Difference between revisions of "2008 AIME II Problems/Problem 14"
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− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
Notice that the given equation implies | Notice that the given equation implies | ||
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We have <math>2by \ge y^2</math>, so <math>2ax \le a^2 \implies x \le \frac {a}{2}</math>. | We have <math>2by \ge y^2</math>, so <math>2ax \le a^2 \implies x \le \frac {a}{2}</math>. | ||
− | Then, notice <math>b^2 + x^2 = a^2 + y^2 \ | + | Then, notice <math>b^2 + x^2 = a^2 + y^2 \ge a^2</math>, so <math>b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}</math>. |
The solution <math>(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)</math> satisfies the equation, so <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3 + 4 = \boxed{007}</math>. | The solution <math>(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)</math> satisfies the equation, so <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3 + 4 = \boxed{007}</math>. | ||
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Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>. | Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>. | ||
− | + | Taking the [[derivative]] shows that <math>-f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0</math>, so the maximum is at the endpoint <math>\theta = 0</math>. We then get | |
− | <center><math>\rho = \frac {\cos{ | + | <center><math>\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}</math></center> |
+ | Then, <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3+4=\boxed{007}</math>. | ||
− | + | (For a non-calculus way to maximize the function above: | |
+ | |||
+ | Let us work with degrees. Let <math>f(x)=\frac{\cos x}{\sin(x+60)}</math>. We need to maximize <math>f</math> on <math>[0,30]</math>. | ||
+ | |||
+ | Suppose <math>k</math> is an upper bound of <math>f</math> on this range; in other words, assume <math>f(x)\le k</math> for all <math>x</math> in this range. Then: <cmath>\cos x\le k\sin(x+60)=k\cdot\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)</cmath> | ||
+ | <cmath>\rightarrow 0\le \left(\frac{\sqrt{3}k}{2}-1\right)\cos x+\frac{k}{2}\sin x\rightarrow 0\le (\sqrt{3}k-2)\cos x+k\sin x</cmath> | ||
+ | <cmath>\rightarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,</cmath> | ||
+ | for all <math>x</math> in <math>[0,30]</math>. In particular, for <math>x=0</math>, <math>\frac{2-\sqrt{3}k}{k}</math> must be less than or equal to <math>0</math>, so <math>k\ge \frac{2}{\sqrt{3}}</math>. | ||
+ | |||
+ | The least possible upper bound of <math>f</math> on this interval is <math>k=\frac{2}{\sqrt{3}}</math>. This inequality must hold by the above logic, and in fact, the inequality reaches equality when <math>x=0</math>. Thus, <math>f(x)</math> attains a maximum of <math>\frac{2}{\sqrt{3}}</math> on the interval.) | ||
=== 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 | ||
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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^{\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^{\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>. | ||
− | <math> | + | Note: |
+ | 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> | ||
− | <math>\ | + | === Solution 4 === |
+ | The problem is looking for an intersection in the said range between parabola <math>P</math>: <math>y = \tfrac{(x-a)^2 + b^2-a^2}{2b}</math> and the hyperbola <math>H</math>: <math>y^2 = x^2 + b^2 - a^2</math>. The vertex of <math>P</math> is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the <math>H</math>, which is <math>\sqrt{a^2 - b^2}</math>. So for the intersection to exist with <math>x<a</math> and <math>y \geq 0</math>, <math>P</math> needs to cross x-axis between <math>\sqrt{a^2 - b^2}</math>, and <math>a</math>, meaning, | ||
+ | <cmath> (\sqrt{a^2 - b^2}-a)^2 + b^2-a^2 \geq 0</cmath> | ||
+ | Divide both side by <math>b^2</math>, | ||
+ | <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 \leq \frac{4}{3}</math> | ||
+ | <math>\boxed{007}</math> | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 13:52, 16 August 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
Solutions
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 |
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