A good solution may involve (the interior of) sets with boundary of fractal dimension. However, using just right triangles one can get around $\frac{4}{5}$ of the possibilities (in a sense made precise below) and very slightly more using rectangles as well.
I'll focus on the squared ratios $\rho_i=r_i^2$ with $\rho_1+\rho_2+\rho_3 \le 1.$ So I will allow $r_i=0$ and allow the sum to actual equal $1$. I find it easier to consider the triples in all $6$ possible orders. So the set of possible $\rho$ triples can be thought of as a body $\mathbf{B}$ comprising part, or perhaps all, of the pyramid $\mathbf{P}$ in $\mathbb{R}^3$ with corners at the origin and the points $(1,0,0)$ , $(0,1,0)$ and $(0,0,1).$ The volume of $\mathbf{P}$ is $\frac{1}{6}.$ My claim is that the volume of $\mathbf{B}$ is over $\frac{4}{5}$ of this. If a point $P=(\rho_1,\rho_2,\rho_3)$ is in $\mathbf{B}$ so are all the points of the box determined by the origin and $P$. So $\mathbf{B}$ could be specified were one able to described the surface made of the points in each radial direction furthest from the origin.
Certainly the the points with $\rho_1+\rho_2+\rho_3=1$ are of interest. As was kindly pointed out to me, we can achieve the triples $(t,1-t,0)$ using (the interior of) a right triangle with legs of lengths $\sqrt{t},\sqrt{1-t}$ and the usual division by the perpendicular to the hypotenuse. Iterating this for one of the sub-triangles (picture at the end) gives us triples of the form $(t,t^2,1-t-t^2)$ where $0 \le t \le \frac{\sqrt{5}-1}{2}.$ Here is a sketch of these points making up the boundary of the equilateral triangle and $6$ internal curves.
I can't really picture the body made up of all the boxes determined by these points and the origin, but the volume , roughly estimated by counting the included points of the form $(\frac{a}{100},\frac{b}{100},\frac{c}{100})$ , is somewhere between $0.83$ and $0.86$ of the total volume of $\mathbf{P}$ depending on if I use the floor or round in estimating. All the points with $\rho_1+\rho_2+\rho_3 \le 0.75$ appear to be achieved however then things start to fall off.
Certainly the point $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is not accounted for. That can be achieved using a rectangle with sides $3 \times \sqrt{3}$ partitioned into three rectangles of with sides $1 \times \sqrt{3}.$ That increases the volume of $\mathbf{B}$ but not by very much.
Below is an sketch of possibly missing points with $\rho_1+\rho_2+\rho_3=0.9.$
The small triangle missing in the center comes from the rectangle just mentioned. So a good goal (as might have been obvious) would be triples like $(0.48.0.48.0.04)$, $(0.94,0.03,0.03)$ and $(0.32,0.32,0.36)$ with two equal components close to but not equal to $\frac{1}{2}$ or $0$ or $\frac{1}{3}$ and sum equal to, or nearly equal to , $1$.
Another interesting rectangle partition is a $4 \times 2\sqrt{2}$ rectangle partitioned in half and then one half again partitioned in half. However this gives the same result $(\frac{1}{2}, \frac{1}{4},\frac{1}{4})$ as an isosceles right triangle.
I thought of a clever (I thought) ways to pack a figure with similar copies which fill it mostly but not completely. However the $\rho$-triple is $(t^2,t^4,t^4)$ for $t=\frac{\sqrt{5}-1}{2}$ so this is not as good as the triangle construction $(1-t^2-t^4 ,t^2,t^4).$ The proportion covered increases and the pieces rotate if we hold one leg fixed and shrink the other, but this seems likely to be turning into the triangle situation. I don't think varying the angle would help.
