Difference between revisions of "2008 AIME II Problems/Problem 14"

Revision as of 23:26, 23 October 2011

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 $a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2$ 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 $m + n$ .

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 $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}$ .

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 $AC^2 = a^2 + y^2 = b^2 + x^2$ From Ptolemy's Theorem, $ax + by = AC(BD)$ , so $AC^2 = (a - x)^2 + (b - y)^2 = a^2 + y^2 + b^2 + x^2 - 2(ax + by) = 2AC^2 - 2AC*BD$ 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 $\frac 2{\sqrt {3}}$ . This squared is $(\frac 2{\sqrt {3}})^2 = \frac4{3}$ , and $4 + 3 = \boxed{007}$ .

@@ 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</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 +
+= \boxed{007}</math>.
 == See also ==

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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