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New theorem based on Ptolemy’s theorem : Preliminary results

publication Thursday 14 August 2014.


In this article we will see a few results we will later use for the new theorem


Proposition 1.1 : If MAB is a triangle et I the foot of the interior bisector form point M, the I belongs to segment [AB] and \frac{MA}{MB} = \frac{IA}{IB}

1.2

This rather basical property won’t be demonstrated here.


Proposition 1.2 (Ptolemy’ theorem) : If A, B, C et D are four points son aligned , then AC \times BD \le AB \times CD + BC \times DA with equality if, and only of, the points are cocyclic and the quadrangle ABCD convex.

This property won’t be demonstrated here.


Proposition 1.3 : If A et B are two distincts points and k a real number, k>0, the set C_{A,B,k} containing the points M such as \frac{MA}{MB}=k is :
- The perpendicular bisector of segment [AB] if k = 1.
- Else the circle of diameter [IJ] where I = bar (A,1); (B, k) et J = bar (A,1); (B, - k).

More, if M is a some point of this circle non lying on line (AB), I et J are the repectives feet of the interior and exterior bisectors from M in triangle MAB.

Demonstration :
- If k = 1 :
\frac{MA}{MB} = k \Leftrightarrow \frac{MA}{MB} = 1 \Leftrightarrow MA = MB \Leftrightarrow M est sur la médiatrice [AB].
- If k \ne 1 :
\frac{MA}{MB} = k \Leftrightarrow MA = kMB \Leftrightarrow M{A^2} = {\left( {kMB} \right)^2} \Leftrightarrow {\left( {\overrightarrow {MA} } \right)^2} = {\left( {k\overrightarrow {MB} } \right)^2}.
So  \frac{MA}{MB} = k \Leftrightarrow {\left( {\overrightarrow {MA} } \right)^2} - {\left( {k\overrightarrow {MB} } \right)^2} = 0 \Leftrightarrow \left( {\overrightarrow {MA}  + k\overrightarrow {MB} } \right).\left( {\overrightarrow {MA}  - k\overrightarrow {MB} } \right) = 0.

Naming I the barycenter of (A,1); (B, k) and J the barycenter of (A,1) et (B, - k) (these barycenters exist because  k > 0 and  k \ne 1 so  1+k \ne 0 et  1-k \ne 0 ) :
For each point M, \overrightarrow {MA}  + k\overrightarrow {MB}  = \left( {1 + k} \right)\overrightarrow {MI} and \overrightarrow {MA}  - k\overrightarrow {MB}  = \left( {1 - k} \right)\overrightarrow {MJ}
So \frac{MA}{MB} = k \Leftrightarrow (1 + k)\overrightarrow {MI} .(1 - k)\overrightarrow {MJ}  = 0 \Leftrightarrow \overrightarrow {MI} .\overrightarrow {MJ}  = 0 because  1+k \ne 0 and  1-k \ne 0.
So \frac{{MA}}{{MB}} = k \Leftrightarrow M is on circle with diameter [IJ].

Considering a point M of this circle which does’nt belong to line (AB) and I’ the foot of the internal bisector of MAB from M. According to proposition 1.1, we know that \frac{I’A}{I’B}=\frac{MA}{MB}=k. So point I’ is on the circle and on segment [AB]. Finally we get I’=I.

We deduct immediatly from this that J is the foot of the external bisector of M in triangle MAB since these two bisectors are perpendicular.

The figure below illustrates this property :


Proposition 1.4 : If i is thel’inversion of center O and ration k, then, for all points M and N with rrepective images M’ et N’ through i, we get : M’N’ = \frac{{\left| k \right|MN}}{{OM\;ON}}

Demonstration :
We are in the complex plane with point O as origin.
If M, N, M’ et N’ have for affixes respectively \alpha, \beta, \alpha’, \beta’ then :
\alpha ’ = \frac{k}{\overline \alpha  } and \beta ’ = \frac{k}{\overline \beta  }.

