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

(→‎Solution 4 (super ultimate desperation (and completely wrong)))
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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==
 
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>
 
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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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>
  
 
==Solution 2==
 
==Solution 2==
  
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Assume <math>AC</math> is the asymptope of the hyperbola, <math>BD</math> in that case is the smallest. 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 wouldn't achieve, so all <math>BD^2</math> would be greater than <math>\boxed{480}</math>
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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>
  
 
~Bluesoul
 
~Bluesoul
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==Solution 3 (ultimate desperation (and wrong) REALLY REALLY BAD )==
 
  
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<math>\textbf{warning: this solution is wrong}</math>
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==Video Solution==
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https://youtu.be/9Fxz50ZMk1E?si=O2y5t0VXAAfPPTbv
  
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The pythagorean theorem in the last step is missing a factor of 2 - this was a lucky "solve".
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~MathProblemSolvingSkills.com
  
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A square is a rhombus. Take B to have coordinates <math>(x,x)</math> and D to have coordinates <math>(-x,-x)</math>. This means that <math>x</math> satisfies the equations <math>\frac{x^2}{20}-\frac{x^2}{24}=1 \rightarrow x^2=120</math>. This means that the distance from <math>B</math> to <math>D</math> is <math>\sqrt{2x^2+2x^2}\rightarrow 2x = \sqrt{480}</math>. So <math>BD^2 = \boxed{480}</math>. We use a square because it minimizes the length of the long diagonal (also because it's really easy).
 
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~amcrunner
 
  
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==Solution 4 (super ultimate desperation (and completely wrong))==
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==Video Solution==
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The only "numbers" provided in this problem are <math>24</math> and <math>20</math>, so the answer must be a combination of some operations on these numbers. The most likely option is <math>24\cdot 20</math>, as this yields <math>\boxed{480}</math> and seems like a plausible answer for this question.
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by OmegaLearn.org https://youtu.be/Ex-IGnoAS48
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~Mathkiddie
 
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==Video Solution 1 by OmegaLearn.org==
 
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https://youtu.be/Ex-IGnoAS48
 
  
 
==Video Solution==
 
==Video Solution==

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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