Problem: A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$

Solution: We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.
Denote $\angle AFE = \theta$ . Thus, $\angle FIB = \angle CEF = \angle EKG = \angle KLC = \theta$
Hence, \begin{align*} AB & = AF + FB \\ & = EF \cos \angle EFA + IF \sin \angle FIB \\ & = 3 \cos \theta + \sin \theta , \end{align*} and \begin{align*} AC & = AE + EK + KC \\ & = EF \sin \angle EFA + EG \cos \angle CEG + KG \cos \angle EKG + KL \sin \angle CLK \\ & = 3 \sin \theta + \cos \theta + \cos \theta + \sin \theta \\ & = 2 \cos \theta + 4 \sin \theta . \end{align*}
Because $AB = AC$ $3 \cos \theta + \sin \theta = 2 \cos \theta + 4 \sin \theta$ .
Hence, $\tan \theta = \frac{1}{3}$ .
Hence, $\sin \theta = \frac{1}{\sqrt{10}}$ and $\cos \theta = \frac{3}{\sqrt{10}}$
Hence, $AB = AC = BD = CD = \sqrt{10}$
Now, we put the graph to a coordinate plane by setting point $A$ as the origin, putting $AB$ in the $x$ -axis and $AC$ on the $y$ -axis.
Hence, $A = \left( 0 , 0 \right)$ $B = \left( \sqrt{10} , 0 \right)$ $C = \left( 0 , \sqrt{10} \right)$ $D = \left( \sqrt{10} , \sqrt{10} \right)$ $E = \left( 0 , \frac{3}{\sqrt{10}} \right)$ $F = \left( \frac{9}{\sqrt{10}} , 0 \right)$ $G = \left( \frac{1}{\sqrt{10}} , \frac{6}{\sqrt{10}} \right)$ $H = \left( \frac{4}{\sqrt{10}} , \frac{7}{\sqrt{10}} \right)$ $I = \left( \sqrt{10} , \frac{3}{\sqrt{10}} \right)$
Denote $P = \left( \frac{10 - u}{\sqrt{10}} , \sqrt{10} \right)$ $Q = \left( \sqrt{10} , \frac{10 - v}{\sqrt{10}} \right)$
Because $HPQJ$ is a rectangle, $HP \perp PQ$ . Hence, $m_{HP} m_{PQ} = -1$ .
We have $m_{HP} = \frac{3}{6 - u}$ and $m_{PQ} = - \frac{v}{u}$ .
Hence, \[ \frac{3}{6 - u} \cdot \left( - \frac{v}{u} \right) = - 1 . \hspace{1cm} (1) \]
Because $HPQJ$ is a rectangle, $x_J + x_P = x_H + x_Q$ and $y_J + y_P = y_H + y_Q$ .
Hence, $J = \left( \frac{4 + u}{\sqrt{10}} , \frac{7 - v}{\sqrt{10}} \right)$
The equation of line $GI$ is \begin{align*} y & = \frac{\frac{3}{\sqrt{10}} - \frac{6}{\sqrt{10}}}{\sqrt{10} - \frac{1}{\sqrt{10}}} \left( x - \frac{1}{\sqrt{10}} \right) + \frac{6}{\sqrt{10}} \\ & = - \frac{x}{3} + \frac{19}{3 \sqrt{10}} . \end{align*} Because point $J$ is on line $GI$ , plugging the coordinates of $J$ into the equation of line $GI$ , we get \[ \frac{7 - v}{\sqrt{10}} = - \frac{\frac{4 + u}{\sqrt{10}}}{3} + \frac{19}{3 \sqrt{10}} . \hspace{1cm} (2) \]
By solving Equations (1) and (2), we get $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$ or $\left( 3 , 3 \right)$
Case 1: $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$
We have $P = \left( \frac{8}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{22}{3 \sqrt{10}} \right)$ .
Thus, $HP = \frac{5}{\sqrt{10}}$ and $PQ = \frac{10}{3\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{5}{3}$
Case 2: $\left( u , v \right) = \left( 3 , 3 \right)$
We have $P = \left( \frac{7}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{7}{ \sqrt{10}} \right)$ .
Thus, $HP = \frac{3 \sqrt{2}}{\sqrt{10}}$ and $PQ = \frac{3 \sqrt{2}}{\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{9}{5}$
Putting these two cases together, the sum of all possible values of the area of $R$ is $\frac{5}{3} + \frac{9}{5} = \frac{52}{15}$
Therefore, the answer is $\boxed{67}$

Answer: E 67