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 == | + | ==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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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
Contents
Problem
Let points = , = , = , and = . Points , , , and are midpoints of line segments and respectively. What is the area of ?
Solution 1
Consider a tetrahedron with vertices at on the -plane. The length of is just one-half of because it is the midsegment of The same concept applies to the other side lengths. and . Then and . The line segments lie on perpendicular planes so quadrilateral is a rectangle. The area is
Solution 2
Computing the points of gives . The vector is , while the vector is also , meaning the two sides and are parallel. Similarly, the vector is , while the vector is also . Again, these are equal in both magnitude and direction, so and are parallel. Thus, figure is a parallelogram.
Computation of vectors and 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 is needed to find the angle between the sides.
Taking the dot product of vector and vector gives , which means the two vectors are perpendicular. (Alternately, as above, note that vector goes directly down on the z-axis, while vector 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 and , giving an area of
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 |