Consider points $O, A, B, C, D$ and $R$ in the diagram below, such that $AR = 4, BR = 2, CR = 2,$ and $DR = 6$:
[asy]
size(250);
import TrigMacros;
import olympiad;
pair O, A, B, C, D, R;
O = (0,0);
A = 3*dir(-40);
B = A + 6*dir(100);
R = 1/3*A + 2/3*B;
C = R - 2dir(10);
D = 4R - 3C;
rr_cartesian_axes(-3, 10, -3, 8, usegrid = false);
draw(A--B);
draw(C--D);
dot("$A$", A, E);
dot("$B$", B, N);
dot("$C$", C, NW);
dot("$D$", D, NE);
dot("$R$",R, SE);
dot("$O$",O, SW);
[/asy]If there exists a real number $t$ such that
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t) \overrightarrow{OC} + t \overrightarrow{OD},\]enter $t$ below. If this is impossible, enter a question mark.
Let's first write the given equation in terms of vectors:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t) \overrightarrow{OC} + t \overrightarrow{OD}.\]
Expanding this equation, we get:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = \overrightarrow{OC} - t\overrightarrow{OC} + t \overrightarrow{OD}.\]
Simplifying further, we have:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1 - t)\overrightarrow{OC} + t(\overrightarrow{OC} + \overrightarrow{OD}).\]
Now let's substitute the given values into this equation:
\[(1-t)(\overrightarrow{OC} + \overrightarrow{AC}) + t(\overrightarrow{OC} + \overrightarrow{BC}) = (1-t)\overrightarrow{OC} + t(\overrightarrow{OC} + \overrightarrow{OD}).\]
Expanding and canceling out terms, we have:
\[(1-t)\overrightarrow{AC} + t\overrightarrow{BC} = t\overrightarrow{OD}.\]
Now let's calculate the magnitudes of both sides of the equation:
\begin{align*}
\|(1-t)\overrightarrow{AC} + t\overrightarrow{BC}\| &= \|t\overrightarrow{OD}\| \\
\|(1-t)\overrightarrow{AC} + t\overrightarrow{BC}\| &= |t|\|\overrightarrow{OD}\| \\
\|(1-t)\overrightarrow{AC} + t\overrightarrow{BC}\| &= |t| \cdot 6 \\
\|(1-t)\overrightarrow{AC} + t\overrightarrow{BC}\| &= 6|t|.
\end{align*}
Since the magnitudes are equal, we can compare the coefficients:
\[(1-t) = 6 \text{ or } -6.\]
Solving for $t$, we have two possible values: $t = -5$ or $t = \frac{7}{5}$.
However, $t$ cannot be negative because it represents a position along the line segment $AB$, and we cannot "go back" along the line segment. Therefore, the only valid value for $t$ is $t = \frac{7}{5}$.
Hence, $t = \boxed{\frac{7}{5}}$.
We have the equation \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t) \overrightarrow{OC} + t \overrightarrow{OD}.\]Expanding this equation gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = \overrightarrow{OC} - t\overrightarrow{OC} + t \overrightarrow{OD}.\]Now, we can express all the vectors in terms of $\overrightarrow{OR}$ since we have the relations $R = \frac{1}{3}A + \frac{2}{3}B$, $C = R - 2\overrightarrow{OR}$, and $D = 4\overrightarrow{OR} - 3C$. Substitute these relations in the above equation.
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t)(\overrightarrow{OR} - 2\overrightarrow{OR}) + t(4\overrightarrow{OR} - 3C).\]Simplifying the equation yields \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t + 2t) \overrightarrow{OR} -3tC.\]Next, we substitute $R = \frac{1}{3}A + \frac{2}{3}B$ and $C = R - 2\overrightarrow{OR}$ to get \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t + 2t) \overrightarrow{OR} -3t(R - 2\overrightarrow{OR}).\]Expanding this equation gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (3t + 1 - t)\overrightarrow{OR} -3t(R - 2\overrightarrow{OR}).\]Simplifying both sides yields \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (2t + 1) \overrightarrow{OR} -3tR + 6t\overrightarrow{OR}.\]Grouping like terms gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (2t + 1 + 6t)\overrightarrow{OR} -3tR.\]Combining like terms gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (8t + 1)\overrightarrow{OR} -3tR.\]
Now, we can substitute $R = \frac{1}{3}A + \frac{2}{3}B$ to get \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (8t + 1)\left(\frac{1}{3}A + \frac{2}{3}B\right) - 3tR.\]Simplifying this equation gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB}= \frac{8t+1}{3}A + \frac{16t + 2}{3}B -3tR.\]Multiplying out the terms gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = \frac{8t+1}{3}A + \frac{16t + 2}{3}B - \frac{9}{3}tA - \frac{18}{3}tB.\]Combining like terms gives \[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = \left(\frac{8t+1}{3} - \frac{9t}{3}\right)A + \left(\frac{16t + 2}{3} - \frac{18t}{3}\right)B.\]Now we can equate the coefficients of $A$ and $B$ on both sides.
For the $A$ coefficient, we have \[\frac{8t+1}{3} - \frac{9t}{3} = 0.\]Multiplying through by 3 gives \[8t + 1 - 9t = 0.\]Combining like terms gives \[-t + 1 = 0.\]Solving for $t$ gives $t = 1$.
For the $B$ coefficient, we have \[\frac{16t + 2}{3} - \frac{18t}{3} = 0.\]Multiplying through by 3 gives \[16t + 2 - 18t = 0.\]Combining like terms gives \[-2t + 2 = 0.\]Solving for $t$ gives $t = 1$.
Therefore, the only value of $t$ that satisfies the equation is $t = 1$.
To solve this problem, we can use the properties of vectors.
We have the equation:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-t) \overrightarrow{OC} + t \overrightarrow{OD}\]
We know that vectors can be represented by the difference of their coordinates. So, we can rewrite this equation as:
\[(1-t)(\overrightarrow{OB} - \overrightarrow{OA}) = (1-t)(\overrightarrow{OD} - \overrightarrow{OC})\]
Next, we can distribute the scalar factor of $(1-t)$ across the vectors:
\[(1-t)(\overrightarrow{OB}) - (1-t)(\overrightarrow{OA}) = (1-t)(\overrightarrow{OD}) - (1-t)(\overrightarrow{OC})\]
Expanding the terms, we get:
\[\overrightarrow{OB} - \overrightarrow{OA} - (1-t)(\overrightarrow{OA}) = \overrightarrow{OD} - \overrightarrow{OC} - (1-t)(\overrightarrow{OC})\]
Now, we have:
\[\overrightarrow{OB} - \overrightarrow{OA} - \overrightarrow{OA} + t\overrightarrow{OA} = \overrightarrow{OD} - \overrightarrow{OC} - \overrightarrow{OC} + t\overrightarrow{OC}\]
Combining like terms, we get:
\[\overrightarrow{OB} - 2\overrightarrow{OA} + t\overrightarrow{OA} = \overrightarrow{OD} - 2\overrightarrow{OC} + t\overrightarrow{OC}\]
To solve for $t$, we can compare the coefficients of the vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$ on both sides of the equation.
We have the equation:
\[-2\overrightarrow{OA} + t\overrightarrow{OA} = -2\overrightarrow{OC} + t\overrightarrow{OC}\]
Since the vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$ are not collinear (as can be seen from the diagram), their coefficients must be equal:
\[-2 + t = 0\]
Solving for $t$, we find:
\[t = 2\]
Therefore, the value of $t$ is $\boxed{2}$.