2018 AMC 10B Problems/Problem 10

Revision as of 18:47, 17 February 2018 by Dogsareawesome123 (talk | contribs) (Solution 1)

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, and label it as b. Note that $bh/2=3$ and we want $bh/3$, so the answer is $\boxed{2}$. (AOPS12142015)

Solution 2

IMPORTANT: This solution assumed that the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution didn't work. Sorry Adarshk.

Solution 3

IMPORTANT: This solution assumed that the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution didn't work. Sorry Archimedes15.

Solution 4 (Vectors)

IMPORTANT: This solution assumed that the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution didn't work. Sorry SS4. .

Solution 5 (slicker method)

IMPORTANT: This solution assumed that the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution didn't work. Sorry MathloverMC

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png