Difference between revisions of "2007 AMC 12B Problems/Problem 25"
(→See also) |
(→See also) |
||
| Line 8: | Line 8: | ||
==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}} | ||
Revision as of 16:06, 1 January 2017
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
