Difference between revisions of "2024 AIME I Problems/Problem 9"
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==Solution== | ==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} | + | 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>m, \frac{1}{m} \in \left(-\frac{\sqrt6}{\sqrt5}, \frac{\sqrt6}{\sqrt5}\right).</math> This gives us <math>\frac{5}{6} < m^2 < \frac{6}{5}.</math> |
| − | 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>\left(\frac{BD}{2}\right)^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> | + | 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>\left(\frac{BD}{2}\right)^2 = x^2 + y^2,</math> so we wish to find a lower bound for <math>x^2 + y^2 = 120\left(\frac{1+m^2}{6-5m^2}\right).</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 17:54, 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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
