Difference between revisions of "2024 AIME I Problems/Problem 9"
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be point on the hyperbola <math>\frac{x^2}{20}- \frac{y^2}{24} = 1</math> such that <math>ABCD</math> is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than <math>BD^2</math> for all such rhombi. | Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be point on the hyperbola <math>\frac{x^2}{20}- \frac{y^2}{24} = 1</math> such that <math>ABCD</math> is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than <math>BD^2</math> for all such rhombi. | ||
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+ | ==Solution== | ||
+ | For any rhombus, the two diagonals bisect each other and are perpendicular. The first condition is satisfied because of the hyperbola's symmetry about the origin. To satisfy the second one, we set <math>BD</math> as the line <math>y = mx</math> and <math>AC</math> as <math>y = -\frac{1}{m}x.</math> Because the hyperbola has asymptotes of slopes <math>\pm \frac{\sqrt6}{\sqrt5},</math> we have <math>-\frac{\sqrt6}{\sqrt5} < m, -\frac{1]{m} < \frac{\sqrt6}{\sqrt5}.</math> This gives us <math>\frac{5}{6} < m^2 < \frac{6}{5}.</math> | ||
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+ | Plugging <math>y = mx</math> into the equation for the hyperbola yields <math>x^2 = \frac{120}{6-5m^2}</math> and <math>y^2 = \frac{120m^2}{6-5m^2}.</math> By symmetry, we know that <math>(\frac{BD}{2})^2 = x^2 + y^2,</math> so we wish to find a lower bound for <math>x^2 + y^2 = 120(\frac{1+m^2}{6-5m^2}).</math> This is equivalent to minimizing <math>\frac{1+m^2}{6-5m^2} = -\frac{1}{5} + \frac{11}{5(6-5m^2)}</math> within the bounds we have for <math>m^2.</math> It's then easy to see that this expression increases with <math>m^2,</math> so we plug in <math>m^2 = \frac{5}{6}</math> to get <math>x^2+y^2 > 120,</math> so <math>BD^2 > \boxed{480}.</math> | ||
==See also== | ==See also== |
Revision as of 17:52, 2 February 2024
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
For any rhombus, the two diagonals bisect each other and are perpendicular. The first condition is satisfied because of the hyperbola's symmetry about the origin. To satisfy the second one, we set as the line and as Because the hyperbola has asymptotes of slopes we have $-\frac{\sqrt6}{\sqrt5} < m, -\frac{1]{m} < \frac{\sqrt6}{\sqrt5}.$ (Error compiling LaTeX. Unknown error_msg) This gives us
Plugging into the equation for the hyperbola yields and By symmetry, we know that so we wish to find a lower bound for This is equivalent to minimizing within the bounds we have for It's then easy to see that this expression increases with so we plug in to get so
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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