2020 AMC 12A Problems/Problem 12
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 1
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 . This 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 ~ShawnX (diagram)
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 (Rotation Matrix)
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 .
Solution 6 (Angle Bisector)
Note is on the line. Construct the perpendicular line . This creates a right triangle that intersects the x-axis at and a distance of apart. The transformation will bisect the right angle. The angle bisector theorem tells us the will split in ratio to the lengths of the sides. These are and and . Thus the x intercept will split the line from to into a ratio of making the x-intercept .
Solution 7 (Complex Numbers)
Converting to the complex plane, we can see that two numbers on the line are and . Translating to the origin, we get and . Multiplying each of them by , we get and . This line has a slope of . Now, back to the cartesian plane. We have a line passing through with slope which gives the equation as which implies the coordinate of the intercept is .
~rocketsri (minor error corrected by kn07)
Solution 8 (quick)
A quick check tells us that falls on the given line. Common sense tells us that if the slope of the original line is , or 45 degrees from the horizontal, a 45 counter clockwise rotation will result in a vertical line, and the x-intercept will be . Thinking of a 45 degree counter clockwise rotation as a 90 degree counter clockwise rotation that is bisected will helps in visualizing this line of reasoning. Therefore, it follows that if the original line is made steeper, then the x-intercept will move away from to the right. If the original line is made lower, then the opposite will happen. Our given line has slope , so the answer must be or . can be eliminated because an x-intercept of can only occur when the original line is horizontal. In conclusion, the answer must be .
~jackshi2006
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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