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Difference between revisions of "2018 AMC 10B Problems/Problem 10"

(→‎Solution 2)
(→‎Solution 2)
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We start by finding side <math>\overline{BE}</math> of base <math>BCHE</math> by using the Pythagorean theorem on <math>\triangle ABE</math>. Doing this, we get
 
We start by finding side <math>\overline{BE}</math> of base <math>BCHE</math> by using the Pythagorean theorem on <math>\triangle ABE</math>. Doing this, we get
  
−
<cmath>\overline{BE}^2 = \overline{AB}^2  + \overline{AE}^2 = 3^2 + 2^2 = 13.</cmath>
+
<cmath>\overline{BE}^2 = \overline{AB}^2  + \overline{AE}^2 = 3^2 + 2^2 = 15.</cmath>
  
 
Taking the square root of both sides of the equation, we get <math>\overline{BE} = \sqrt {13}</math>. We can then find the area of rectangle <math>BCHE</math>, noting that  
 
Taking the square root of both sides of the equation, we get <math>\overline{BE} = \sqrt {13}</math>. We can then find the area of rectangle <math>BCHE</math>, noting that  
  
−
<cmath>[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {13} \cdot 1 = \sqrt {13}.</cmath>
+
<cmath>[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {15} \cdot 1 = \sqrt {15}.</cmath>
  
  
Line 60: Line 60:
 
Multiplying both sides of the equation by 2 and substituting in known values, we get
 
Multiplying both sides of the equation by 2 and substituting in known values, we get
  
−
<cmath>2 \cdot 3 = FK \cdot \sqrt {13} \Rightarrow FK = \frac{6\sqrt {13}}{13}.</cmath>
+
<cmath>2 \cdot 3 = FK \cdot \sqrt {15} \Rightarrow FK = \frac{6\sqrt {15}}{15}.</cmath>
  
−
Deducing that the altitude from vertex <math>M</math> to base <math>BCHE</math> is <math>\frac{6\sqrt {13}}{13}</math> and calling the point of intersection between the altitude and the base as point <math>N</math>, we get the area of the rectangular pyramid to be  
+
Deducing that the altitude from vertex <math>M</math> to base <math>BCHE</math> is <math>\frac{6\sqrt {15}}{15}</math> and calling the point of intersection between the altitude and the base as point <math>N</math>, we get the area of the rectangular pyramid to be  
  
−
<cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {13} \cdot \frac{6\sqrt {13}}{13}\right) = \frac{6}{3} = \boxed{2}.</cmath>
+
<cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {15} \cdot \frac{6\sqrt {15}}{15}\right) = \frac{6}{3} = \boxed{2}.</cmath>
  
 
Written by: Adharshk
 
Written by: Adharshk

Revision as of 19:36, 16 February 2018

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);[/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

\[\overline{BE}^2 = \overline{AB}^2 + \overline{AE}^2 = 3^2 + 2^2 = 15.\]

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

\[[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {15} \cdot 1 = \sqrt {15}.\]


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

\[\frac{\overline{BF} \cdot \overline{FE}}{2} = \frac{\overline{BE} \cdot FK}{2}.\]

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

\[2 \cdot 3 = FK \cdot \sqrt {15} \Rightarrow FK = \frac{6\sqrt {15}}{15}.\]

Deducing that the altitude from vertex $M$ to base $BCHE$ is $\frac{6\sqrt {15}}{15}$ 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

\[\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {15} \cdot \frac{6\sqrt {15}}{15}\right) = \frac{6}{3} = \boxed{2}.\]

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

See Also

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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