Difference between revisions of "2012 AMC 10A Problems/Problem 21"

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<math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3} </math>
 
<math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3} </math>
  
==Solution ==
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==Solution 1==
  
 
Consider a tetrahedron with vertices at <math>A,B,C,D</math> on the <math>xyz</math>-plane. The length of <math>EF</math> is just one-half of <math>AD</math> because it is the midsegment of <math>\triangle ABD.</math> The same concept applies to the other side lengths. <math>AD=3</math> and <math>BC=\sqrt{1^2+2^2}=\sqrt{5}</math>. Then <math>EF=HG=\frac32</math> and <math>EH=FG=\frac{\sqrt{5}}{2}</math>. The line segments lie on perpendicular planes so quadrilateral <math>EFGH</math> is a rectangle. The area is
 
Consider a tetrahedron with vertices at <math>A,B,C,D</math> on the <math>xyz</math>-plane. The length of <math>EF</math> is just one-half of <math>AD</math> because it is the midsegment of <math>\triangle ABD.</math> The same concept applies to the other side lengths. <math>AD=3</math> and <math>BC=\sqrt{1^2+2^2}=\sqrt{5}</math>. Then <math>EF=HG=\frac32</math> and <math>EH=FG=\frac{\sqrt{5}}{2}</math>. The line segments lie on perpendicular planes so quadrilateral <math>EFGH</math> is a rectangle. The area is
  
 
<cmath>EF \cdot FG = \frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}</cmath>  
 
<cmath>EF \cdot FG = \frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}</cmath>  
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==Solution 2==
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Computing the points of <math>EFGH</math> gives <math>E(0.5, 0, 1.5), F(0.5, 0, 0), G(0,1,0), H(0,1,1.5)</math>.  The vector <math>EF</math> is <math>(0,0,-1.5)</math>, while the vector <math>HG</math> is also <math>(0,0,-1.5)</math>, meaning the two sides <math>EF</math> and <math>GH</math> are parallel.  Similarly, the vector <math>FG</math> is <math>(-0.5, 1, 0)</math>, while the vector <math>EH</math> is also <math>(-0.5, 1, 0)</math>.  Again, these are equal in both magnitude and direction, so <math>FG</math> and <math>EH</math> are parallel.  Thus, figure <math>EFGH</math> is a parallelogram.
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Computation of vectors <math>EF</math> and <math>HG</math> is sufficient evidence that the figure is a parallelogram, since the vectors are not only point in the same direction, but are of the same magnitude, but the other vector <math>FG</math> is needed to find the angle between the sides.
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Taking the dot product of vector <math>EF</math> and vector <math>FG</math> gives <math>0 \cdot -0.5 + 0 \cdot 1 + -1.5 \cdot 0 = 0</math>, which means the two vectors are perpendicular.  (Alternately, as above, note that vector <math>EF</math> goes directly down on the z-axis, while vector <math>FG</math> has no z-component and lie completely in the xy plane.)  Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.  With the distance formula in three dimensions, we find that <math>EF = \frac{3}{2}</math> and <math>FG = \frac{\sqrt{5}}{2}</math>, giving an area of <math>\frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2012|ab=A|num-b=20|num-a=22}}

Revision as of 00:49, 10 February 2012

Problem

Let points $A$ = $(0 ,0 ,0)$, $B$ = $(1, 0, 0)$, $C$ = $(0, 2, 0)$, and $D$ = $(0, 0, 3)$. Points $E$, $F$, $G$, and $H$ are midpoints of line segments $\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$?

$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3}$

Solution 1

Consider a tetrahedron with vertices at $A,B,C,D$ on the $xyz$-plane. The length of $EF$ is just one-half of $AD$ because it is the midsegment of $\triangle ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$. Then $EF=HG=\frac32$ and $EH=FG=\frac{\sqrt{5}}{2}$. The line segments lie on perpendicular planes so quadrilateral $EFGH$ is a rectangle. The area is

\[EF \cdot FG = \frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}\]

Solution 2

Computing the points of $EFGH$ gives $E(0.5, 0, 1.5), F(0.5, 0, 0), G(0,1,0), H(0,1,1.5)$. The vector $EF$ is $(0,0,-1.5)$, while the vector $HG$ is also $(0,0,-1.5)$, meaning the two sides $EF$ and $GH$ are parallel. Similarly, the vector $FG$ is $(-0.5, 1, 0)$, while the vector $EH$ is also $(-0.5, 1, 0)$. Again, these are equal in both magnitude and direction, so $FG$ and $EH$ are parallel. Thus, figure $EFGH$ is a parallelogram.

Computation of vectors $EF$ and $HG$ is sufficient evidence that the figure is a parallelogram, since the vectors are not only point in the same direction, but are of the same magnitude, but the other vector $FG$ is needed to find the angle between the sides.

Taking the dot product of vector $EF$ and vector $FG$ gives $0 \cdot -0.5 + 0 \cdot 1 + -1.5 \cdot 0 = 0$, which means the two vectors are perpendicular. (Alternately, as above, note that vector $EF$ goes directly down on the z-axis, while vector $FG$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle. With the distance formula in three dimensions, we find that $EF = \frac{3}{2}$ and $FG = \frac{\sqrt{5}}{2}$, giving an area of $\frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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