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2008 AIME II Problems/Problem 14

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Problem

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Contents

1 Problem
2 Solution
3 See also

Solution

Solution 1

Notice that the given equation implies

$a^2 + y^2 = b^2 + x^2 = 2(ax + by)$

We have $2by \ge y^2$ , so $2ax \le a^2 \implies x \le \frac {a}{2}$ .

Then, notice $b^2 + x^2 = a^2 + y^2 \ge a^2$ , so $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \ge \frac {4}{3}$ .

The solution $(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)$ satisfies the equation, so $\rho^2 = \frac {4}{3}$ , and the answer is $3 + 4 = \boxed{007}$ .

Solution 2

Consider the points $(a,y)$ and $(x,b)$ . They form an equilateral triangle with the origin. We let the side length be $1$ , so $a = \cos{\theta}$ and $b = \sin{\left(\theta + \frac {\pi}{3}\right)}$ .

Thus $f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}$ and we need to maximize this for $0 \le \theta \le \frac {\pi}{6}$ .

A quick differentiation shows that $f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0$ , so the maximum is at the endpoint $\theta = \frac {\pi}{6}$ . We then get

$\rho = \frac {\cos{\frac {\pi}{6}}}{\sin{\frac {\pi}{2}}} = \frac {\sqrt {3}}{2}$

Then, $\rho^2 = \frac {3}{4}$ , and the answer is $3+4=\boxed{007}$ .

Solution 3

Consider a cyclic quadrilateral $ABCD$ with $\angle B = \angle D = 90$ , and $AB = y, BC = a, CD = b, AD = x$ . Then From Ptolemy's Theorem, $ax + by = AC(BD)$ , so Simplifying, we have $BD = AC/2$ .

Note the circumcircle of $ABCD$ has radius $r = AC/2$ , so $BD = r$ and has an arc of $60$ degrees, so $\angle C = 30$ . Let $\angle BDC = \theta$ .

$\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150 - \theta)}$ , where both $\theta$ and $150 - \theta$ are $\leq 90$ since triangle $BCD$ must be acute. Since $\sin$ is an increasing function over $(0, 90)$ , $\frac{\sin \theta}{\sin(150 - \theta)}$ is also increasing function over $(60, 90)$ .

$\frac ab$ maximizes at $\theta = 90 \Longrightarrow \frac ab$ maximizes at $\frac 2{\sqrt {3}}$ .

See also

2008 AIME II (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_II_Problems/Problem_14"

Category: Intermediate Algebra Problems

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