle whose center is to the infinty. This perpendicular bisector is invariant through i.

Circle C being invariant through i, it is natural to get interested in it’s image through T

Proposition 4.6 : $\Gamma$being the circumscribed circle of a triangle ABC, and k a real number strictly positive, let us name M1 and M2 the two intersection points of CA,B,k with the circle $\Gamma$ and M’1 et M’2 their respective images through T = TABC. The image through T of CA,B,k is the segment [M’1M’2].

Particular case : If k = 1, C is the perpendicular bisector of [AB] and if M lies on C, M’=T(M) is on line (CP), P being the midpoint of [AB].

Demonstration : Le us name $\Gamma_1$ one of the two circle arcs of C = CA,B,k of summits M1 and M2 and let us assume that M is a point of $\Gamma_1$. By definition of circle C , the foot of the bisector of M1 in AM1B is the same as the foot of the bisector of M in AMB. Assuming that P is the symmetric point of Ithrough the midpoint of [AB]. We know the that M’=T(M) is on line (CP).

We set M’1=T(M1) and M’2=T(M2).

M1 et M2 beeing on $\Gamma$, we know that M’1 and M’2 are on the sides of triangle A’B’C’. Moreover we know that M’ is inside this triangle, so M’ is on segment [M’1 M’2].

Remain to be proved that the image through T of $\Gamma_1$ is the entire segment [M’1 M’2].

Fo this we will use a bit of topology.

The function f : C ----> C associated to T defined by :
$f(z)=\frac{\left| z-a \right|a+\left| z-b \right|b+\left| z-c \right|c}{\left| z-a \right|+\left| z-b \right|+\left| z-c \right|}$ being continuous, the image through f of a connex set is a connex set.

The circle arc $\Gamma_1$ being connex and included in segment [M’1M’2] can only be the segment [M’1M’2].

Moreover this proposition proves that the image through T of the entire plane is the inside of triangle A’B’C’.

Indeed let us assume that point M ’is a given point inside triangle A’B’C’. We consider P, intersection point of (CM’) with segment [AB] and I the symmetric point of P through the midpoint of segment [AB]. What precedes proves that, setting $k = \frac{IA}{IB} = \frac{PB}{PA}$, M’ is on the image through T of CA,B,k.

Case where k $\ne$1 Cas où k = 1

Thorem 4.7 : If ABC is a triangle, the image through T=TABC of the entire plane is the inside of triangle A’B’C’, where A’, B’ et C’ are the repective images of A, B et C through T.

Moreover we have a method to construct the antecdent points of a point M’ through T.

We only have to do the preceeding construction with two circles C as on the left figure above where M2 and M3 are the antecedent points of M’ through T. Moreover we know that line(M2M3) is going through O,center of the circumscribed circle $\Gamma$to ABC, since they are images odf each other through the inversion of center O letting A, Band C invariant. Of course, if we draw the third circle C relative to points C et A, it goes through M2 and M3 as well.

If point M’ lies on the side of triangle A’B’C’, the three circles are tangent in a point located on the circumscribed circle $\Gamma$(since M’ has got an unique antecedent point through T) as you can see on the rght figure below.

5. A simpler construction of TABC(M) when M lies on the circumscribed circle of ABC.

Let us assume that M lies on the cricle arc ofe $\Gamma$ with summits B and C which does’nt contain point A.

The line (A’M) re-intersects the circle $\Gamma$ in a point N.

We are going to demonstrate that M’ = T(M) is on line (AN).

We will name R the intersection point of line (AN) with segment [BC].

M’ = bar {(A, MA) ; (B, MB) ; (C, MC)}.

According to the inscribed angle theorem we get that triangles A’MB and A’CN are similar.
So we get $\frac{MC}{CA’} = \frac{NB}{BA’}$.
In the same way, A’MC et A’NB are imilar. So we get $\frac{MC}{CA’} = \frac{NB}{BA’}$.
Therefore M’ = bar {(A, MA) ; (B,$\frac{CN \times BA’}{NA’}$) ; (C,$\frac{NB \times CA’}{BA’}$)}.

We know that R = bar {(B, RC) ; (C, RB)}, using the partial barycenter theorem, we now have to prove that $\frac{CN \times BA’}{NA’ \times RC} = \frac{NB \times CA’}{BA’ \times RB} \Leftrightarrow \frac{CN \times RB}{BN \times RC} = \frac{CA’}{BA’}$ (1).

Knowing that A’ = T(A) = bar {(B, BA) ; (C, CA)}, we get $\frac{CA’}{BA’} = \frac{AB}{AC}$ so (1) $ \Leftrightarrow \frac{CN \times RB}{BN \times RC} = \frac{AB}{AC}$ (2).

Triangles RBN et RAC being similar, we get : $\frac{RB}{BN} = \frac{RA}{AC}$

Triangles RBN et RAC being similar, we have : $\frac{RB}{BN} = \frac{RA}{AC}$

Triangles RAB et RCN being similar, we get : $\frac{CN}{RC} = \frac{AB}{RA}$.

We deduct from these two equalities that (2) est true, so M ’$\in$ (AR)=(AN).

Continuation of the demonstration

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