Difference between revisions of "2025 AIME I Problems/Problem 9"
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Revision as of 20:47, 13 February 2025
Problem
The parabola with equation 
is rotated 
counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has 
-coordinate 
, where 
, 
, and 
are positive integers, and and are relatively prime. Find 
.
Solution 1
To begin with notice, a 
rotation counterclockwise about the origin on the 
axis is the same as a reflection over the line 
Since the parabola 
is symmetric about the axis as well, we can simply reflect it over the line. In addition any point of intersection between the line and parabola will also be on the rotated parabola. So we solve for the intersection, ![\[-x\sqrt{3}=x^2-4.\]](http://latex.artofproblemsolving.com/d/1/c/d1cedc97909d606e2dac9b6a88d169eb2b14fc4e.png)
![\[x^2+x\sqrt{3}-4=0.\]](http://latex.artofproblemsolving.com/2/2/c/22cb9348cfa6cf26a030a37e6863ed9eeb9ad610.png)
![\[x=\frac{-\sqrt{3} \pm \sqrt{19}}{2}.\]](http://latex.artofproblemsolving.com/a/8/a/a8ad5d137875681f8510f9368521a27be830b3b8.png)
Since we want the point in the fourth quadrant we only care about the negative case giving us, ![\[y=x^2-4=\left(\frac{-\sqrt{3} - \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.\]](http://latex.artofproblemsolving.com/c/d/8/cd890c011adbb406d395220e2e50c239b92bb9ee.png)
See also
| 2025 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
