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 exist real numbers $t$ and $u$ such that
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) \overrightarrow{OC} + u \overrightarrow{OD},\]enter $t$ and $u$ in that order. If this is impossible, enter a question mark for each number.
Well, let's see what we can find out.
We have $\overrightarrow{OA}$ and $\overrightarrow{OB}$ as given vectors. We need to find $\overrightarrow{OC}$ and $\overrightarrow{OD}$.
Let's start with $\overrightarrow{OC}$. We know that $OC = OR - RC$, so we can write $\overrightarrow{OC}$ in terms of $\overrightarrow{OR}$ and $\overrightarrow{RC}$.
Similarly, we have $\overrightarrow{OD} = \overrightarrow{OR} - \overrightarrow{RD}$.
Now, we can substitute these expressions into the equation:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) (\overrightarrow{OR} - \overrightarrow{RC}) + u (\overrightarrow{OR} - \overrightarrow{RD}).\]
Expanding and grouping like terms, we have:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) \overrightarrow{OR} - (1-u) \overrightarrow{RC} + u \overrightarrow{OR} - u \overrightarrow{RD}.\]
Let's look at the $OR$ terms on the right side. We can combine them to get:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u+u) \overrightarrow{OR} - (1-u) \overrightarrow{RC} - u \overrightarrow{RD}.\]
Simplifying further, we have:
\[(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = \overrightarrow{OR} - (1-u) \overrightarrow{RC} - u \overrightarrow{RD}.\]
Now, we know that $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OR}$ are given vectors, and we want the equation to hold for all vectors $\overrightarrow{RC}$ and $\overrightarrow{RD}$.
This means that the coefficients of $\overrightarrow{RC}$ and $\overrightarrow{RD}$ on the right side must be zero.
Let's set up the equations based on this requirement:
$(1-u) = 0$
$-u = 0$
Solving these equations, we find that $u=0$ and $u=1$. However, both values of $u$ lead to contradictions.
Therefore, it is impossible to find real numbers $t$ and $u$ that satisfy the given equation. So the answer is ??.
I hope this helped clarify things for you! If you have any more questions, feel free to ask!
We are given that $(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) \overrightarrow{OC} + u \overrightarrow{OD}.$
Expanding the vectors using their component form, we have
\[(1-t) \begin{pmatrix} 3 \\ -4 \end{pmatrix} + t \begin{pmatrix} -6 \\ 6 \end{pmatrix} = (1-u) \begin{pmatrix} -1 \\ 4 \end{pmatrix} + u \begin{pmatrix} 2 \\ 8 \end{pmatrix}.\]This simplifies to
\begin{align*}
\begin{pmatrix} 3 - 3t \\ -4 + 4t \end{pmatrix} + \begin{pmatrix} -6t \\ 6t \end{pmatrix} &= \begin{pmatrix} -1 + u \\ 4 - 4u \end{pmatrix} + \begin{pmatrix} 2u \\ 8u \end{pmatrix} \\
\begin{pmatrix} -6t + 3 - 3t \\ 6t - 4 + 4t \end{pmatrix} &= \begin{pmatrix} 2u - 1 + u \\ 8u + 4 - 4u \end{pmatrix} \\
\begin{pmatrix} -9t + 3 \\ 10t - 4 \end{pmatrix} &= \begin{pmatrix} 3u - 1 \\ 4u + 4 \end{pmatrix}.
\end{align*}Matching corresponding entries, we get
\begin{align*}
-9t + 3 &= 3u - 1, \\
10t - 4 &= 4u + 4.
\end{align*}From the first equation, $3u = -9t + 4,$ so $u = -3t + \frac{4}{3}.$ Substituting into the second equation, we get
\[10t - 4 = 4(-3t + 4/3) + 4.\]This simplifies to $40t = 40,$ so $t = 1.$ Then $u = -3(1) + \frac{4}{3} = \frac{1}{3}.$
Therefore, $t = \boxed{1}$ and $u = \boxed{\frac{1}{3}}.$
To solve this problem, we will use the fact that vectors are equal if and only if their corresponding components are equal.
Let $\overrightarrow{v} = \overrightarrow{OA}$ and $\overrightarrow{w} = \overrightarrow{OB}$. Similarly, let $\overrightarrow{x} = \overrightarrow{OC}$ and $\overrightarrow{y} = \overrightarrow{OD}$.
Given the equation $(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) \overrightarrow{OC} + u \overrightarrow{OD}$, we can rewrite it as:
$(1-t)\overrightarrow{v} + t\overrightarrow{w} = (1-u)\overrightarrow{x} + u\overrightarrow{y}$.
Now, let's express these vectors in terms of their coordinates. Let $O$ be the origin, so we have:
$\overrightarrow{v} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$, $\overrightarrow{w} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, $\overrightarrow{x} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, and $\overrightarrow{y} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
Substituting these vectors into the equation gives us:
$(1-t)\begin{pmatrix} 3 \\ -4 \end{pmatrix} + t\begin{pmatrix} 0 \\ 0 \end{pmatrix} = (1-u)\begin{pmatrix} 1 \\ 2 \end{pmatrix} + u\begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
Now, we can expand and simplify the equation:
$\begin{pmatrix} 3-3t \\ -4+4t \end{pmatrix} = \begin{pmatrix} 1-u+3u \\ 2-2u+4u \end{pmatrix}$.
Comparing the corresponding components, we get the following system of equations:
$3-3t = 1-u+3u$ and $-4+4t = 2-2u+4u$.
Simplifying each equation gives us:
$3t + u = 2$ and $4t - 2u = -6$.
To solve these equations, we can use any method of solving linear equations, such as substitution or elimination.
Let's solve this system using elimination. Multiplying the first equation by 2 and the second equation by 3, we get:
$6t + 2u = 4$ and $12t - 6u = -18$.
Now, we can subtract the first equation from the second equation:
$(12t - 6u) - (6t + 2u) = -18 - 4$.
This simplifies to:
$6t - 8u = -22$.
Dividing this equation by 2 gives us:
$3t - 4u = -11$.
Now, we have two equations in two variables:
$3t + u = 2$ and $3t - 4u = -11$.
We can solve this system using any method, such as substitution or elimination. For this example, we'll use substitution.
Solving the first equation for $u$, we get:
$u = 2 - 3t$.
Substituting this value of $u$ into the second equation gives:
$3t - 4(2 - 3t) = -11$.
Expanding and simplifying, we have:
$3t - 8 + 12t = -11$.
Combining like terms, we get:
$15t - 8 = -11$.
Solving for $t$, we have:
$15t = -3$.
Dividing both sides by 15, we get:
$t = -\frac{1}{5}$.
Substituting this value of $t$ into the first equation, we can solve for $u$:
$3(-\frac{1}{5}) + u = 2$.
Simplifying, we have:
$-\frac{3}{5} + u = 2$.
Adding $\frac{3}{5}$ to both sides gives:
$u = \frac{13}{5}$.
Therefore, the solution to the equation $(1-t) \overrightarrow{OA} + t \overrightarrow{OB} = (1-u) \overrightarrow{OC} + u \overrightarrow{OD}$ is $t = -\frac{1}{5}$ and $u = \frac{13}{5}$.
So the answer is $\boxed{(-\frac{1}{5}, \frac{13}{5})}$.