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Difference between revisions of "2024 AIME I Problems/Problem 9"
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==Problem==
 
==Problem==
  
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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.
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Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> be points 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==
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==Solution 1==
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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>
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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 <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>m^2 \in \left(\frac{5}{6}, \frac{6}{5}\right).</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 of the hyperbola, 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>. 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> giving <math>BD^2 > \boxed{480}.</math>
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==Solution 2==
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Assume that <math>AC</math> is the asymptote of the hyperbola, in which case <math>BD</math> is minimized. The expression of <math>BD</math> is <math>y=-\sqrt{\frac{5}{6}}x</math>. Thus, we could get <math>\frac{x^2}{20}-\frac{y^2}{24}=1\implies x^2=\frac{720}{11}</math>. The desired value is <math>4\cdot \frac{11}{6}x^2=480</math>. This case can't be achieved, so all <math>BD^2</math> would be greater than <math>\boxed{480}</math>
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~Bluesoul
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==Video Solution==
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https://youtu.be/9Fxz50ZMk1E?si=O2y5t0VXAAfPPTbv
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 +
~MathProblemSolvingSkills.com
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 +
 
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==Video Solution==
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by OmegaLearn.org https://youtu.be/Ex-IGnoAS48
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/HsTmPBPd6N4
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
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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>\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==

Latest revision as of 13:01, 24 December 2024

Problem

Let $A$, $B$, $C$, and $D$ be points 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 1

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 $BD$ 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 $m^2 \in \left(\frac{5}{6}, \frac{6}{5}\right).$


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 of the hyperbola, 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 $\frac{1+m^2}{6-5m^2} = -\frac{1}{5} + \frac{11}{5(6-5m^2)}$. 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 $BD^2 > \boxed{480}.$

Solution 2

Assume that $AC$ is the asymptote of the hyperbola, in which case $BD$ is minimized. The expression of $BD$ is $y=-\sqrt{\frac{5}{6}}x$. Thus, we could get $\frac{x^2}{20}-\frac{y^2}{24}=1\implies x^2=\frac{720}{11}$. The desired value is $4\cdot \frac{11}{6}x^2=480$. This case can't be achieved, so all $BD^2$ would be greater than $\boxed{480}$

~Bluesoul

Video Solution

https://youtu.be/9Fxz50ZMk1E?si=O2y5t0VXAAfPPTbv

~MathProblemSolvingSkills.com


Video Solution

by OmegaLearn.org https://youtu.be/Ex-IGnoAS48

Video Solution

https://youtu.be/HsTmPBPd6N4

~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
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All AIME Problems and Solutions

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