2018 AMC 10B Problem 10

The base $BCHE$ is a rectangle: $BC = 1$, and $BE$ is a diagonal of the $3 \times 2$ face $ABFE$, so $BE = \sqrt{3^2 + 2^2} = \sqrt{13}$. Its area is $\sqrt{13}$.

For the height we need the distance from $M$ to the plane $BCHE$. Segment $FG$ is parallel to $BC$, hence parallel to that plane, so every point of $FG$ is the same distance from it; in particular $M$ is as far from the plane as $F$ is.

Look at the face $ABFE$: the plane $BCHE$ cuts it along $BE$, and the plane is perpendicular to this face (it contains $BC$, which is perpendicular to the face). So the distance from $F$ to the plane is just the distance from $F$ to the line $BE$ inside the face, i.e. the altitude from the right angle in triangle $EFB$ with legs $EF = 3$ and $FB = 2$:
$$h = \frac{3 \cdot 2}{\sqrt{13}} = \frac{6}{\sqrt{13}}.$$

Volume:
$$V = \frac{1}{3} \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2.$$

The answer is $\boxed{\textbf{(E)}\ 2}$.

A pyramid's height is a perpendicular distance to the base plane, not an edge length. Two facts make it computable: a point on a line parallel to the plane can be slid along that line for free, and when the plane is perpendicular to a face, the 3-D distance becomes a 2-D distance in that face.

The plane $BCHE$ splits the box into two triangular prisms of volume $3$ each. The pyramid $F$-$BCHE$ (same height as $M$-$BCHE$) has as base the rectangular face of the prism $EFB$-$HGC$ and apex on the opposite edge, so its volume is $\frac{2}{3}$ of the prism: $\frac{2}{3} \cdot 3 = 2$. Coordinates also work: $M = (3, \tfrac12, 2)$, plane $2x + 3z = 6$, distance $\frac{6}{\sqrt{13}}$.

• Using the box height 2 (or the distance from M to line BC) as the pyramid's height instead of the perpendicular distance from M to the slanted plane BCHE.
• Taking the base area as $3 \cdot 1$ or $2 \cdot 1$ instead of $\sqrt{13} \cdot 1$.