Triangle $ABC$ has vertices $A(0, 8)$, $B(2, 0)$, $C(8, 0)$.  A horizontal line with equation $y=t$ intersects line segment $ \overline{AB} $ at $T$ and line segment $ \overline{AC} $ at $U$, forming $\triangle ATU$ with area 13.5.  Compute $t$.
The line through $A$ and $B$ has slope $\frac{0-8}{2-0}=-4$ and passes through $(0,8)$, so thus has equation $y=-4x+8$. The line through $A$ and $C$ has slope $\frac{0-8}{8-0}=-1$ and passes through $(0,8)$, so thus has equation $y=-x+8$.

The point $T$ is the point on the line $y=-4x+8$ with $y$-coordinate $t$.  To find the $x$-coordinate, we solve $t=-4x+8$ to get $4x = 8-t$ or $x = \frac{1}{4}(8-t)$.  The point $U$ is the point on the line $y=-x+8$ with $y$-coordinate $t$.  To find the $x$-coordinate, we solve $t=-x+8$ to get $x = 8-t$.

Therefore, $T$ has coordinates $(\frac{1}{4}(8-t),t)$, $U$ has coordinates $(8-t,t)$, and $A$ is at $(0,8)$.

$TU$ is horizontal and has length $(8-t)-\frac{1}{4}(8-t)=\frac{3}{4}(8-t)$ and the distance from $TU$ to $A$ is $8-t$, so the area in terms of $t$ is \[\frac{1}{2}\left(\frac{3}{4}(8-t)\right)(8-t) = \frac{3}{8}(8-t)^2.\]Since this equals $13.5$, we have $\frac{3}{8}(8-t)^2 = 13.5$ or $(8-t)^2 = \frac{8}{3}(13.5)=36$.  Because line segment $TU$ is below $A$, $t<8$, and so $8-t>0$.  Therefore, $8-t=6 \Rightarrow t=8-6=\boxed{2}$.
2
Level 5
Geometry
Problem: Triangle $ABC$ has vertices $A(0, 8)$, $B(2, 0)$, $C(8, 0)$.  A horizontal line with equation $y=t$ intersects line segment $ \overline{AB} $ at $T$ and line segment $ \overline{AC} $ at $U$, forming $\triangle ATU$ with area 13.5.  Compute $t$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( t \), we first need to determine the equations of the lines \( \overline{AB} \) and \( \overline{AC} \). 

The slope of line segment \( \overline{AB} \) can be calculated using the coordinates of points \( A(0, 8) \) and \( B(2, 0) \). The slope \( m_{AB} \) is given by:

\[
m_{AB} = \frac{0 - 8}{2 - 0} = -4
\]

Using point-slope form, the equation of line \( \overline{AB} \) is:

\[
y - 8 = -4(x - 0) \implies y = -4x + 8
\]

Next, we find the slope of line segment \( \overline{AC} \) using points \( A(0, 8) \) and \( C(8, 0) \):

\[
m_{AC} = \frac{0 - 8}{8 - 0} = -1
\]

The equation of line \( \overline{AC} \) is:

\[
y - 8 = -1(x - 0) \implies y = -x + 8
\]

Now, we find the intersection points \( T \) and \( U \) where the horizontal line \( y = t \) intersects \( \overline{AB} \) and \( \overline{AC} \).

For line \( \overline{AB} \):

\[
t = -4x + 8 \implies 4x = 8 - t \implies x = \frac{8 - t}{4}
\]

Thus, the coordinates of point \( T \) are:

\[
T\left(\frac{8 - t}{4}, t\right)
\]

For line \( \overline{AC} \):

\[
t = -x + 8 \implies x = 8 - t
\]

Thus, the coordinates of point \( U \) are:

\[
U(8 - t, t)
\]

Next, we calculate the area of triangle \( ATU \) using the formula for the area of a triangle given by vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substituting the coordinates of points \( A(0, 8) \), \( T\left(\frac{8 - t}{4}, t\right) \), and \( U(8 - t, t) \):

\[
\text{Area} = \frac{1}{2} \left| 0(t - t) + \frac{8 - t}{4}(t - 8) + (8 - t)(8 - t) \right|
\]

This simplifies to:

\[
\text{Area} = \frac{1}{2} \left| \frac{8 - t}{4}(t - 8) + (8 - t)(8 - t) \right|
\]

Calculating the terms:

1. The first term: 
\[
\frac{8 - t}{4}(t - 8) = \frac{(8 - t)(t - 8)}{4} = \frac{-(8 - t)^2}{4}
\]

2. The second term:
\[
(8 - t)(8 - t) = (8 - t)^2
\]

Combining these:

\[
\text{Area} = \frac{1}{2} \left| \frac{-(8 - t)^2}{4} + (8 - t)^2 \right| = \frac{1}{2} \left| \frac{3(8 - t)^2}{4} \right| = \frac{3(8 - t)^2}{8}
\]

Setting the area equal to 13.5:

\[
\frac{3(8 - t)^2}{8} = 13.5
\]

Multiplying both sides by 8:

\[
3(8 - t)^2 = 108
\]

Dividing by 3:

\[
(8 - t)^2 = 36
\]

Taking the square root:

\[
8 - t = 6 \quad \text{or} \quad 8 - t = -6
\]

Solving these gives:

1. \( t = 2 \)
2. \( t = 14 \)

Since \( t \) must be within the bounds of the triangle, we discard \( t = 14 \) as it exceeds the maximum \( y \)-coordinate of point \( A \). Thus, the only valid solution is:

\[
t = 2
\] 

The final answer is \( 2 \).<|END_OF_LATENT|> The final answer is $2$.