Expected area of a triangle where 1 point is within another triangleSuppose we have triangle ABC with area k with a point P chosen inside ABC. What is the expected area of triangle PBC?I'm pretty sure if we let P be the centroid we get k/3. Also, how would you solve this question for an n-sided polygon?

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Expected area of a triangle where 1 point is within another triangleSuppose we have triangle ABC with area k with a point P chosen inside ABC. What is the expected area of triangle PBC?I&#039;m pretty sure if we let P be the centroid we get k/3. Also, how would you solve this question for an n-sided polygon?

Marlene Welch · Accepted Answer

Step 1Let the vertices of the triangle be a, b, c. Then the map                    (1)                              g                :                (        u        ,        v        )        ↦                  {                                                    x                (                u                ,                v                )                                            =                                  a                  1                                +                u                (                                  b                  1                                −                                  a                  1                                )                +                v                (                                  c                  1                                −                                  a                  1                                )                                                                    y                (                u                ,                v                )                                            =                                  a                  2                                +                u                (                                  b                  2                                −                                  a                  2                                )                +                v                (                                  c                  2                                −                                  a                  2                                )                                 ,                                                                  maps the arbitrary triangle from  (  x  ,  y  )  =  (      a    1    ,      a    2    )  ,  (      b    1    ,      b    2    )  ,  (      c    1    ,      c    2    ) bijectively onto a right triangle with   (  u  ,  v  )  =  (  0  ,  0  )  ,  (  1  ,  0  )  ,  (  0  ,  1  ) (make a sketch for visualization). Therefore the transformation formula for multiple integrals can be used, and we obtain      ∫    T    f  (  x  ,  y  )        d    (  x  ,  y  )  =      ∫                  T        ′              f            (        x  (  u  ,  v  )  ,  y  (  u  ,  v  )            )                    |            J          g        (  u  ,  v  )            |               d    (  u  ,  v  )     .The Jacobian determinant is a constant: From (1) we obtain      J          g        (  u  ,  v  )  =  det      [                                        x            u                                                x            v                                                            y            u                                                y            v                                ]    =  (      b    1    −      a    1    )  (      c    2    −      a    2    )  −  (      c    1    −      a    1    )  (      b    2    −      a    2    )     .Step 2Therefore we can write      ∫    T    f  (  x  ,  y  )        d    (  x  ,  y  )  =            |            J          g                  |              ∫    0    1          ∫    0          1      −      u        f            (        x  (  u  ,  v  )  ,  y  (  u  ,  v  )            )        d  v     d  u     .The area of an arbitrary triangle a,b,p is   f  =  1      /    2  ⋅  base  ⋅  height. In u,v coordinates the base is 1 and the height is v. For normalization the integral has to be divided by the total area that is 1/2. The expected area of a random triangle a,b,p is therefore                    (2)                              E                [                  A                                    a                        ,                          b                        ,                          p                                      ]        =                              1            2                                1            2                                                |                                    J                      g                                    |                          ∫          0          1                          ∫          0                      1            −            u                          v        d        v                 d        u                 =                  1          6                                      |                          (                  b          1                −                  a          1                )        (                  c          2                −                  a          2                )        −        (                  c          1                −                  a          1                )        (                  b          2                −                  a          2                )                              |                              As the area of the triangle a, b, c is      A                  a            ,              b            ,              c              =      1    2              |        (      b    1    −      a    1    )  (      c    2    −      a    2    )  −  (      c    1    −      a    1    )  (      b    2    −      a    2    )            |      the expected area of the random triangle is 1/3 of the area of the original triangle, independently which side is used as a base.

Expected area of a triangle where 1 point is within another triangle

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