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: [asy] size(8cm); label("D",(0,0),SW); label("A",(0,10),NW); label("B",(10,10),NE); label("C",(10,0),SE); label("H",(1,5.7),SE); label("O",(0,6),W); label("I",(0,3),W); 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)); draw((1,6)--(0,6)); 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); draw((0.3,6)--(0.3,5.7)--(0,5.7)); 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); label("E",(0,9),W); label("F",(3,10),N); label("P",(4,10),N); label("L",(8,10),N); draw((4,7)--(4,10)); draw((4,9.7)--(4.3,9.7)--(4.3,10)); label("G",(3.9,6.8),S); label("N",(6.2,4.1),S); label("M",(10,22/3),E); label("Q",(10,13/3),E); draw((10,13/3)--(6,13/3)); draw((9.7,13/3)--(9.7,4.033333333333333333333333)--(10,4.03333333333333333333333333)); label("K",(10,3),E); label("J",(9,0),S); [/asy] We will scale every number up by a factor of $\sqrt{10}.$ This implies our final area will be $\frac{1}{10}$ of the answer we receive.
We have \[FAE\sim EOH \sim IOH\sim JDI\sim KCJ \sim NQK\sim GPF.\] Let $AE=a$ and $FA=b.$ We have \[FP=AE=OH=JC=\frac13ID=a\] and \[PG=AF=EO=OI=KC=\frac13 DJ=b\] As $ABCD$ is a square, we have $AD=DC$ or \[a+2b+3a=3b+a \Rightarrow 3a=b.\] Since $a^2+b^2=10,$ we have \[a=1,b=3.\] We have $\triangle GPL\cong \triangle MQN$ which implies \[MQ=GP=3.\] Denote $QK=x.$ As $\triangle NQK\sim \triangle JDI,$ we have $NQ=3x.$
We have \begin{align*}BM&=BC-(CK+QK+MQ)\\ &=4-x.\end{align*}
In addition, \begin{align*}LB&=AB-(AF+FP+PL)\\&=6-3x.\end{align*}
Since $\triangle LBM\sim \triangle MQN,$ we have \[\frac{LB}{BM}=\frac{MQ}{QN}\Rightarrow \frac{6-3x}{4-x}=\frac{3}{3x}=\frac{1}{x}.\] Simplifying we have \[3x^2-7x+4=0 \Rightarrow x=\frac43, 1.\] We have \begin{align*}[GLMN]&=MN\cdot LM\\ &= 3\sqrt{x^2+1}\cdot \sqrt{10x^2-44x+52}.\end{align*}
Plugging in $x=1,$ we have $[GLMN]=18.$
Plugging in $x=\frac43,$ we have $[GLMN]=\frac{50}{3}.$
The sum of the two possible $R$ s is \[\frac{1}{10}\cdot\frac{104}{3}=\frac{52}{15}.\] Hence, $52+15=\boxed{67}.$

Answer: E 67