Difference between revisions of "2008 AIME II Problems/Problem 14"
(→Solution 1) |
Danielguo94 (talk | contribs) (→Solution 3) |
||
Line 39: | Line 39: | ||
<math>\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150 - \theta)}</math>, where both <math>\theta</math> and <math>150 - \theta</math> are <math>\leq 90</math> since triangle <math>BCD</math> must be [[acute triangle|acute]]. Since <math>\sin</math> is an increasing function over <math>(0, 90)</math>, <math>\frac{\sin \theta}{\sin(150 - \theta)}</math> is also increasing function over <math>(60, 90)</math>. | <math>\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150 - \theta)}</math>, where both <math>\theta</math> and <math>150 - \theta</math> are <math>\leq 90</math> since triangle <math>BCD</math> must be [[acute triangle|acute]]. Since <math>\sin</math> is an increasing function over <math>(0, 90)</math>, <math>\frac{\sin \theta}{\sin(150 - \theta)}</math> is also increasing function over <math>(60, 90)</math>. | ||
− | <math>\frac ab</math> maximizes at <math>\theta = 90 \Longrightarrow \frac ab</math> maximizes at <math>\frac 2{\sqrt {3}}</math>. | + | <math>\frac ab</math> maximizes at <math>\theta = 90 \Longrightarrow \frac ab</math> maximizes at <math>\frac 2{\sqrt {3}}</math>. This squared is <math>(\frac 2{\sqrt {3}})^2 = \frac4{3}</math>, and <math>4 + |
+ | 3 = \boxed{007}</math>. | ||
== See also == | == See also == |
Revision as of 23:26, 23 October 2011
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
.
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
.
A quick differentiation shows that , so the maximum is at the endpoint
. We then get
![$\rho = \frac {\cos{\frac {\pi}{6}}}{\sin{\frac {\pi}{2}}} = \frac {\sqrt {3}}{2}$](http://latex.artofproblemsolving.com/8/6/2/862c6fbac5788827b818c9f12d0ba3bbdb7ee1cb.png)
Then, , and the answer is
.
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
.
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 |