So M’N’=\left| \beta’-\alpha’ \right|=\left| \frac{k}{\overline{\beta}}-\frac{k}{\overline {\alpha}} \right|=\left| k \right|\times \left| \frac{1}{\overline {\beta}} - \frac{1}{\overline {\alpha }}\right|=\left| k \right|\times \left| \frac{\overline{\alpha}-\overline{\beta}}{\overline{\alpha}\overline{\beta}} \right|=\left| k \right|\times \frac{\left| \overline{\alpha-\beta} \right|}{\left| \alpha \beta  \right|}=\left| k \right|\times \frac{\left| \alpha -\beta  \right|}{\left| \alpha \beta  \right|}

So we have M’N’=\frac{\left| k \right|\times \left| \alpha -\beta  \right|}{\left| \alpha  \right|\times \left| \beta  \right|}=\frac{\left| k \right|\times MN}{OM\times ON}


Proposition 1.5 : Let us consider a triangle ABC non isosceles in A, I and J the fett of the two bisectors of angle in A and K the midpoint of [IJ].
Then line (AK) est tangent in A to the circumscribed to ABC.


Démonstration :
We suppose that triangle ABC is not isosceles in A.

Let’s call\Gamma the circumscribed circle of ABC, M the second intersection point of the internal bisector (AI) with the circle \Gamma eand N the second intersection point of de la external bisecto (AJ) with circle \Gamma.

The internal and externla bisectors being perpendicular, the triangle AMN is rectangle in A so it’s circumscribed circle (which is \Gamma) has for diameter segment [MN]. So segment [MN] is a diameter of \Gamma.

Applying the inscribed angle theorem we deduct from it that \widehat {BMN} = \widehat {BAN} =  \widehat {BAI} and \widehat {CMN} = \widehat {CAN} =  \widehat {CAI}

(AI) is he internal bisector of angle A in so \widehat {BMN} = \widehat {BAI} =  \widehat {CAI}.
We deduct that \widehat {BMN} = \widehat {CMN}

Having OM = OB = OC, triangles OMN et OMC are isometric so MB = MC. Knowing that OB = OC, (MO) is the perpendicualr bisector of [BC].

To demonstrate that (AO) is perpendicualr to (AK) we will demonstrate that \overrightarrow {AO} .\overrightarrow {AK}  = 0.

K being the midpoint of [IJ], \overrightarrow {AK}  = \frac{1}{2}\left( {\overrightarrow {AI}  + \overrightarrow {AJ} } \right).

O being the midpoint of [MN], \overrightarrow {AO}  = \frac{1}{2}\left( {\overrightarrow {AM}  + \overrightarrow {AN} } \right).
So we have \overrightarrow {AO} .\overrightarrow {AK}  = \frac{1}{4}\left( {\overrightarrow {AM}  + \overrightarrow {AN} } \right).\left( {\overrightarrow {AI}  + \overrightarrow {AJ} } \right) = \frac{1}{4}\left( {\overrightarrow {AM} .\overrightarrow {AI}  + \overrightarrow {AN} .\overrightarrow {AI}  + \overrightarrow {AM} .\overrightarrow {AJ}  + \overrightarrow {AN} .\overrightarrow {AJ} } \right).
Yet \overrightarrow {AM} .\overrightarrow {AI}  = 0and \overrightarrow {AN} .\overrightarrow {AJ}  = 0.
Moreover \overrightarrow {AN} .\overrightarrow {AI}  = \left( {\overrightarrow {AM}  + \overrightarrow {MN} } \right).\overrightarrow {AI}  = \overrightarrow {AM} .\overrightarrow {AI}  + \overrightarrow {MN.} \overrightarrow {AI}  = \overrightarrow {MN.} \overrightarrow {AI} car (AM) \bot \left( {AI} \right).
And \overrightarrow {AM} .\overrightarrow {AJ}  = \left( {\overrightarrow {AN}  + \overrightarrow {NM} } \right).\overrightarrow {AJ}  = \overrightarrow {AN} .\overrightarrow {AJ}  + \overrightarrow {NM} .\overrightarrow {AJ}  = 0 + \overrightarrow {NM} .\overrightarrow {AJ}  = \overrightarrow {NM} .\left( {\overrightarrow {AI}  + \overrightarrow {IJ} } \right) = \overrightarrow {NM} .\overrightarrow {AI}  + \overrightarrow {NM} .\overrightarrow {IJ}  = \overrightarrow {NM} .\overrightarrow {AI} .
We deduct \overrightarrow {AO} .\overrightarrow {AK}  = \frac{1}{4}\left( {\overrightarrow {MN} .\overrightarrow {AI}  + \overrightarrow {NM} .\overrightarrow {AI} } \right) = 0 so (AO) \bot (AK).


