Difference between revisions of "2020 AMC 12A Problems/Problem 12"
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Revision as of 15:14, 1 February 2020
Problem
Line 
in the coordinate plane has equation 
. This line is rotated 
counterclockwise about the point 
to obtain line 
. What is the 
-coordinate of the -intercept of line 
Solution
The slope of the line is 
. We must transform it by .
creates an isosceles right triangle since the sum of the angles of the triangle must be 
and one angle is 
which means the last leg angle must also be .
In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of slope on graph paper. That line with slope starts at 
and will go to 
, the vector 
.
Construct another line from to 
, the vector 
. This is 
and equal to the original line segment. The difference between the two vectors is 
, which is the slope 
, and that is the slope of line .
Furthermore, the equation passes straight through since 
, which means that any rotations about would contain . We can create a line of slope through . The -intercept is therefore 
~lopkiloinm
See Also
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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. 
