2012 AMC 10A Problems/Problem 21
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 | |
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All AMC 10 Problems and Solutions |