Logiciel libre de géométrie, d'analyse et de simulation multiplateforme par Yves Biton
 
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New theorem : The case of points Ai cocyclic. Image through T of the circumscribed circle.

publication jeudi 14 août 2014.


V) The case of points Ai cocyclic Image through T dof 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 A<sub1 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 whcih 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) wherr 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 $1 \le i \le n$, $A{’_i} = {T_{{A_1}{A_2}...{A_n}}}({A_i})$.

Construction for a quadrangle.

When point M is on the arc A1A2, we can use the preceeding construction creating first the barycenter g og (A1, MA1), (A2, MA2), (A3, MA3).

We construct point A’1, barycenter of (A2, A1A2), (A3, A1A3), (A4, A1A4) and point A’2, barycentre de (A1, A2A1), (A3, A2A3), (A4, A2A4).

We create first point g, barycenter of (A1, MA1), (A2, MA2) et (A3, MA3) applying the method already studied for a triangle case.

Point M’, barycentrer of (A1, MA1), (A2, MA2), (A3, MA3) et (A4, MA4) is therefore the intersection point between the lines (gA4) et (A’1A’2).

Continuation of the demonstration

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