Difference between revisions of "2024 AIME I Problems/Problem 9"

m (See also)
Line 2: Line 2:
  
 
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.
 +
 +
==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>
 +
 +
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 $A$, $B$, $C$, 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

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 $BD$ 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 $-\frac{\sqrt6}{\sqrt5} < m, -\frac{1]{m} < \frac{\sqrt6}{\sqrt5}.$ (Error compiling LaTeX. Unknown error_msg) This gives us $\frac{5}{6} < m^2 < \frac{6}{5}.$

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 $(\frac{BD}{2})^2 = x^2 + y^2,$ so we wish to find a lower bound for $x^2 + y^2 = 120(\frac{1+m^2}{6-5m^2}).$ 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 $x^2+y^2 > 120,$ so $BD^2 > \boxed{480}.$

See also

2024 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png