Difference between revisions of "2008 AIME II Problems/Problem 14"
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=== Solution 5 === | === Solution 5 === | ||
− | The given system is equivalent to the points <math>(a, | + | The given system is equivalent to the points <math>(a,y)</math> and <math>(x,b)</math> forming an equilateral triangle with the origin. WLOG let this triangle have side length <math>1</math>, so <math>x=\sqrt{1-a^2}</math>. The condition <math>x<a</math> implies <math>(x,b)</math> lies to the left of <math>(a,y)</math>, so <math>(x,b)</math> is the top coordinate. Now we can compute (by complex numbers, or the sine angle addition identity) that <math>b = \frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}</math>, so <math>\frac{a}{b} = \frac{a}{\frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}} = \frac{1}{\frac{\sqrt{3}}{2} + \frac{1}{2a}\sqrt{1-a^2}}</math>. Minimizing this is equivalent to minimizing the denominator, which happens when <math>\sqrt{1-a^2} = 0</math> and thus <math>a=1</math>, resulting in <math>\rho = \frac{2}{\sqrt{3}}</math>, so <math>\rho^2 = \frac{4}{3}</math> and the answer is <math>\boxed{007}</math>. |
As a remark, expressing the condition that the triangle is equilateral purely algebraically instead of using trig eliminates the need for calculus or analyzing the behavior of sine. | As a remark, expressing the condition that the triangle is equilateral purely algebraically instead of using trig eliminates the need for calculus or analyzing the behavior of sine. |
Revision as of 20:08, 26 December 2023
Contents
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
.
Solutions
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
, 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 ,
,
and
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
Solution 5
The given system is equivalent to the points and
forming an equilateral triangle with the origin. WLOG let this triangle have side length
, so
. The condition
implies
lies to the left of
, so
is the top coordinate. Now we can compute (by complex numbers, or the sine angle addition identity) that
, so
. Minimizing this is equivalent to minimizing the denominator, which happens when
and thus
, resulting in
, so
and the answer is
.
As a remark, expressing the condition that the triangle is equilateral purely algebraically instead of using trig eliminates the need for calculus or analyzing the behavior of sine.
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.