2016 AMC 10B Problem 9

1. $\overline{BC}$ is flat (parallel to the $x$-axis), so $B$ and $C$ are at the same height.
2. The graph of $y = x^2$ is the same on the left and right of the $y$-axis. For example, $(-3, 9)$ and $(3, 9)$ are both on it. So two points at the same height are mirror images. Call them $B = (-b, b^2)$ and $C = (b, b^2)$, with $b$ positive.
3. The base $BC$ goes from $x = -b$ to $x = b$, so its length is $2b$.
4. The height is the vertical distance from $A = (0,0)$ up to the line $BC$, which is $b^2$.
5. Area $= \frac12 \cdot \text{base} \cdot \text{height} = \frac12 \cdot 2b \cdot b^2 = b^3$.
6. Set $b^3 = 64$. Then $b = 4$, and $BC = 2b = 8$.

Check: the points are $(-4, 16)$ and $(4, 16)$, the base is $8$, the height is $16$, and the area is $\frac12 \cdot 8 \cdot 16 = 64$.

The answer is $\boxed{\textbf{(C)}\ 8}$.

When a shape sits on a graph, use the graph's symmetry to name the points with as few letters as possible. Here one letter, $b$, describes the whole triangle, and the area formula gives one equation for it.

Test the choices. If $BC = 8$, then $B$ and $C$ are at $x = \pm 4$, so they sit at height $16$. The area is $\frac12 \cdot 8 \cdot 16 = 64$, which matches. (For example, $BC = 6$ gives only $\frac12 \cdot 6 \cdot 9 = 27$.)

Since $\overline{BC}$ is horizontal, $B$ and $C$ have the same $y$-coordinate. On the parabola $y = x^2$, two points with the same height are mirror images across the $y$-axis, so let $B = (-b, b^2)$ and $C = (b, b^2)$ with $b > 0$.

Then $BC = 2b$, and the height from $A = (0,0)$ to line $BC$ is $b^2$. The area is
$$\frac12 \cdot 2b \cdot b^2 = b^3 = 64,$$
so $b = 4$ and $BC = 2b = 8$.

The answer is $\boxed{\textbf{(C)}\ 8}$.