Difference between revisions of "2008 AIME II Problems/Problem 14"
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Divide both side by <math>b^2</math>, | Divide both side by <math>b^2</math>, | ||
<cmath> (\sqrt{\rho^2 - 1}-\rho)^2 + 1-\rho^2 \geq 0</cmath> | <cmath> (\sqrt{\rho^2 - 1}-\rho)^2 + 1-\rho^2 \geq 0</cmath> | ||
− | which can be easily solved by moving <math>1-\rho^2</math> to RHS and taking square roots. Final answer <math>\rho^2 \ | + | which can be easily solved by moving <math>1-\rho^2</math> to RHS and taking square roots. Final answer <math>\rho^2 \leq \frac{4}{3}</math> |
<math>\boxed{007}</math> | <math>\boxed{007}</math> | ||
Revision as of 05:54, 28 January 2019
Problem
Let and
be positive real numbers with
. Let
be the maximum possible value of
for which the system of equations
has a solution in
satisfying
and
. Then
can be expressed as a fraction
, where
and
are relatively prime positive integers. Find
.
Contents
Solution
Solution 1
Notice that the given equation implies
![$a^2 + y^2 = b^2 + x^2 = 2(ax + by)$](http://latex.artofproblemsolving.com/1/f/e/1fe34e0d0d9d0690ea7bf57f621161bf4f92ee54.png)
We have , so
.
Then, notice , so
.
The solution satisfies the equation, so
, and the answer is
.
Solution 2
Consider the points and
. They form an equilateral triangle with the origin. We let the side length be
, so
and
.
Thus and we need to maximize this for
.
Taking the derivative shows that , so the maximum is at the endpoint
. We then get
![$\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}$](http://latex.artofproblemsolving.com/3/9/5/395b0af817addfad6c00ea4b877ba45e24562973.png)
Then, , and the answer is
.
(For a non-calculus way to maximize the function above:
Let us work with degrees. Let . We need to maximize
on
.
Suppose is an upper bound of
on this range; in other words, assume
for all
in this range. Then:
for all
in
. In particular, for
,
must be less than or equal to
, so
.
The least possible upper bound of on this interval is
. This inequality must hold by the above logic, and in fact, the inequality reaches equality when
. Thus,
attains a maximum of
on the interval.)
Solution 3
Consider a cyclic quadrilateral with
, and
. Then
From Ptolemy's Theorem,
, so
Simplifying, we have
.
Note the circumcircle of has radius
, so
and has an arc of
degrees, so
. Let
.
, where both
and
are
since triangle
must be acute. Since
is an increasing function over
,
is also increasing function over
.
maximizes at
maximizes at
. This squared is
, and
.
Note:
None of the above solutions point out clearly the importance of the restriction that a, b, x and y be positive. Indeed, larger values of p are obtained when the lower vertex of the equilateral triangle in Solution 2 dips below the x-axis. Take for example . This yields
Solution 4
The problem is looking for an intersection in the said range between parabola :
and the hyperbola
:
. The vertex of
is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the
, which is
. So for the intersection to exist with
and
,
needs to cross x-axis between
, and
, meaning,
Divide both side by
,
which can be easily solved by moving
to RHS and taking square roots. Final answer
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.