Fall 2021 AMC 10 B

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 $$△FAE\cong △EOH\cong △IOH\cong △KCJ\cong △FPG,$$ it follows that $EO=IO=KC=y$ and $CJ=AE=x$. Furthermore, $△JDI$ 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 $△AEF$ 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 $△GPL\sim △LBM$, it follows that $BM=\frac{1}{3}abx$. Furthermore, $△NQK\sim △KCJ$, 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 $$\begin{array}{rl}LG\cdot LM& =\sqrt{(ax{)}^2+(3x{)}^2}\cdot \sqrt{(bx{)}^2+{\left(\frac{1}{3}abx\right)}^2}\\ & =\sqrt{(a^2+9)x^2}\cdot \sqrt{\left(b^2+\frac{1}{9}{a}^2{b}^2\right)x^2}\\ & =x^2\sqrt{a^2+9}\cdot \sqrt{b^2+\frac{1}{9}{a}^2{b}^2}\\ & =x^2\sqrt{a^2+9}\cdot \sqrt{b^2\left(1+\frac{1}{9}{a}^2\right)}\\ & =x^2\sqrt{a^2+9}\cdot b\sqrt{1+\frac{1}{9}{a}^2}\\ & =x^{2}b\sqrt{(a^2+9)\left(1+\frac{1}{9}{a}^2\right)}.\end{array}$$ 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 $52+15=67$.