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

Revision as of 18:44, 17 February 2018

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)

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

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)

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)

@@ Line 60: / Line 60: @@
 ==Solution 4 (Vectors)==
-IMPORTANT: This is assuming the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution doesn't work.
+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.
+.
-By the Pythagorean theorem, <math>EB=\sqrt{13}</math>. Because <math>EH=1</math>, the area of the base is <math>\sqrt{13}</math>. Now, we need to find the height.
-Define <math>X</math> as the midpoint of <math>BC</math> and <math>Y</math> as the midpoint of <math>EH</math>. Consider a vector coordinate system with origin <math>X</math> with <math>x, y,</math> and <math>z</math> axes parallel to <math>AB, BC,</math> and <math>CG</math> respectively (positive <math>x</math> direction is towards <math>A</math>, positive <math>y</math> direction is towards <math>C</math>, positive <math>z</math> direction is towards <math>G</math>). Then,
-<cmath>
-M=\begin{bmatrix}
-\\
-\\
-\\
-         \end{bmatrix},
-Y=\begin{bmatrix}
-\\
-\\
-\\
-         \end{bmatrix}
-</cmath>
-The dot product of <math>M</math> and <math>Y</math> is the length of the projection of <math>M</math> onto <math>Y</math> multiplied by the length of <math>Y</math>, so dividing the dot product of <math>M</math> and <math>Y</math> by the length of <math>Y</math> should give the length of the projection of <math>M</math> onto <math>Y</math>. Doing this calculation, we get that the length of the projection is <math>\frac4{\sqrt{13}}</math>. Notice that this projection onto <math>Y</math> is the same as projecting <math>M</math> onto the plane.
-Denote <math>P</math> as the foot of the projection of <math>M</math> onto <math>Y</math>. Then <math>\angle MPY</math> is right, so <math>\triangle MPY</math> is a right triangle. Applying the Pythagorean theorem on <math>\triangle MPY</math> and calling <math>MP</math> (which is actually the height of the pyramid) <math>x</math>, we get <math>x^2+\frac4{\sqrt{13}}^2=2^2</math>. Therefore, <math>x=\sqrt{4-\frac{16}{13}}=\frac6{\sqrt{13}}</math>.
-Now since we have the base and the height of the pyramid, we can find its volume. <math>\dfrac{\sqrt{13}\times\dfrac6{\sqrt{13}}}3=2</math>, so the answer is <math>\boxed{\text{(E)}}</math>.
-Written by: SS4
 ==Solution 5 (slicker method)==

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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