Proposition 1.6 : If T is the tangent in A to a circle \Gamma of center O, then each circle with it’s center on T and going through A is unvariant by the inversion i of center O et ratio OA^2.

Demonstration : We are in the complex plane with an orthonormal frame whose origin is O.
we will set OA=R, AB=r and b will stand for B affix.
The inversion i being involutive (i o i=Id), to demonstrate that i\left( \Gamma  \right) = \Gamma it’s enough to show that i\left( \Gamma  \right) \subset \Gamma .
A some point M of the circle of centre B and going through A has an affix as follows : z = b + re^{i\theta } awith \theta real number.
It’s image through the inversion of center O and ration R^2 has an affix of this form : z’=\frac{R^2}{\overline {z}}.

To show that M’ belongs to \Gamma we just have to show that \left| z’ - b \right| = r.

\left| z’-b \right|=\left| \frac{R^2}{\overline {z}}-b \right|=\left| \frac{R^2-b\overline {z}}{\overline {z}} \right| with \overline {z}=\overline {b+r\mathrm{e}^{i\theta }}=\overline {b}+r\mathrm{e}^{-i\theta }.
.

So z’-b=\frac{R^2-b\times \left( \overline {b}+r\mathrm{e}^{-i\theta } \right)}{\overline {b}+r\mathrm{e}^{-i\theta }}=\frac{R^2-b\overline {b}-br\mathrm{e}^{-i\theta}}{\overline {b}+r\mathrm{e}^{i\theta}}=\frac{R^2-{\left| b \right|}^2-br\mathrm{e}^{-i\theta}}{\overline {b}+r\mathrm{e}^{i\theta}}.

Yet {\left| b \right|^2} = O{B^2} = O{A^2} + A{B^2} = {R^2} + {r^2} because triangle OAB is rectangle in A (Pythagora theorem).

So \left| z’-b \right|=\left| \frac{R^2-\left( R^2+r^2 \right)-br\mathrm{e}^{-i\theta }}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=\left| \frac{-r\left( r+b\mathrm{e}^{-i\theta} \right)}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=\left| \frac{-r\mathrm{e}^{i\theta}\left( r\mathrm{e}^{i\theta}+b \right)}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=r\left| \frac{r\mathrm{e}^{-i\theta}+b}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|.

Yet \left| \frac{r\mathrm{e}^{i\theta}+b}{\overline b+r\mathrm{e}^{-i\theta }} \right|=\left| \frac{r\mathrm{e}^{i\theta}+b}{\overline{r\mathrm{e}^{i\theta}+b}} \right|=\frac{\left| r\mathrm{e}^{i\theta }+b \right|}{\left| \overline{r\mathrm{e}^{i\theta}+b} \right|}=\frac{\left| r\mathrm{e}^{i\theta}+b \right|}{\left| r\mathrm{e}^{i\theta}+b \right|}=1.

So \left| z’-b \right| = r and point M’ really belongs to the circle\Gamma.

Continuation of the demonstration

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