2018 AMC 10B Problem 10

P1-10~4 minPrint

Solid Geometry Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

In the rectangular parallelepiped shown, $AB=3$ , $BC=1$ , and $CG=2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$ ?

Show solution without answering

Key insight

The base is a slanted 1-by-square-root-of-13 rectangle. The height is the distance from M to that slanted rectangle, which you can measure from F in the front face.

Solution

A pyramid's volume is $\frac13 \times (\text{base area}) \times (\text{height})$ , where the height is measured straight to the base. The base here is tilted, so be careful with the height.

1. The base. $BCHE$ is a rectangle. One side is $BC = 1$ . The other side $BE$ runs diagonally across the face $ABFE$ , which is $3$ by $2$ , so
$BE = \sqrt{3^2 + 2^2} = \sqrt{13}.$
The base area is $1 \times \sqrt{13} = \sqrt{13}$ .
2. Move the apex to make the height easier. The edge $FG$ runs parallel to $BC$ , which lies in the base. So every point of $FG$ is the same distance from the base. That means $M$ is exactly as far from the base as $F$ is.
3. Find the distance from $F$ in a flat picture. Look straight at the face $ABFE$ . The slanted base shows up as the line $BE$ , and $F$ is the corner of the right triangle $EFB$ with legs $EF = 3$ and $FB = 2$ . (This works because the base rectangle stands straight out from this face, along $BC$ .) The distance from $F$ to the base is the height of this right triangle drawn to its long side $BE$ .
4. Compute that height. Find the triangle's area two ways. Using the legs, the area is $\frac12 \times 3 \times 2 = 3$ . Using $BE$ as the base, the area is $\frac12 \times \sqrt{13} \times h$ . So
$\tfrac12 \sqrt{13}\, h = 3, \qquad h = \frac{6}{\sqrt{13}}.$
5. Volume.
$V = \frac13 \times \sqrt{13} \times \frac{6}{\sqrt{13}} = \frac13 \times 6 = 2.$

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

Why this works

Sliding the apex along a line parallel to the base does not change the height, so you can pick the easiest point. When the tilted base stands straight out from one face of the box, the 3-D distance becomes a 2-D distance in that face, and the "area two ways" trick finds it.

Alternative approach

Cut the box along the rectangle $BCHE$ . The box has volume $3 \times 1 \times 2 = 6$ , so each half is a wedge of volume $3$ . The half containing $F$ and $G$ is a wedge with triangle ends $EFB$ and $HGC$ . A pyramid inside this wedge, with the rectangle $BCHE$ as base and its tip on the opposite edge $FG$ , takes up $\frac23$ of the wedge (the rest is a smaller pyramid of $\frac13$ ). So the volume is $\frac23 \times 3 = 2$ . Since $M$ is on edge $FG$ , the pyramid with tip $M$ has this same volume.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 2}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 15: Curved Surfaces