Expected value of the area of a triangle

Question

A point $M$ is chosen randomly within a triangle $ABC$ whose sides are equal. The area of the triangle is 1. Find the expected value of the area of triangle $ABM$.

I have one possible way to solve this problem as following: graph the triangle $ABC$, set the $x$-coordinate of $M$ and express its $y$-coordinate . Then set up the double integral to find the expected value, but I feel like it's a tedious way and there must be a more elegant way to solve it.

Answer 1

HINT.

Area is proportional to the distance $x$ of point $M$ from base $AB$. If $h$ is the height of triangle $ABC$ with respect to $AB$ then it is quite obvious that the expected value of $x$ is $$ \langle x\rangle = {1\over3}h. $$

Notice that the result holds for any triangle, not necessarily equilateral.

Answer 2

Let $S_1,S_2,S_3$ be the areas of ABM, ACM and BCM, respectively. Then:

1. $S_1+S_2+S_3=1$.
2. $S_1,S_2,S_3$ are identically distributed, hence $E S_1 = E S_2 =E S_3$.

Thus, $E S_1 = \frac 13$.

Answer 3

This proof does not suppose that $\triangle(ABC)$ is equilateral.

Note that $$\{M: {\rm Area}(AMB)>t\}=\triangle (A'CB')$$ Where $A'$, $B'$ are the points where the line parallele to $AB$ at distance equal to $2t/AB$ from $AB$ cuts the sides $CA$ and $CB$ respectively. Now, if the height from $C$ of $\triangle(ABC)$ is denoted by $h$ then the height of $\triangle(A'B'C)$ is given by $ h-2t/AB$. Therefore, $$\frac{{\rm Area}(A'CB')}{{\rm Area}(ACB)}=\left(\frac{h-2t/AB}{h}\right)^2=\left(\frac{h\cdot AB/2-t}{h\cdot AB/2}\right)^2=(1-t)^2$$ This means that for $t\in [0,1]$ we have $$\mathbb{P}({\rm Area}(AMB)>t)=(1-t)^2$$ It follows that $$\mathbb{E}({\rm Area}(AMB))=\int_0^1\mathbb{P}({\rm Area}(AMB)>t)dt= \int_0^1(1-t)^2dt=\frac13.$$