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

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

modification lundi 14 août 2014.

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

We suppose that A1A2….An 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), … (An, MAn).

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),…,(An, A1An).

If M = A2, then M’ = A’2 où A’2 is the barycenterof the n - 1 balanced points (A1, A2A1), (A3, A2A3),…,(An, A2An).

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, ….MA2AnA1, 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), … (An-1, MAn-1), i.e. $P = T_{A_1A_2...A_{n - 1}}(M)$, then M’ is the barycenter of (P,MA1+ MA2 + …. + MAn-1) et de (An, MAn).

So M’ is the intersection point of segments [PAn] 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…AnAn+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), …, (An+1, A1An+1).

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

A’2 is barycenter of (A1, A2A1), (A3, A2A3), … (An+1, A2An+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), …, (An, A1An).

B2 is the barycenter of (A1, A2A1), (A3, A2A3), … (An, A2An).

A’1 is the barycenter of (B1, A1A2+A1A3+...+A1An) et (An+1, A1An+1).

A’2 is the barycenter of (B2, A2A1+A2A3+...+A2An ) et (An+1, A2An+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 (An+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 (An+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 (An+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 An+1 with positive coefficients. So M’ lies on segment [PAn+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+...+MAn) and of (An+1, MAn+1).

So M" is on segment [PAn+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...An 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 $1 \le i \le n$, $A{’_i} = {T_{{A_1}{A_2}...{A_n}}}({A_i})$.