A point M is chosen at random (uniformly) inside an equilateral triangle ABC of area 1. How to find the expected area of the triangle ABM?
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Hint: $$ {\bf E}[ \verb/Area/(\triangle ABM) ] = {\bf E}[ \verb/Area/(\triangle BCM) ] = {\bf E}[ \verb/Area/(\triangle CAM) ]$$ – achille hui May 12 '15 at 11:19

@achille hui But why is it so? – Koncopd May 12 '15 at 12:05

It is "obvious" once you remember $\triangle ABC$ is equilateral! Actually, we can drop the "equilateral" requirement. Under linear transform of the form $$ \begin{bmatrix}x \\ y\end{bmatrix} \mapsto \begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix} a & b\\c & d\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} + \begin{bmatrix}e \\ f\end{bmatrix} $$ the ratio of areas of any two geometric shapes are preserved. Since one can always find a linear transform which map a general triangle to an equilateral one, the statement is true for any triangle. – achille hui May 12 '15 at 13:46
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Try to prove that the expected area is equal to the volume of the right pyramid with base $ABC$ and height 1. The answer is $\frac{1}{3}$.
Hint: show that the height above any point $M\in\Delta ABC$ in the right pyramid is the area of $ABM$.
Alex Fok
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Can't get this geometric reasoning. Can you help me with building the probability space? – Koncopd May 12 '15 at 10:42

Ok, i got this finally. Joint density is 1 everywhere, so we can calculate the expected value as an iterated integral over the triangle with the linear function that changes from 0 at A to 1 at C. It is a right pyramid, and the integral is it's volume. – Koncopd May 13 '15 at 07:50


A proof that the area is 1/3 can be found here: https://math.stackexchange.com/questions/3525061 – granular bastard Feb 01 '20 at 06:31