Difference between revisions of "2007 AMC 12B Problems/Problem 25"
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<math>\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}</math> | <math>\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Let <math>A=(0,0,0)</math>, and <math>B=(2,0,0)</math>. Since <math>EA=2</math>, we could let <math>C=(2,0,2)</math>, <math>D=(2,2,2)</math>, and <math>E=(2,2,0)</math>. Now to get back to <math>A</math> we need another vertex <math>F=(0,2,0)</math>. 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 <math>FA</math>. Now we can bend these three sides into an equilateral triangle, and the coordinates change: <math>A=(0,0,0)</math>, <math>B=(2,0,0)</math>, <math>C=(2,0,2)</math>, <math>D=(1,\sqrt{3},2)</math>, and <math>E=(1,\sqrt{3},0)</math>. Checking for all the requirements, they are all satisfied. Now we find the area of triangle <math>BDE</math>. | + | [[File:2007 AMC 12B Problem 25.png|center]] |
+ | Link to graph: https://www.math3d.org/pHFSD6vRi | ||
+ | |||
+ | |||
+ | Let <math>A=(0,0,0)</math>, and <math>B=(2,0,0)</math>. Since <math>EA=2</math>, we could let <math>C=(2,0,2)</math>, <math>D=(2,2,2)</math>, and <math>E=(2,2,0)</math>. Now to get back to <math>A</math> we need another vertex <math>F=(0,2,0)</math>. 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 <math>FA</math>. Now we can bend these three sides into an equilateral triangle, and the coordinates change: <math>A=(0,0,0)</math>, <math>B=(2,0,0)</math>, <math>C=(2,0,2)</math>, <math>D=(1,\sqrt{3},2)</math>, and <math>E=(1,\sqrt{3},0)</math>. Checking for all the requirements, they are all satisfied. Now we find the area of triangle <math>BDE</math>. The side lengths of this triangle are <math>2, 2, 2\sqrt{2}</math>, which is an isosceles right triangle. Thus the area of it is <math>\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Similar to solution 1, we allow | ||
+ | <math>A=(0,0,0)</math>, <math>B=(2,0,0)</math>, and <math>C=(0,2,0)</math>. This creates the isosceles right triangle on the plane of <math>z=0</math> | ||
+ | |||
+ | Now, note that <math>\angle CDE=\angle DEA=90^o</math>. This means that there exists some vector <math>DE</math> parallel to the plane of <math>ABC</math> that forms two right angles with <math>AE</math> and <math>CD</math>. By definition, this is the cross product of the two vectors <math>AE</math> and <math>CD</math>. Finding this cross product, we take the determinant of vectors | ||
+ | |||
+ | <math>AE=<x_1,y_1,z></math> and | ||
+ | |||
+ | <math>CD=<x_2,y_2,z></math> *Note that z is constant because the line is parallel to the plane* | ||
+ | |||
+ | to get <math>(y_1-y_2)zi+(x_1-x_2)zj+(x_1y_2-y_1x_2)k</math> | ||
+ | |||
+ | Because there can be no movement in the <math>z</math> 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 <math>DE</math> is simply <math>2tj+2k</math> | ||
+ | |||
+ | From this, you can figure out that line <math>BE=2</math>, and the area of <math>BDE=\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}</math>. | ||
==See also== | ==See also== | ||
− | {{AMC12 box|year=2007|ab=B|num-b=24|after=Problem | + | {{AMC12 box|year=2007|ab=B|num-b=24|after=Last Problem}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:Area Problems]] | [[Category:Area Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:46, 17 July 2023
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
Link to graph: https://www.math3d.org/pHFSD6vRi
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 |
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