Problem: O is the circumcenter of △ A B C, which is not a right triangle. | A B ⋅ C O | | A C ⋅ B O | = | A B ⋅ B O | | A C ⋅ C O | = 3.Find tan ⁡ A. Here A B ⋅ C O represents the dot product of vector A B → and C O →

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Problem:   O is the circumcenter of   △  A  B  C, which is not a right triangle.                |      A      B      ⋅      C      O          |                  |      A      C      ⋅      B      O          |        =                |      A      B      ⋅      B      O          |                  |      A      C      ⋅      C      O          |        =  3.Find   tan  ⁡  A. Here   A  B  ⋅  C  O represents the dot product of vector             A      B        →   and             C      O        →

Trey Ross · Accepted Answer

Let us record as in the OP the values for the scalar products, then start the computation using the formulas   a  =  2  R  sin  ⁡  A (and the similar ones for   b  ,  c) to express the appearing sines, and       b    2    +      c    2    −      a    2    =  2  b  c  cos  ⁡  A (and the similar ones) to express the appearing cosines.                          |        A        B        ⋅        C        O              |                            =        A        B        ⋅        C        O        ⋅        cos        ⁡        (                                    A            B            ,            C            O                    ^                )        =        c        R                cos        ⁡                  (                      π            2                    −          A          +          B          )                =        c        R        sin        ⁡        (        A        −        B        )                 ,                                  |        A        C        ⋅        B        O              |                            =        A        C        ⋅        B        O        ⋅        cos        ⁡        (                                    A            C            ,            B            O                    ^                )        =        b        R                cos        ⁡                  (                      π            2                    −          A          +          C          )                =        b        R        sin        ⁡        (        A        −        C        )                 ,                                  |        A        B        ⋅        B        O              |                            =        A        B        ⋅        B        O        ⋅        cos        ⁡        (                                    A            B            ,            B            O                    ^                )        =        c        R                cos        ⁡                  (                      π            2                    −          C          )                =        c        R        sin        ⁡        C        =        2                  R          2                          sin          2                ⁡        C                 ,                                  |        A        C        ⋅        C        O              |                            =        A        C        ⋅        C        O        ⋅        cos        ⁡        (                                    A            C            ,            C            O                    ^                )        =        b        R                cos        ⁡                  (                      π            2                    −          B          )                =        b        R        sin        ⁡        B        =        2                  R          2                          sin          2                ⁡        B                 .            From the last two equalities, and the given second proportion, we get  3  =                |      A      B      ⋅      B      O          |                  |      A      C      ⋅      C      O          |        =                    sin        2            ⁡      C                      sin        2            ⁡      B        =            c      2              b      2           .So   c  =  b      3  . We may want to norm   b  =  2, so then   c  =  2      3  . Let us use the first condition...                    ±        3                            =                                        |            A            B            ⋅            C            O                      |                                          |            A            C            ⋅            B            O                      |                                                            =                              c            R            sin            ⁡            (            A            −            B            )                                b            R            sin            ⁡            (            A            −            C            )                          =                  c          b                ⋅                              sin            ⁡            A            cos            ⁡            B            −            cos            ⁡            A            sin            ⁡            B                                sin            ⁡            A            cos            ⁡            C            −            cos            ⁡            A            sin            ⁡            C                                                            =                  c          b                ⋅                              a            ⋅                                                            a                  2                                +                                  c                  2                                −                                  b                  2                                                            2                a                c                                      −                                                            b                  2                                +                                  c                  2                                −                                  a                  2                                                            2                b                c                                      ⋅            b                                a            ⋅                                                            a                  2                                +                                  b                  2                                −                                  c                  2                                                            2                a                b                                      −                                                            b                  2                                +                                  c                  2                                −                                  a                  2                                                            2                b                c                                      ⋅            c                                                            =                              (                          a              2                        +                          c              2                        −                          b              2                        )            −            (                          b              2                        +                          c              2                        −                          a              2                        )                                (                          a              2                        +                          b              2                        −                          c              2                        )            −            (                          b              2                        +                          c              2                        −                          a              2                        )                                                            =                                            a              2                        −                          b              2                                                          a              2                        −                          c              2                                               .            Recall the norming making       b    2    =  4,       c    2    =  3      b    2    =  12. The two chances for       a    2   are now   10 and   16  =  4  +  12, where the above becomes   −  3  =            10      −      4              10      −      12       and   +  3  =            16      −      4              16      −      12      . The second case gives rise to a triangle with   A  =      90    ∘  , which was excluded in the title. So we have to work with the triangle with sides:                            a          =                      10                               ,                     b          =          2                     ,                     c          =          2                      3                    =                      12                               .                        Let us compute   tan  ⁡  A by explicitly computing   cos  ⁡  A,   S,   R,   sin  ⁡  A in this order:                    cos        ⁡        A                            =                                            b              2                        +                          c              2                        −                          a              2                                            2            b            c                          =                              4            +            12            −            10                                2            ⋅            2            ⋅            2                          3                                      =                  6                      8                          3                                      =                                            3                        4                                   ,                                      S          2                                    =        s        (        s        −        a        )        (        s        −        b        )        (        s        −        c        )                (Heron)                                          =                  1          16                (        a        +        b        +        c        )        (        b        +        c        −        a        )        (        a        +        c        −        b        )        (        a        +        b        −        c        )        =                  1          16                                  (                (        b        +        c                  )          2                −                  a          2                                  )                                  (                          a          2                −        (        b        −        c                  )          2                                  )                                                  =                  1          16                                  (                2        b        c        +        (                  b          2                +                  c          2                −                  a          2                )                          )                                  (                2        b        c        −        (                  b          2                +                  c          2                −                  a          2                )                          )                =                  1          16                                  (                4                  b          2                          c          2                −        (                  b          2                +                  c          2                −                  a          2                          )          2                                  )                                                  =                  1          16                                  (                4        ⋅        4        ⋅        12        −                  6          2                                  )                =                  39          4                         ,                            S                            =                              39                    2                         ,                            R                            =                              a            b            c                                4            S                          =        2                              10            13                                   ,                            sin        ⁡        A                            =                  a                      2            R                          =                              10                                4                          10                          /              13                                      =                                            13                        4                                   ,                            tan        ⁡        A                            =                              sin            ⁡            A                                cos            ⁡            A                          =                                            13                                  /            4                                              3                                  /            4                          =                                                                                                                             13                    3                                                                                                          .

Problem: O is the circumcenter of <mi mathvariant="normal">△ A B

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