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

Revision as of 19:59, 23 February 2018

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 $0 \le \theta \le \frac {\pi}{6}$ .

Taking the derivative 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 $\theta = 0$ . We then get

$\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}$

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

(For a non-calculus way to maximize the function above:

Let us work with degrees. Let $f(x)=\frac{\cos x}{\sin(x+60)}$ . We need to maximize $f$ on $[0,30]$ .

Suppose $k$ is an upper bound of $f$ on this range; in other words, assume $f(x)\le k$ for all $x$ in this range. Then: $\cos x\le k\sin(x+60)=k\cdot\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)$ $\rightarrow 0\le \left(\frac{\sqrt{3}}{2}-1\right)\cos x+\frac{k}{2}\sin x\rightarrow 0\le (\sqrt{3}k-2)\cos x+k\sin x$ $\rightarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,$ for all $x$ in $[0,30]$ . In particular, for $x=0$ , $\frac{2-\sqrt{3}k}{k}$ must be less than or equal to $0$ , so $k\ge \frac{2}{\sqrt{3}}$ .

The least possible upper bound of $f$ on this interval is $k=\frac{2}{\sqrt{3}}$ . This inequality must hold by the above logic, and in fact, the inequality reaches equality when $x=0$ . Thus, $f(x)$ attains a maximum of $\frac{2}{\sqrt{3}}$ on the interval.)

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 25: / Line 25: @@
 <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}}{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 ===

2008 AIME II (Problems • Answer Key • Resources)
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