Difference between revisions of "2008 AIME II Problems/Problem 14"
(→Solution 4) |
m (→Solution 3) |
||
Line 39: | Line 39: | ||
=== Solution 3 === | === Solution 3 === | ||
Consider a [[cyclic quadrilateral]] <math>ABCD</math> with | Consider a [[cyclic quadrilateral]] <math>ABCD</math> with | ||
− | <math>\angle B = \angle D = 90</math>, and <math>AB = y, BC = a, CD = b, AD = x</math>. Then | + | <math>\angle B = \angle D = 90^{\circ}</math>, and <math>AB = y, BC = a, CD = b, AD = x</math>. Then |
<cmath>AC^2 = a^2 + y^2 = b^2 + x^2</cmath> | <cmath>AC^2 = a^2 + y^2 = b^2 + x^2</cmath> | ||
From [[Ptolemy's Theorem]], <math>ax + by = AC(BD)</math>, so | From [[Ptolemy's Theorem]], <math>ax + by = AC(BD)</math>, so | ||
Line 45: | Line 45: | ||
Simplifying, we have <math>BD = AC/2</math>. | Simplifying, we have <math>BD = AC/2</math>. | ||
− | Note the [[circumcircle]] of <math>ABCD</math> has [[radius]] <math>r = AC/2</math>, so <math>BD = r</math> and has an arc of <math>60</math> | + | Note the [[circumcircle]] of <math>ABCD</math> has [[radius]] <math>r = AC/2</math>, so <math>BD = r</math> and has an arc of <math>60^{\circ}</math>, so |
− | <math>\angle C = 30</math>. Let <math>\angle BDC = \theta</math>. | + | <math>\angle C = 30^{\circ}</math>. Let <math>\angle BDC = \theta</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 = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150^{\circ} - \theta)}</math>, where both <math>\theta</math> and <math>150^{\circ} - \theta</math> are <math>\leq 90^{\circ}</math> since triangle <math>BCD</math> must be [[acute triangle|acute]]. Since <math>\sin</math> is an increasing function over <math>(0, 90^{\circ})</math>, <math>\frac{\sin \theta}{\sin(150^{\circ} - \theta)}</math> is also increasing function over <math>(60^{\circ}, 90^{\circ})</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 + | + | <math>\frac ab</math> maximizes at <math>\theta = 90^{\circ} \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>. | 3 = \boxed{007}</math>. | ||
Note: | 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 <math>-15= \theta</math>. This yields <math>p = (1 + \sqrt{3})/2 > 4/3</math> | + | None of the above solutions point out clearly the importance of the restriction that <math>a</math>, <math>b</math>, <math>x</math> and <math>y</math> 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 <math>-15= \theta</math>. This yields <math>p = (1 + \sqrt{3})/2 > 4/3</math> |
=== Solution 4 === | === Solution 4 === |
Revision as of 21:39, 4 June 2020
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
[hide]Solution
Solution 1
Notice that the given equation implies

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

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
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.