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

(→‎Problem: = the mistakenly added of the term 'rectangle' IN the problem statement. Though true, it is not in the original problem statement (based on Mr. R's video), indicating it was meant to be figured out by the solver!!)
m (→‎Solution 1)
 
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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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