2007 AMC 12B Problems/Problem 25
Contents
Problem
Points and
are located in 3-dimensional space with
and
. The plane of
is parallel to
. What is the area of
?
Solution 1
Let
, and
. Since
, we could let
,
, and
. Now to get back to
we need another vertex
. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw
. Now we can bend these three sides into an equilateral triangle, and the coordinates change:
,
,
,
, and
. Checking for all the requirements, they are all satisfied. Now we find the area of triangle
. The side lengths of this triangle are
, which is an isosceles right triangle. Thus the area of it is
.
Solution 2
Similar to solution 1, we allow
,
, and
. This creates the isosceles right triangle on the plane of
Now, note that . This means that there exists some vector
parallel to the plane of
that forms two right angles with
and
. By definition, this is the cross product of the two vectors
and
. Finding this cross product, we take the determinant of vectors
and
*Note that z is constant because the line is parallel to the plane*
to get
Because there can be no movement in the direction, the k unit vector must be zero. Also, because the i unit vector must be orthogonal and also 0. Thus, the vector of line
is simply
From this, you can figure out that line , and the area of
.
See also
2007 AMC 12B (Problems • Answer Key • Resources) | |
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