I found this in a textbook without a solution and I wasnt able to solve it myself.Let ABCD be a tetrahedron with all faces acute. Let E be the mid point of the longer arc AB on a circle ABD. Let F be the mid point of the longer arc BC on a circle BCD. Let G be the mid point of the longer arc AC on a circle ACD.Show that points D,E,F,G lie on a circle.My approach to that was to try to show these point were co-planear. Since they all lie on one sphere (the one with inscribed tetrahedron ABCD) that would solve the problem. Needles to say I failed at that.

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boomzwamhc · Accepted Answer

Let             a      →        =                    D        A            →        ,            b      →        =                    D        B            →        ,            c      →        =                    D        C            →       and   a  ,  b  ,  c be corresponding magnitudes.Let us look at what happens on the plane holding circle ABD. Let X be the circle&#039;s center and       E    ′   be the mid point of the shorter arc AB. It is not hard to see  ∠      E    ′    D  B  =      1    2    ∠      E    ′    X  B  =      1    2    ∠  A  X      E    ′    =  ∠  A  D      E    ′  This implies   D      E    ′   is the angular bisector of   ∠  A  D  B and                     D                  E          ′                    →       is pointing along the direction                     a        →              a    +                    b        →              b  . Since DE is perpendicular to   D      E    ′  ,                     D        E            →       is pointing along the direction                     a        →              a    −                    b        →              b  .To proceed, we will re-express this fact in terms of barycentric coordinates.For any   P  ∈            R        3  , the barycentric coorindates of P with respect to tetrahedron ABCD is a 4-tuple   (      α    P    ,      β    P    ,      γ    P    ,      δ    P    ) which satisfies:      α    P    +      β    P    +      γ    P    +      δ    P    =  1     and               P      →        =      α    P              A      →        +      β    P              B      →        +      γ    P              C      →        +      δ    P              D      →      In particular, the barycentric coordinates for D is (0,0,0,1).Let&#039;s look at point E. Since E lies on the plane holding ABD,       γ    E    =  0. Since DE is pointing along the direction                     a        →              a    −                    b        →              b  , we find       α    E    :      β    E    =      1    a    :  −      1    b  . From this, we can deduce there is a       λ    E   such that  (      α    E    ,      β    E    ,      γ    E    ,      δ    E    )  =      (                  λ        E            a        ,    −                  λ        E            b        ,    0    ,    1    +          λ      E                      a        −        b                    a        b              )  By a similar argument, we can find       λ    F   and       λ    G   such that  (      α    F    ,      β    F    ,      γ    F    ,      δ    F    )  =      (    0    ,                  λ        F            b        ,    −                  λ        F            c        ,    1    +          λ      F                      b        −        c                    b        c              )      (      α    G    ,      β    G    ,      γ    G    ,      δ    G    )  =      (    −                  λ        G            a        ,    0    ,                  λ        G            c        ,    1    +          λ      G                      c        −        a                    c        a              )  In terms of barycentric coordinates,   D  ,  E  ,  F  ,  G are coplanar when and only when following determinant evaluates to zero.      D                      =                    d        e        f                  |                                        α            E                                                β            E                                                γ            E                                                δ            E                                                            α            F                                                β            F                                                γ            F                                                δ            F                                                            α            G                                                β            G                                                γ            G                                                δ            G                                                            α            D                                                β            D                                                γ            D                                                δ            D                                |    =      |                                        α            E                                                β            E                                                γ            E                                                δ            E                                                            α            F                                                β            F                                                γ            F                                                δ            F                                                            α            G                                                β            G                                                γ            G                                                δ            G                                                0                          0                          0                          1                      |    =      |                                        α            E                                                β            E                                                γ            E                                                            α            F                                                β            F                                                γ            F                                                            α            G                                                β            G                                                γ            G                                |  Substitute above expression of barycentric coordinates of   E  ,  F  ,  G into last determinant, we find      D    =      λ    E        λ    F        λ    G        |                                        1            a                                    −                      1            b                                    0                                      0                                      1            b                                    −                      1            c                                                −                      1            a                                    0                                      1            c                                |    =  0as the rows of determinant on RHS sum to zero.From this, we can conclude   D  ,  E  ,  F  ,  G are coplanar. Since   D  ,  E  ,  F  ,  G lie on the intersection of a sphere and a plane, they lie on a circle.

I found this in a textbook without a solution and I wasnt able to solve it myself. Let ABCD be

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