Expected area of a random triangle with fixed perimeterI'm trying to calculate the expected area of a random triangle with a fixed perimeter of 1.My initial plan was to create an ellipse where one point on the ellipse is moved around and the triangle that is formed with the foci as the two other vertices (which would have a fixed perimeter) would have all the varying areas. But then I realized that I wouldn't account for ALL triangles using that method. For example, an equilateral triangle with side lengths one third would not be included.

Question

Expected area of a random triangle with fixed perimeterI&#039;m trying to calculate the expected area of a random triangle with a fixed perimeter of 1.My initial plan was to create an ellipse where one point on the ellipse is moved around and the triangle that is formed with the foci as the two other vertices (which would have a fixed perimeter) would have all the varying areas. But then I realized that I wouldn&#039;t account for ALL triangles using that method. For example, an equilateral triangle with side lengths one third would not be included.

Cullen Petersen · Accepted Answer

Step 1Since the Heron&#039;s formula gives:   A  =      1    4        (          a      2        +          b      2        +          c      2              )      2        −    2    (          a      4        +          b      4        +          c      4        )    ,, assuming that the random choice is made accordingly to the broken stick model (notice that not every choice for the side lengths satisfies the triangle inequality, hence for such cases we set the area as zero):                              E                [        A        ]                            =                              1          2                          ∫                      0                                1                                    ∫                      x                                1                                    max                      [            0            ,            (                          x              2                        +            (            y            −            x                          )              2                        +            (            1            −            y                          )              2                                      )              2                        −            2            (                          x              4                        +            (            y            −            x                          )              4                        +            (            1            −            y                          )              4                        )            ]                                  d        y                d        x                                          =                    0.00747998        …            Step 2A random model that makes more sense is to choice a random point (with respect to the uniform measure) on the set   T  =  {  (  u  ,  v  ,  w  )  :  u  ,  v  ,  w  ≥  0  ,  u  +  v  +  w  =      1    2    }, then pick the side lengths as   a  =  v  +  w  ,  b  =  u  +  w  ,  c  =  u  +  v. In such a way the triangle inequality is always fullfilled and the Heron&#039;s formula gives simply:  A  =                    u        v        w            2        ,so:      E    [  A  ]  =      8          6            ∫          T            u    v    w      d  μ  =      1          3            ∫          0              π              /            2            ∫          0              π              /            2            sin    5    ⁡  θ      cos    2    ⁡  θ      sin    4    ⁡  ϕ    d  θ    d  ϕ  =      π          70              3              .

klasyvea · Accepted Answer

Step 1Let   0  &amp;lt;  x  &amp;lt;  y  &amp;lt;  1 be the points at which the &quot;stick is broken&quot;, and so   x  ,  y  −  x  ,  1  −  y are the lengths of the three segments. For a triangle to be formed, the sum of any two sides must be greater than the third side. Therefore we get the following inequalities:  x  +  (  y  −  x  )  &amp;gt;  1  −  y    (  y  −  x  )  +  (  1  −  y  )  &amp;gt;  x    (  1  −  y  )  +  x  &amp;gt;  y  −  xStep 2Plotting these on a coordinate system gives a triangular region with vertices (0,1/2),(1/2,1/2),(1/2,1). So any pair (x,y) contained within that region results in a triangle of perimeter 1. I parameterize these pairs:      (                  a        1            2        ,                  1        +                  a          2                    2        )    ,, for   0  &amp;lt;      a    2    &amp;lt;      a    1    &amp;lt;  1. Now these can be plugged in Heron&#039;s formula (and simplified):  A  (      a    1    ,      a    2    )  =      1    4        (    1    −          a      1        )    (          a      1        −          a      2        )    (          a      2        )  Taking the integral for the average value:  E  (  A  )  =      1          A      (      R      )            ∫    0    1              ∫    0                  a        1              A  (      a    1    ,      a    2    )    d      a    2    d      a    1    =  0.0299199

Calculate the expected area of a random triangle with a fixed perimeter of 1.

Answered question

Answer & Explanation

New Questions in High school geometry