2016/17 AHSMC Part 1 Problem 15

Using the notations from the above diagram and the conditions from the problem one obtains: $$3(r+x) = r+y \iff y = 3x + 2r$$ and $$(x+y)^2 = 4r^2 + \bigl((r+y)-(r+x)\bigr)^2 \iff xy = r^2.$$ Hence $$x(3x+2r) = r^2 \iff x = \frac{r}{3}$$ and consequently $y = 3r$. On the other hand the area of the trapezoid $ABCD$ is $4$, thus $$(r+x + r+y)r = 4.$$ Substituting for $x = \dfrac{r}{3}$ and $y = 3r$ we get $r = \dfrac{\sqrt{3}}{2}$. The answer is (D).