New theorem : The case of points A_i cocyclic. Image through T of the circumscribed circle.

V) The case of points A_i cocyclic Image through T of the circumscribed circle.

We suppose that A₁A₂….A_n is a convex polygon inscribed in a circle $\Gamma$.

We also suppose that M lies on the circle arc of $\Gamma$ with summits A₁ and A₂ which does’nt contain the other points.

We name M’ the barycenter of the n balanced points(A₁, MA₁), (A₂, MA₂), … (A_n, MA_n).

In other words, $M’ = T_{A_1A_2...A_n}(M)$.

If M = A₁, then M’ = A’₁ où A’₁ is the barycenter of the n - 1 balanced points (A₂, A₁A₂), (A₃, A₁A₃),…,(A_n, A₁A_n).

If M = A₂, then M’ = A’₂ où A’₂ is the barycenterof the n - 1 balanced points (A₁, A₂A₁), (A₃, A₂A₃),…,(A_n, A₂A_n).

For each point Q :
$\left( A_1A_2 + A_1A_3 + ... + A_1A_n \right)\overrightarrow {QA’_1} = A_1A_2 \overrightarrow {QA_2} + A_1A_3 \overrightarrow {QA_3} + ... + A_1A_n \overrightarrow {QA_n}$.

For Q = A₂, we get that A₂ is the barycenter of $(A’_1,A_1A_2 + A_1A_3 + ... + A_1A_n)$, $(A_3,- A_1A_3)$, …, $(A_n, - A_1A_n)$ (sum of coefficients : A₁A₂).

In the same way, A₁ is the barycenter of $(A’_2,A_2A_1 + A_2A_3 + ... + A_2A_n)$, $(A_3, - A_2A_3)$, …, $(A_n, - A_2A_n)$ (sum of coefficients : A₁A₂).

Then we get :

A₁ is the barycenter of $(A’_2,\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_1A_2} \times MA_1)$,
$({A_3},- \frac{A_2A_3}{A_1A_2} \times MA_1)$,..., $({A_n},- \frac{A_2A_n}{A_1A_2} \times MA_1)$ (sum of coefficients : MA₁).

A₂ is the barycenter of $(A’_1,\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2} \times MA_2)$, $(A_3,- \frac{A_1A_3}{A_1A_2} \times M{A_2})$, …, $(A_n,- \frac{A_1A_n}{A_1A_2} \times MA_2)$ (sum of coefficients : MA₂).

Applying the partial barycenter theorem we get that M’ is the barycenter of $(A’_1,\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2} \times MA_2)$, $(A’_2,\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_1A_2} \times M{A_1})$, $(A_3,- \frac{A_2A_3}{A_1A_2} \times M{A_1})$, $(A_3,- \frac{A_1A_3}{A_1A_2} \times MA_2)$, ....., $(A_n, - \frac{A_2A_n}{A_1A_2} \times MA_1)$, $(A_n,- \frac{A_1A_n}{A_1A_2} \times M{A_2})$, $(A_2,MA_2)$, $(A_3,MA_3)$, ..., $(A_n,MA_n)$.

Therefore M’ is the barycenter of $(A{’_1},\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2} \times MA_2)$, $(A’_2,\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_1A_2} \times M{A_1})$, $({A_3},\frac{MA_3 \times A_1A_2 - MA_1 \times A_2A_3 - MA_2 \times A_1A_3}{A_1A_2})$, ..., $({A_n},\frac{MA_n \times A_1A_2 - MA_1 \times A_2A_n - MA_2 \times A_1An}{A_1A_2})$.

Applying n-2 times the Ptomemy’s theorem to convexes quadranges inscribed in $\Gamma$ : MA₂A₃A₁, ….MA₂A_nA₁, we see that the n-2 last coefficients ar null and so M’$\in$ [A’₁A’₂].

Construction of point M’ :

If P is the barycenter of the n balanced points (A₁ ; MA₁), (A₂ ; MA₂), … (A_n-1, MA_n-1), i.e. $P = T_{A_1A_2...A_{n - 1}}(M)$, then M’ is the barycenter of (P,MA₁+ MA₂ + …. + MA_n-1) et de (A_n, MA_n).

So M’ is the intersection point of segments [PA_n] and [A’₁A’₂].

Injectivity

We assume that two points M et N are located on the circle arc of $\Gamma$ with summits A₁ and A₂ which does’nt contain the other points and that M and N have the same image M’ through the transformation $T_{A_1A_2...A_n}$.

The we deduct from the preceeding issues that the barycenters of $(A’_1,\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2} \times MA_2)$, $(A’_2,\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_1A_2} \times MA_1)$ et de $(A’_1,\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2} \times NA_2)$, $(A’_2,\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_1A_2} \times NA_1)$ are the same and their coefficents are proportionnal.

So we get $\frac{MA_1}{MA_2} = \frac{NA_1}{NA_2} = k$ where k > 0. The points M et N are therefore located at the intersection of cercle $\Gamma$ with the circle arc of summits A₁ and A₂ of $\Gamma$. But this circle can only have one intersection point with this arc.

Therefore we have M = N.

