2024 AIME I Problems/Problem 9
Contents
Problem
Let ,
,
, and
be point on the hyperbola
such that
is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than
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 as the line
and
as
Because the hyperbola has asymptotes of slopes
we have
This gives us
Plugging into the equation for the hyperbola yields
and
By symmetry of the hyperbola, we know that
so we wish to find a lower bound for
This is equivalent to minimizing
. It's then easy to see that this expression increases with
so we plug in
to get
giving
Solution 2
Assume is the asymptope of the hyperbola,
in that case is the smallest. The expression of
is
. Thus, we could get
. The desired value is
. This case wouldn't achieve, so all
would be greater than
~Bluesoul
Solution 3 (ultimate desperation (and wrong) \b REALLY REALLY BAD \b)
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 and D to have coordinates
. This means that
satisfies the equations
. This means that the distance from
to
is
. So
. 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
Video Solution
~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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