, 2021

Label the diagram as shown below, where $O$, $P$, and $Q$ are the feet of the perpendicular segments to the respective sides. Let $x = AE$ and $y = AF$.

Because
$$\triangle FAE \cong \triangle EOH \cong \triangle IOH \cong \triangle KCJ \cong \triangle FPG,$$
it follows that $EO = IO = KC = y$ and $CJ = AE = x$. Furthermore, $\triangle JDI$ is similar to all these triangles with scale factor $3$, so $ID = 3x$ and $DJ = 3y$. Therefore
$$3y + x = DC = DA = 2y + 4x,$$
so $y = 3x$. Applying the Pythagorean Theorem to $\triangle AEF$ yields $x = \frac{1}{\sqrt{10}}$.

Now set $PL = ax$ and $LB = bx$ for some positive real numbers $a$ and $b$. Note that $MQ = PG = y = 3x$, and because $\triangle GPL \sim \triangle LBM$, it follows that $BM = \frac{1}{3}abx$. Furthermore, $\triangle NQK \sim \triangle KCJ$, and because $NQ = PL = ax$, it follows that $QK = \frac{1}{3}ax$. Therefore
$$3x + \frac{1}{3}ax + 3x + \frac{1}{3}abx = BC = AD = 10x,$$
which simplifies to $a(b + 1) = 12$. Because $PB = AB - AP = 6x$, it follows that $a + b = 6$. Solving these equations simultaneously for $a$ and $b$ shows that $(a, b)$ equals either $(3, 3)$ or $(4, 2)$.

Finally, observe that the area of $R$ is
$$\begin{aligned} LG \cdot LM &= \sqrt{(ax)^2 + (3x)^2} \cdot \sqrt{(bx)^2 + \left(\frac{1}{3}abx\right)^2} \\ &= \sqrt{(a^2 + 9)x^2} \cdot \sqrt{\left(b^2 + \frac{1}{9}a^2b^2\right)x^2} \\ &= x^2 \sqrt{a^2 + 9} \cdot \sqrt{b^2 + \frac{1}{9}a^2b^2} \\ &= x^2 \sqrt{a^2 + 9} \cdot \sqrt{b^2\left(1 + \frac{1}{9}a^2\right)} \\ &= x^2 \sqrt{a^2 + 9} \cdot b \sqrt{1 + \frac{1}{9}a^2} \\ &= x^2 b \sqrt{(a^2 + 9)\left(1 + \frac{1}{9}a^2\right)}. \end{aligned}$$
Since $x = \frac{1}{\sqrt{10}}$, $x^2 = \frac{1}{10}$, so the area is
$$\frac{b}{10} \sqrt{(a^2 + 9)\left(1 + \frac{1}{9}a^2\right)}.$$
When $(a, b) = (3, 3)$, this area equals $\frac{9}{5}$; and when $(a, b) = (4, 2)$, this area equals $\frac{5}{3}$. The sum of these two areas is $\frac{9}{5} + \frac{5}{3} = \frac{27}{15} + \frac{25}{15} = \frac{52}{15}$, and the requested sum of numerator and denominator is $52 + 15 = 67$.