The transformation $T_{A_1A_2...A_n}$ is injective on the circle $\Gamma$.

Reciprocal problem : Given a point M’ of segment [A’₁A’₂], let us prove that there is only one unique point M of arc A₁A₂ which does’nt contain the other quadrangle’s summits such as getting $T_{A_1A_2....A_n}\left( M \right) = M’$.

For this we will use a recurrent demonstration. The result has already been proved when n=2 et n=3.

Let us suppose it has been demonstrated for a convex polygon with n summits.

Nous suppose that A₁A₂…A_nA_n+1 is a convex polygon inscribed in the circle $\Gamma$.

We note $A{’_1} = T_{A_1A_2...A_{n + 1}}(A_1)$.

A’₁ is the barycenter of (A₂, A₁A₂), (A₃, A₁A₃), …, (A_n+1, A₁A_n+1).

We note $A{’_2} = T_{A_1A_2...A_{n + 1}}(A_2)$.

A’₂ is barycenter of (A₁, A₂A₁), (A₃, A₂A₃), … (A_n+1, A₂A_n+1).

We note $B_1 = T_{A_1A_2...A_n}(A_1)$ et $B_2 = T_{A_1A_2...A_n}(A_2)$.

B₁ is the barycenter of (A₂, A₁A₂), (A₃, A₁A₃), …, (A_n, A₁A_n).

B₂ is the barycenter of (A₁, A₂A₁), (A₃, A₂A₃), … (A_n, A₂A_n).

A’₁ is the barycenter of (B₁, A₁A₂+A₁A₃+...+A₁A_n) et (A_n+1, A₁A_n+1).

A’₂ is the barycenter of (B₂, A₂A₁+A₂A₃+...+A₂A_n ) et (A_n+1, A₂A_n+1).

A’₁ is barycenter of (B₁,$\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2 + A_1A_3 + ... + A_1A_n + A_1A_{n + 1}}$) et (A_n+1,$\frac{A_1A_{n + 1}}{A_1A_2 + A_1A_3 + ... + A_1A_n + A_1A_{n + 1}}$) (sum of coefficients 1).

A’₂ is barycenter of (B₂,$\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_2A_1 + A_2A_3 + ... + A_2A_n + A_2A_{n + 1}}$) and (A_n+1,$\frac{A_2A_{n + 1}}{A_2A_1 + A_2A_3 + ... + A_2A_n + A_2A_{n + 1}}$) (sum of coefficients 1).

If M’ is a point of segment [A’₁A’₂], M’ can be seen as barycenter of (A’₁, k₁) and (A’₂, k₂) where k₁ et k₂ are two real positive numbers.

M’ is therefore barycenter of (B₁,$\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2 + A_1A_3 + ... + A_1A_n + A_1A_{n + 1}} \times {k_1}$), (B₂,$\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_2A_1 + A_2A_3 + ... + A_2A_n + A_2A_{n + 1}} \times {k_2}$) and of (A_n+1,$\frac{k_1A_1A_{n + 1}}{A_1A_2 + A_1A_3 + ... + A_1A_n + A_1A_{n + 1}} + \frac{k_2A_2A_{n + 1}}{A_2A_1 + A_2A_3 + ... + A_2A_n + A_2A_{n + 1}}$).

No let us consider P, barycenter of (B₁,$\frac{A_1A_2 + A_1A_3 + ... + A_1A_n}{A_1A_2 + A_1A_3 + ... + A_1A_n + A_1A_{n + 1}} \times k_1$) et de (B₂,$\frac{A_2A_1 + A_2A_3 + ... + A_2A_n}{A_2A_1 + A_2A_3 + ... + A_2A_n + A_2A_{n + 1}} \times {k_2}$).

M’ is therefore barycenter of P and A_n+1 with positive coefficients. So M’ lies on segment [PA_n+1].

The barycenter’s coefficients being positive, we can say that P is on segment [B₁B₂].

According to our hypothesis of recurrence, we can say that there is a unique point M of arc A₁A₂ not containing the other points such as $P = {T_{{A_1}{A_2}...{A_n}}}(M)$.

Let us note $M" = {T_{{A_1}{A_2}...{A_n}{A_{n + 1}}}}(M)$.

M" is then barycenter of (P,MA₁+MA₂+...+MA_n) and of (A_n+1, MA_n+1).

So M" is on segment [PA_n+1].

Moreover we have demonstrated that M" is on segment [A’₁A’₂].

Point M’ assuming also these two consitions we deduct that M" = M’.

The unicity of the point M of arc A₁A₂ is set since we have already got that $T_{A_1A_2...A_n A_{n + 1}}$ is injective on the circle.

So the recurrence is achieved.

Theorem 5.1 : A₁A₂...A_n being a convex polygon inscribed in a cercle $\Gamma$, the transformation $T_{A_1A_2...A_n}$ sets a bijection from $\Gamma$ into the polygon A’₁A’₂…A’_n where, for $1 \le i \le n$, $A{’_i} = {T_{{A_1}{A_2}...{A_n}}}({A_i})$.

Construction for a quadrangle.

Continuation of the demonstration

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