Distance between incenter and orthocenter

["Luis Lopes" <llopes@xxxxxxxxxxxxx>]

Sauda,c~oes,

Uma discussão interessante de uma outra lista.

[ ]'s
Luís

>Dear Gilles,
>
> >>  But the following result it unknow to me:For a point M :
> >>
> >>
> >> a MA^2 + b MB^2 + c MC^2 = 2p MI^2 + a IA^2 + b IB^2 + c IC^2
> >>
>
> >In France this result is "Leibniz's formula"
>
> >( u : v : w ) are barycentric coordinates of the point P relatively to
>ABC,
>thus :
>
>         -->    -->    -->
> >     u PA + v PB + w PC = 0
>
> >So for a point M :
>
>         -->                       -->  -->
> >     u MA^2 = u MP^2 +u PA^2+ 2u MP . PA
>
> >similarly with B and C, and the result by addition :
>
> >u MA^2 + v MB^2 + w MC^2 = (u+v+w) MP^2 + u PA^2 + b PB^2 + w PC^2
>(*)
>
> >In your problem, P = I, then u=a, v=b, w=c.
>
> >This result is the base of euclidean barycentric geometry :
>
> >Put X = u PA^2 + b PB^2 + w PC^2, for M =A, next B, next C, thus
>
>
> >     v AB^2 + w AC^2 = (u+v+w) AP^2 + X
>
> >     u BA^2 + w BC^2 = (u+v+w) BP^2 + X
>
> >     u CA^2 + v CB^2 = (u+v+w) CP^2 + X
>
> >and, combining these equations with coefficients u,v,w :
>
> >2 uv AB^2 + 2 vw BC^2 + 2 uw CA^2 = 2 (u+v+w) X
>
> >The sidelengths a, b, c are sufficient to calculate X and
> >"Leibniz's formula" gives a relation between MA, MB, MC and MP.
>
>Thank your for this useful explaination.
>By means of "Leibnitz's formula" I have proved the relations:
>
>               1
>       GI^2 = --- (p^2 - 16Rr + 5r^2)
>               9
>
>                R
>       NI^2 = (--- - r)^2     <==>    Feuerbach's theorem  (N = 9ptC)
>                2
>
>with the same technique by which you have found IH^2 .
>
>By (*) we can prove also the Euler's formula
>
>       OI^2 = R^2 - 2Rr
>
>For M = O :
>
>
>a OA^2 + b OB^2 + c MC^2 = 2p OI^2 + a IA^2 + b IB^2 + c IC^2
>
>
>                      (p - a)bc   (p - b)ac   (p - c)ab
>2p R^2 = 2p OI^2 + a --------- + --------- + ---------
>                         p           p           p
>
>
>2p R^2 = 2p OI^2 + abc
>
>
>2p R^2 = 2p OI^2 + 4rRP    ===>   OI^2 = R^2 - 2Rr
>
>
>I apologize for my bad english.
>
>Best regards
>
>Ercole
>
>P.S.
>
>In  I.F. Sharygin - Problems in plane geometry, MIR Publisher (pag 97),
>is called Leibnitz's theorem :
>
>Let M be an arbitrary point in the plane, G the centre of mass of a
>triangle ABC. Then the following equality is fulfilled:
>
>                                1
>3 MG^2 = MA^2 + MB^2 + MC^2 - --- ( AB^2 + BC^2 + CA^2 )     (**)
>                                3
>
>
>We can obtain (**) from (*) with P = G (1/3 : 1/3 : 1/3).
>
===
>This formula is very efficient : you can also calculate OK^2, GK^2 HK^2 and
>more other.
>
>I write perhaps a note in FG.
>
>Best regards
>
>Gilles
===