Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AIME I Problems/Problem 9"
(Incorrect solutions can only misguide the reader. It provides no useful content for improving upon mistakes, but rather encourages it.)
 
(22 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
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 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.
 +
 
 +
==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>
 +
 
 +
 
 +
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==
 +
 
 +
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
 +
 
 +
==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)
  
−
==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 <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 (\frac{5}{6}, \frac{6}{5}).</math>
 
−
\newline
 
−
\newline
 
−
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> giving <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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_9&oldid=237730"