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

New theorem : Case of three cocyclic points. Image of the circumscribed circle of these points.

publication Thursday 14 August 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\neA and M\neB 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\inE. 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 : \Gammabeing 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 \ne1 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 \Gammato 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

Return to the main page of the new theorem