From any point P within a given
triangle, say ABC lines are drawn parallel to the two sides, meeting the 3rd side on points
D, E.
Show that the mean
value of triangle PDE's area is one-sixth ABC'S area.
The solver may assume that each point within ABC may be chosen with the same probability as others (i.e. uniform distribution).
The probability that a given point is at a given height relative to the height of the original triangle is proportional to the width at that height.
Integ{0 to 1} (1-h) dh = 1/2
gives the area of the triangle, including all points within the triangle. Multiply the integrand by the relative area at that point:
Integ{0 to 1} ((1-h) * h^2) dh
=
Integ{0 to 1} (h^2 - h^3) dh
= h^3 / 3 - h^4 / 4 | h=1
= 1/3 - 1/4 = 1/12
and of course divide by the total area:
1/12
----
1/2
= 1/6
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Posted by Charlie
on 2017-02-24 11:18:28 |
solution
