What is the expected value of the area of a triangle inside a triangle?

Question

One random point is chosen randomly from each of the sides of a triangle with an area A. These lines are connected to be a new triangle inside of the original triangle. Calculate the expected area of the new triangle.

I don't even know how to start thinking this problem. If I choose the random variable to be $X= \text{Area of the new triangle}$, can I assume it to be uniformly distributed?

Update: The answer is given and it is supposed to be $\frac{1}{4}A.$

Answer 1

Following @SharkyKesa's hint, observe that if you apply a linear transformation to your triangle (and the three selected points), the ratio between the area of the subtriangle and the main triangle remains invariant. So you can begin by transforming the main triangle to any one that's convenient to work with. I propose one with vertices $A = (0,0), B = (1, 0), C = (0, 1)$, whose area is $T = \frac12$.

Picking vertices for the subtriangle in each edge in turn gives $$ v_1 = (0, t)\\ v_2 = (u, 0)\\ v_3 = (1-s, s). $$ where $s,t,u$ are uniform random variables on $[0, 1]$.

Note that $v_1v_2v_3$ is a counterclockwise traversal of the triangle.

Following the discussion here, the area of the triangle $v_1v_2v_3$ is \begin{align} S &= \frac12\left(-ut + us + (1-s)t \right)\\ 2S&= u(s-t) + t - st\\ \end{align} To compute the expected value of $2S$, we integrate over $[0, 1]^3$ to get \begin{align} avg(2S) &= \int_0^1 \int_0^1 \int_0^1 u(s-t) + t - st ~du ~dt ~ds\\ &= \int_0^1 \int_0^1 \left. (\frac{u^2}{2}(s-t) + u(t - st)) \right|_{u=0}^1 ~dt ~ds\\ &= \int_0^1 \int_0^1 \frac{1}{2}(s-t) + (t - st) ~dt ~ds\\ &= \int_0^1 \int_0^1 \frac{1}{2}s + \frac12 t - st ~dt ~ds\\ &= \int_0^1 \left. \frac{1}{2}st + \frac12 \frac{t^2}{2} - s\frac{t^2}{2} \right|_{t=0}^1 ~ds\\ &= \int_0^1 \frac{1}{2}s + \frac14 - \frac{s}{2} ~ds\\ &= \int_0^1 \frac14~ds\\ &= \frac14 \end{align} so that the average of $S$ is $\frac{1}{8}$, which is $\frac{1}{4}T$, where $T = \frac12$ is the area of the original outer triangle $ABC$.

Answer 2

Here's one way you might start thinking about the problem.

Suppose that the original triangle has vertices at $(x_i,y_i)$ for $i=1,2,3$.

Let $U$, $V$ and $W$ be uniform $[0,1]$ random variables. Then the new triangle has vertices at points

$(U x_1 + (1-U)x_2, Uy_1 + (1-U)y_2)$, $(V x_2 + (1-V)x_3, Vy_2 + (1-V)y_3)$ and $(W x_3 + (1-W)x_1, Wy_3 + (1-W)y_1)$. Can you see what I've done here? Do you know a way to work out the area of a triangle from just the coordinates of it's vertices?