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

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==Problem==
 
==Problem==
  
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Let points <math>A</math> = <math>(0 ,0 ,0)</math>, <math>B</math> = <math>(1, 0, 0)</math>, <math>C</math> = <math>(0, 2, 0)</math>, and <math>D</math> = <math>(0, 0, 3)</math>. Points <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> are midpoints of line segments <math>\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},</math> and <math>\overline{DC}</math> respectively. What is the area of rectangle <math>EFGH</math>?
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Let points <math>A</math> = <math>(0 ,0 ,0)</math>, <math>B</math> = <math>(1, 0, 0)</math>, <math>C</math> = <math>(0, 2, 0)</math>, and <math>D</math> = <math>(0, 0, 3)</math>. Points <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> are midpoints of line segments <math>\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},</math> and <math>\overline{DC}</math> respectively. What is the area of <math>EFGH</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>
 
<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>
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==Solution 1==
 
==Solution 1==
  
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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> in the <math>xyz</math>-space. 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 = \frac 32 \cdot \frac{\sqrt{5}}{2} = \frac{3\sqrt 5} 4\implies \boxed{\textbf C}.</cmath>
 
<cmath>EF \cdot FG = \frac 32 \cdot \frac{\sqrt{5}}{2} = \frac{3\sqrt 5} 4\implies \boxed{\textbf C}.</cmath>

Latest revision as of 18:55, 22 September 2024

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 $A,B,C,D$ in the $xyz$-space. 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

\[EF \cdot FG = \frac 32 \cdot \frac{\sqrt{5}}{2} = \frac{3\sqrt 5} 4\implies \boxed{\textbf C}.\]

[asy] import three; draw((0,0,0)--(1,0,0)--(0,0,3)--cycle); draw((0,0,0)--(0,2,0)); draw((0,2,0)--(0,0,3)); //EFGH draw((0.5,0,1.5)--(0.5,0,0)--(0,1,0)--(0,1,1.5)--(0.5,0,1.5),red); //Points label("$E$",(0.5,0,1.5),NW); label("$F$",(0.5,0,0),S); label("$G$",(0,1,0),S); label("$H$",(0,1,1.5),NE); label("$A$",(0,0,0),NE); label("$B$",(1,0,0),S); label("$C$",(0,2,0),S); label("$D$",(0,0,3),N); [/asy]

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 $\frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc10a/249

~dolphin7

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
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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