Difference between revisions of "2020 AMC 12A Problems/Problem 12"
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+ | ==Solution 5 (Linear Algebra)== | ||
+ | First note that the given line goes through <math>(20,20)</math> with a slope of <math>\frac{3}{5}</math>. This means that <math>(25,23)</math> is on the line. Now consider translating the graph so that <math>(20,20)</math> goes to the origin, then <math>(25,23)</math> becomes <math>(5,3)</math>. We now rotate the line <math>45^\circ</math> about the origin using a rotation matrix. This maps <math>(5,3)</math> to | ||
+ | <cmath>\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}</cmath> | ||
+ | The line through the origin and <math>(\sqrt{2}, 4\sqrt{2})</math> has slope <math>4</math>. Translating this line so that the origin is mapped to <math>(20,20)</math>, we find that the equation for the new line is <math>4x-60</math>, meaning that the <math>x</math>-intercept is <math>x=\frac{60}{4}=\boxed{\textbf{(B) }15}</math>. | ||
==See Also== | ==See Also== |
Revision as of 20:26, 2 February 2020
Contents
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
Solution 2
Since the slope of the line is , and the angle we are rotating around is x, then
Hence, the slope of the rotated line is . Since we know the line intersects the point , then we know the line is . Set to find the x-intercept, and so
~Solution by IronicNinja
Solution 3
Let be and be and respectively. Since the slope of the line is we know that Segments and represent the before and after of rotating by 45 counterclockwise. Thus, and by tangent addition formula. Since is 5 and the sidelength of the square is 20 the answer is
Solution 4 (Cheap)
Using the protractor you brought, carefully graph the equation and rotate the given line counter-clockwise about the point . Scaling everything down by a factor of 5 makes this process easier.
It should then become fairly obvious that the x intercept is (only use this as a last resort).
~Silverdragon
Solution 5 (Linear Algebra)
First note that the given line goes through with a slope of . This means that is on the line. Now consider translating the graph so that goes to the origin, then becomes . We now rotate the line about the origin using a rotation matrix. This maps to The line through the origin and has slope . Translating this line so that the origin is mapped to , we find that the equation for the new line is , meaning that the -intercept is .
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.