Difference between revisions of "2007 AMC 12B Problems/Problem 25"
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==Solution== | ==Solution== | ||
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>. | 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== |
Latest revision as of 16:28, 20 September 2020
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 |
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