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>. It is a <math>2-2-2\sqrt{2}</math> triangle, which is an isosceles right triangle. Thus the area of it is <math>\frac{2 | + | 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>. It is a <math>2-2-2\sqrt{2}</math> triangle, which is an isosceles right triangle. Thus the area of it is <math>\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}</math>. |
==See also== | ==See also== |
Revision as of 16:17, 2 May 2015
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 . It is a triangle, which is an isosceles right triangle. Thus the area of it is .
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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