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About Brocard point in a triangle




            Mih´aly Bencze     1



                In this paper we present some geometrical inequalities related to the Brocard point in a
            triangle.

            Definition 1. In any triangle ABC there exists a unique point P for each ∢PAB = ∢PBC =
            ∢PCA = ω. The point P is the Brocard point, they are named after Henri Brocard (1845-1922),
            a French mathematician.

                This point P is called the first Brocard point, and ω the Brocard angle.
            Theorem 1. (Brocard). In any triangle ABC holds the identity:

                                             ctg ω = ctg A + ctg B + ctg C.
                                                                                                 3
            Proof. The given relation is equivalent to sin(A − ω) · sin(B − ω) · sin(C − ω) = sin ω. But we
                  sin(A − ω)     CP sin(B − ω)       AP sin(C − ω)       BP
            have              =      ,            =      ,            =       and after multiplication holds
                     sin ω       AP      sin ω       BP       sin ω      CP
            the identity.                                                                                  □
            Remark 1. We have the following relations

                                                             P   2       Q
                                                               a     1 +    cos A
                                                X
                                        ctg ω =     ctg A =        =   Q
                                                             4sr          sin A
                                                  P     2      P
                                                     sin A        a sin A
                                               =    Q       = P          ,
                                                  2   sin A       a cos A
                                                                         2
                                                        2
                                                                 2
                                               2
                                            csc ω = csc A + csc B + csc C,
                                                              2sr
                                                                     .
                                                   sin ω = pP
                                                                  2 2
                                                                 a b
            Corollary 1. Denote P the Brocard point in the triangle ABC, then
                                              AP     BP     CP
                                                   +      +      = 2 cos ω.
                                                c      a      b
            Proof. In the triangle APC we have ∢CAP = A − ω, ∢CPA = π − A and using the sine rule
            we can write
                                                  sin(A − ω)     sin A
                                                              =       .
                                                      CP           b
            Similarly hold
                                       sin(B − ω)     sin B sin(C − ω)      sin C
                                                   =        ,            =       .
                                           AP           c        PB           a
            Expanding the expressions sin(A − ω), sin(B − ω), sin(C − ω) we get

                                                CP
                                                    = cos ω − sin ωctg A,
                                                 b
               1
                Profesor dr., Bras , ov, benczemihaly@gmail.com

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