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

Revision as of 12:27, 4 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

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 $AC$ is the asymptope of the hyperbola, $BD$ in that case is the smallest. 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 wouldn't achieve, so all $BD^2$ would be greater than $\boxed{480}$

~Bluesoul

Solution 3 (ultimate desperation (and wrong))

$\textbf{warning: this solution is wrong}$

The pythagorean theorem in the last step is missing a factor of 2 - this was a lucky "solve".

A square is a rhombus. Take B to have coordinates $(x,x)$ and D to have coordinates $(-x,-x)$ . This means that $x$ satisfies the equations $\frac{x^2}{20}-\frac{x^2}{24}=1 \rightarrow x^2=120$ . This means that the distance from $B$ to $D$ is $\sqrt{2x^2+2x^2}\rightarrow 2x = \sqrt{480}$ . So $BD^2 = \boxed{480}$ . We use a square because it minimizes the length of the long diagonal (also because it's really easy). ~amcrunner

Video Solution 1 by OmegaLearn.org

https://youtu.be/Ex-IGnoAS48

Video Solution

https://youtu.be/HsTmPBPd6N4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 14: / Line 14: @@
 ~Bluesoul
-==Solution 3 (ultimate desperation)==
+==Solution 3 (ultimate desperation (and wrong))==
+<math>\textbf{warning: this solution is wrong}</math>
+The pythagorean theorem in the last step is missing a factor of 2 - this was a lucky "solve".
 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).
 ~amcrunner

2024 AIME I (Problems • Answer Key • Resources)
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