Difference between revisions of "2007 AMC 12B Problems/Problem 25"
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+ | Also refer to the [https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_25#Solution| 2007 AMC 10B #25] (same problem) | ||
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==Problem== | ==Problem== | ||
Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overline{DE}</math>. What is the area of <math>\triangle BDE</math>? | Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overline{DE}</math>. What is the area of <math>\triangle BDE</math>? |
Revision as of 01:37, 25 December 2020
Also refer to the 2007 AMC 10B #25 (same problem)
Contents
Problem
Points and
are located in 3-dimensional space with
and
. The plane of
is parallel to
. What is the area of
?
Solution
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) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.