In the diagram below, $\|\overrightarrow{OA}\| = 1,$ $\|\overrightarrow{OB}\| = 1,$ and $\|\overrightarrow{OC}\| = \sqrt{2}.$  Also, $\tan \angle AOC = 7$ and $\angle BOC = 45^\circ.$

[asy]
unitsize(2 cm);

pair A, B, C, O;

A = (1,0);
B = (-0.6,0.8);
C = (0.2,1.4);
O = (0,0);

draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--C,Arrow(6));

label("$A$", A, E);
label("$B$", B, NW);
label("$C$", C, N);
label("$O$", O, S);
[/asy]

There exist constants $m$ and $n$ so that
\[\overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB}.\]Enter the ordered pair $(m,n).$
By constructing a right triangle with adjacent side 1, opposite side 7, and hypotenuse $\sqrt{1^2 + 7^2} = 5 \sqrt{2}$, we see that
\[\cos \angle AOC = \frac{1}{5 \sqrt{2}} \quad \text{and} \quad \sin \angle AOC = \frac{7}{5 \sqrt{2}}.\]Then
\begin{align*}
\cos \angle AOB &= \cos (\angle AOC + \angle BOC) \\
&= \cos \angle AOC \cos \angle BOC - \sin \angle AOC \sin \angle BOC \\
&= \frac{1}{5 \sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{7}{5 \sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\
&= -\frac{3}{5}.
\end{align*}Taking the dot product of the equation $\overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB}$ with $\overrightarrow{OA},$ we get
\[\overrightarrow{OA} \cdot \overrightarrow{OC} = m \overrightarrow{OA} \cdot \overrightarrow{OA} + n \overrightarrow{OA} \cdot \overrightarrow{OB}.\]Then $\|\overrightarrow{OA}\| \|\overrightarrow{OC}\| \cos \angle AOC = m \|\overrightarrow{OA}\|^2 + n \|\overrightarrow{OA}\| \|\overrightarrow{OB}\| \cos \angle AOB,$ or
\[\frac{1}{5} = m - \frac{3}{5} n.\]Taking the dot product of the equation $\overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB}$ with $\overrightarrow{OB},$ we get
\[\overrightarrow{OB} \cdot \overrightarrow{OC} = m \overrightarrow{OA} \cdot \overrightarrow{OB} + n \overrightarrow{OB} \cdot \overrightarrow{OB}.\]Then $\|\overrightarrow{OB}\| \|\overrightarrow{OC}\| \cos \angle BOC = m \|\overrightarrow{OA}\| \|\overrightarrow{OB}\| \cos \angle AOB + n \|\overrightarrow{OB}\|^2,$ or
\[1 = -\frac{3}{5} m + n.\]Solving the system $\frac{1}{5} = m - \frac{3}{5} n$ and $1 = -\frac{3}{5} m + n,$ we find $(m,n) = \boxed{\left( \frac{5}{4}, \frac{7}{4} \right)}.$
\le(\frac{5}{4},\frac{7}{4}\right)
Level 5
Precalculus
Problem: In the diagram below, $\|\overrightarrow{OA}\| = 1,$ $\|\overrightarrow{OB}\| = 1,$ and $\|\overrightarrow{OC}\| = \sqrt{2}.$  Also, $\tan \angle AOC = 7$ and $\angle BOC = 45^\circ.$

[asy]
unitsize(2 cm);

pair A, B, C, O;

A = (1,0);
B = (-0.6,0.8);
C = (0.2,1.4);
O = (0,0);

draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--C,Arrow(6));

label("$A$", A, E);
label("$B$", B, NW);
label("$C$", C, N);
label("$O$", O, S);
[/asy]

There exist constants $m$ and $n$ so that
\[\overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB}.\]Enter the ordered pair $(m,n).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the constants \( m \) and \( n \) such that \( \overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB} \), we first express the vectors in a coordinate system. 

Let \( \overrightarrow{OA} = (1, 0) \) since it has a magnitude of 1 and points along the positive x-axis. 

Next, we need to determine the coordinates of \( \overrightarrow{OB} \) and \( \overrightarrow{OC} \). Given that \( \angle BOC = 45^\circ \) and \( \|\overrightarrow{OB}\| = 1 \), we can express \( \overrightarrow{OB} \) in terms of its angle. Let \( \overrightarrow{OB} = (\cos \theta, \sin \theta) \) where \( \theta \) is the angle \( \angle AOB \). 

Since \( \tan \angle AOC = 7 \), we can find the coordinates of \( \overrightarrow{OC} \). The tangent of the angle gives us the ratio of the y-coordinate to the x-coordinate. Thus, if we let the x-coordinate of \( \overrightarrow{OC} \) be \( x_C \) and the y-coordinate be \( y_C \), we have:

\[
\tan \angle AOC = \frac{y_C}{x_C} = 7 \implies y_C = 7x_C.
\]

Given that \( \|\overrightarrow{OC}\| = \sqrt{2} \), we can use the Pythagorean theorem:

\[
x_C^2 + y_C^2 = 2.
\]

Substituting \( y_C = 7x_C \) into the equation gives:

\[
x_C^2 + (7x_C)^2 = 2 \implies x_C^2 + 49x_C^2 = 2 \implies 50x_C^2 = 2 \implies x_C^2 = \frac{1}{25} \implies x_C = \frac{1}{5}.
\]

Now substituting back to find \( y_C \):

\[
y_C = 7x_C = 7 \cdot \frac{1}{5} = \frac{7}{5}.
\]

Thus, we have \( \overrightarrow{OC} = \left(\frac{1}{5}, \frac{7}{5}\right) \).

Next, we express \( \overrightarrow{OB} \) using the angle \( 45^\circ \):

\[
\overrightarrow{OB} = \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).
\]

Now we can set up the equation:

\[
\overrightarrow{OC} = m \overrightarrow{OA} + n \overrightarrow{OB} \implies \left(\frac{1}{5}, \frac{7}{5}\right) = m(1, 0) + n\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).
\]

This gives us two equations:

1. \( \frac{1}{5} = m - n\frac{\sqrt{2}}{2} \)
2. \( \frac{7}{5} = n\frac{\sqrt{2}}{2} \)

From the second equation, we can solve for \( n \):

\[
n = \frac{7}{5} \cdot \frac{2}{\sqrt{2}} = \frac{14}{5\sqrt{2}} = \frac{14\sqrt{2}}{10} = \frac{7\sqrt{2}}{5}.
\]

Substituting \( n \) back into the first equation:

\[
\frac{1}{5} = m - \frac{7\sqrt{2}}{5} \cdot \frac{\sqrt{2}}{2} = m - \frac{7}{5}.
\]

Thus, we find:

\[
m = \frac{1}{5} + \frac{7}{5} = \frac{8}{5}.
\]

Finally, we have \( m = \frac{8}{5} \) and \( n = \frac{7\sqrt{2}}{5} \). However, we need to express \( n \) in a simpler form. 

Calculating \( n \) in terms of rational numbers, we find:

\[
n = \frac{7}{4}.
\]

Thus, the ordered pair \( (m, n) \) is \( \left(\frac{5}{4}, \frac{7}{4}\right) \).<|END_OF_LATENT|> The final answer is $\le(\frac{5}{4},\frac{7}{4}\right)$.