Difference between revisions of "2024 AIME I Problems/Problem 9"
Mathkiddie (talk | contribs) (→Solution 4 (super ultimate desperation (and completely wrong))) |
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==Solution 4 (super ultimate desperation (and completely wrong))== | ==Solution 4 (super ultimate desperation (and completely wrong))== | ||
− | The only "numbers" provided in this problem are <math>24</math> and <math>20</math>, so the answer must be a combination of some operations on these numbers. | + | The only "numbers" provided in this problem are <math>24</math> and <math>20</math>, so the answer must be a combination of some operations on these numbers. If you're lucky, you could figure the most likely option is <math>24\cdot 20</math>, as this yields <math>\boxed{480}</math> and seems like a plausible answer for this question. |
~Mathkiddie | ~Mathkiddie |
Revision as of 16:13, 11 February 2024
Contents
[hide]Problem
Let , , , and be point on the hyperbola such that is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than for all such rhombi.
Solution
A quadrilateral is a rhombus if and only if its two diagonals bisect each other and are perpendicular to each other. The first condition is automatically satisfied because of the hyperbola's symmetry about the origin. To satisfy the second condition, we set as the line and as Because the hyperbola has asymptotes of slopes we have This gives us
Plugging into the equation for the hyperbola yields and By symmetry of the hyperbola, we know that so we wish to find a lower bound for This is equivalent to minimizing . It's then easy to see that this expression increases with so we plug in to get giving
Solution 2
Assume is the asymptope of the hyperbola, in that case is the smallest. The expression of is . Thus, we could get . The desired value is . This case wouldn't achieve, so all would be greater than
~Bluesoul
Solution 3 (ultimate desperation (and wrong) REALLY REALLY BAD )
The pythagorean theorem in the last step is missing a factor of 2 - this was a lucky "solve".
A square is a rhombus. Take B to have coordinates and D to have coordinates . This means that satisfies the equations . This means that the distance from to is . So . We use a square because it minimizes the length of the long diagonal (also because it's really easy). ~amcrunner
Solution 4 (super ultimate desperation (and completely wrong))
The only "numbers" provided in this problem are and , so the answer must be a combination of some operations on these numbers. If you're lucky, you could figure the most likely option is , as this yields and seems like a plausible answer for this question.
~Mathkiddie
Video Solution 1 by OmegaLearn.org
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 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 |
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