uniform random point in triangle

Question

Suppose you have an arbitrary triangle with vertices $A$, $B$, and $C$. This paper (section 4.2) says that you can generate a random point, $P$, uniformly from within triangle $ABC$ by the following convex combination of the vertices:

$P = (1 - \sqrt{r_1}) A + (\sqrt{r_1} (1 - r_2)) B + (r_2 \sqrt{r_1}) C$

where $r_1, r_2 \sim U[0, 1]$.

How do you prove that the sampled points are uniformly distributed within triangle $ABC$?

Answer 1

I would argue that if it is true for any triangle, it is true for all of them, as we can find an affine transformation between them. So I would pick my favorite triangle, which is $A=(0,0), B=(1,0), C=(0,1)$. Then the point is $(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$ and we need to prove it is always within the triangle and evenly distributed. To be in the triangle we need $x,y\ge 0, x+y\le 1$, which is clear. Then show that the probability to be within an area $(0,x) \times (0,y)$ is $2xy$ by integration.