2024 AIME I Problems/Problem 9

Problem

Let $A$ , $B$ , $C$ , and $D$ be point on the hyperbola $\frac{x^2}{20}- \frac{y^2}{24} = 1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than $BD^2$ for all such rhombi.

Solution

A quadrilateral is a rhombus if and only if its two diagonals bisect each other and are perpendicular. The first condition is automatically satisfied because of the hyperbola's symmetry about the origin. To satisfy the second condition, we set $BD$ as the line $y = mx$ and $AC$ as $y = -\frac{1}{m}x.$ Because the hyperbola has asymptotes of slopes $\pm \frac{\sqrt6}{\sqrt5},$ we have $m, -\frac{1}{m} \in \left(-\frac{\sqrt6}{\sqrt5}, \frac{\sqrt6}{\sqrt5}\right).$ This gives us $m^2 \in (\frac{5}{6}, \frac{6}{5}).$

Plugging $y = mx$ into the equation for the hyperbola yields $x^2 = \frac{120}{6-5m^2}$ and $y^2 = \frac{120m^2}{6-5m^2}.$ By symmetry, we know that $\left(\frac{BD}{2}\right)^2 = x^2 + y^2,$ so we wish to find a lower bound for $x^2 + y^2 = 120\left(\frac{1+m^2}{6-5m^2}\right).$ This is equivalent to minimizing $\frac{1+m^2}{6-5m^2} = -\frac{1}{5} + \frac{11}{5(6-5m^2)}$ within the bounds we have for $m^2.$ It's then easy to see that this expression increases with $m^2,$ so we plug in $m^2 = \frac{5}{6}$ to get $x^2+y^2 > 120,$ giving $BD^2 > \boxed{480}.$

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
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2024 AIME I Problems/Problem 9

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