Right triangle rotation

The loci of the vertex at $90\degree$ will be two circles of radii $a$ and $a\sqrt 3$ , where $a$ is the length of the side opposite to the $30\degree$ vertex. There intersection encloses an area of

$a^2(\frac{π}{2}-\frac{3\sqrt 3}{4})+a^2(\frac{π}{3}-\frac{\sqrt 3}{4})$

$=a^2(\frac{5π}{6}-\sqrt 3)$ .

Area of the triangle is $\dfrac{\sqrt 3a^2}{2}$ .

Therefore the required ratio is

$\dfrac{2}{\sqrt 3}(\frac{5π}{6}-\sqrt 3)\approx \boxed {1.023}$ .

Since the sides of a 30-60-90 triangle are in the ratio of 1:sqrt(3):2, it is no loss of generality to choose a coordinate system with the 90 degree angle at the origin, one vertex at point P(1,0) and another vertex at point Q(0, sqrt(3)). When we place the thumbtack and P and rotate, we obtain the circle (x-1)^2+y^2 = 1. When we move the thumbtack to Q and rotate, we obtain the circle x^2 + (y-sqrt(3))^2 = 3. Solving this system of equations, we obtain intersection points of (0,0) and (3/2, sqrt(3) / 2). The area between these circles is then the integral from x = 0 to x = 3/2 of (1-(x-1)^2)^.5 - sqrt(3) + (3-x^2)^.5. This integral, which can be evaluated by trigonometric substitution, is equal to 5 Pi/6 - sqrt(3) = .8859 (approximately). Since the area of the triangle is sqrt(3) / 2, the required quotient is (5 Pi/6 - sqrt(3)) / (sqrt(3) / 2) = 1.023 (approximately)