2018 AMC 10B Problems/Problem 10

Problem

In the rectangular parallelpiped 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 $M$ ?

$[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G = (5.95,4.2); pair C = (5.95,1.2); pair D = (1.2,1.2); pair H= (1.2,4.2); pair M = ((4.75+5.95)/2,3.6); draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H); draw(B--C); draw(F--G); draw(A--D--H--C--D,dashed); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,W); label("$E$",E1,W); label("$F$",F,SW); label("$G$",G,NE); label("$H$",H,NW); label("$M$",M,N); dot(A); dot(B); dot(E1); dot(F); dot(G); dot(C); dot(D); dot(H); dot(M); label("3",A/2+B/2,S); label("2",C/2+G/2,E); label("1",C/2+B/2,SE); [/asy]$

$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2$

Solution 1

Consider the cross-sectional plane. Note that $bh/2=3$ and we want $bh/3$ , so the answer is $\boxed{2}$ . (AOPS12142015)

Solution 2

We start by finding side $\overline{BE}$ of base $BCHE$ by using the Pythagorean theorem on $\triangle ABE$ . Doing this, we get

$\overline{BE}^2 = \overline{AB}^2 + \overline{AE}^2 = 3^2 + 2^2 = 13.$

Taking the square root of both sides of the equation, we get $\overline{BE} = \sqrt {13}$ . We can then find the area of rectangle $BCHE$ , noting that

$[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {13} \cdot 1 = \sqrt {13}.$

Taking the vertical cross-sectional plane of the rectangular prism, we see that the distance from point $M$ to base $BCHE$ is the same as the distance from point $F$ to side $\overline{BE}$ . Calling the point where the altitude from vertex $F$ touches side $\overline{BE}$ as point $K$ , we can easily find this altitude using the area of right $\triangle BFE$ , as

$\frac{\overline{BF} \cdot \overline{EF}}{2} = \frac{\overline{BE} \cdot FK}{2}.$

Multiplying both sides of the equation by 2 and substituting in known values, we get

$2 \cdot 3 = FK \cdot \sqrt {13} \Rightarrow FK = \frac{6\sqrt {13}}{13}.$

Deducing that the altitude from vertex $M$ to base $BCHE$ is $\frac{6\sqrt {13}}{13}$ and calling the point of intersection between the altitude and the base as point $N$ , we get the area of the rectangular pyramid to be

$\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {13} \cdot \frac{6\sqrt {13}}{13}\right) = \frac{6}{3} = \boxed{2}.$

Written by: Adharshk

Solution 3

We can start by finding the total volume of the parallelepiped. It is $2 \cdot 3 \cdot 1 = 6$ , because a rectangular parallelepiped is a rectangular prism.

Next, we can consider the wedge-shaped section made when the plane $BCHE$ cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is $\frac{1}{2} \cdot 2 \cdot 3 = 3$ . Since BC is given to be $1$ , we have that FM is $\frac{1}{2}$ . Using the formula for the volume of a triangular pyramid, we have $V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}$ . Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume $\frac{1}{2}$ as well.

The original wedge we considered in the last step has volume $3$ , because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have $3 - \frac{1}{2} \cdot 2 = 2$ . Thus, the volume of the figure we are trying to find is $2$ . This means that the correct answer choice is $\boxed{E}$ .

Written by: Archimedes15

Solution 4 (Vectors)

By the Pythagorean theorem, $EB=\sqrt{13}$ . Because $EH=1$ , the area of the base is $\sqrt{13}$ . Now, we need to find the height.

Define $X$ as the midpoint of $BC$ and $Y$ as the midpoint of $EH$ . Consider a vector coordinate system with origin $X$ with $x, y,$ and $z$ axes parallel to $AB, BC,$ and $CG$ respectively (positive $x$ direction is towards $A$ , positive $y$ direction is towards $C$ , positive $z$ direction is towards $G$ ). Then, $M=\begin{bmatrix} 0\\ 0\\ 2\\ \end{bmatrix}, Y=\begin{bmatrix} 3\\ 0\\ 2\\ \end{bmatrix}$ The dot product of $M$ and $Y$ is the length of the projection of $M$ onto $Y$ multiplied by the length of $Y$ , so dividing the dot product of $M$ and $Y$ by the length of $Y$ should give the length of the projection of $M$ onto $Y$ . Doing this calculation, we get that the length of the projection is $\frac4{\sqrt{13}}$ . Notice that this projection onto $Y$ is the same as projecting $M$ onto the plane.

Denote $P$ as the foot of the projection of $M$ onto $Y$ . Then $\angle MPY$ is right, so $\triangle MPY$ is a right triangle. Applying the Pythagorean theorem on $\triangle MPY$ and calling $MP$ (which is actually the height of the pyramid) $x$ , we get $x^2+\frac4{\sqrt{13}}^2=2^2$ . Therefore, $x=\sqrt{4-\frac{16}{13}}=\frac6{\sqrt{13}}$ .

Now since we have the base and the height of the pyramid, we can find its volume. $\dfrac{\sqrt{13}\times\dfrac6{\sqrt{13}}}3=2$ , so the answer is $\boxed{\text{(E)}}$ .

Written by: SS4

Solution 5 (slicker method)

Rotate the rectangular pyramid so that rectangle $GFBC$ is the base of our rectangular pyramid. Now our height becomes $GH=3.$ We know that the volume of our rectangular pyramid is $\dfrac{1}{3} \cdot [GFBC] \cdot GH= \dfrac{1}{3} \cdot 2 \cdot 3= \boxed{2}.$

(MathloverMC)

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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