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*2}{2}=2\Rightarrow \mathrn{(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>. 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*2}{2}=2\Rightarrow \mathrn{(C)}</math>. | ||
| + | ==Solution 2== | ||
| + | guessing is awesome typing fast is awesome scratch is aeosme go and paly dodgeball in the yard!{SODifg)S(DUGjG )(SD | ||
==See also== | ==See also== | ||
Revision as of 19:28, 17 December 2013
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 
. It is a 
triangle, which is an isosceles right triangle. Thus the area of it is $\frac{2*2}{2}=2\Rightarrow \mathrn{(C)}$ (Error compiling LaTeX. Unknown error_msg).
Solution 2
guessing is awesome typing fast is awesome scratch is aeosme go and paly dodgeball in the yard!{SODifg)S(DUGjG )(SD
See also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Final Question |
| 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. 
