2008 AIME II Problems/Problem 14

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 $m + n$ .

Solutions

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: $\cos x\le k\sin(x+60)=k\cdot\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)$ $\rightarrow 0\le \left(\frac{\sqrt{3}k}{2}-1\right)\cos x+\frac{k}{2}\sin x\rightarrow 0\le (\sqrt{3}k-2)\cos x+k\sin x$ $\rightarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,$ 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 $\angle B = \angle D = 90^{\circ}$ , 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 $60^{\circ}$ , so $\angle C = 30^{\circ}$ . Let $\angle BDC = \theta$ .

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

$\frac ab$ maximizes at $\theta = 90^{\circ} \Longrightarrow \frac ab$ maximizes at $\frac 2{\sqrt {3}}$ . This squared is $(\frac 2{\sqrt {3}})^2 = \frac4{3}$ , and $4 + 3 = \boxed{007}$ .

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 $-15= \theta$ . This yields $p = (1 + \sqrt{3})/2 > 4/3$

Solution 4

The problem is looking for an intersection in the said range between parabola $P$ : $y = \tfrac{(x-a)^2 + b^2-a^2}{2b}$ and the hyperbola $H$ : $y^2 = x^2 + b^2 - a^2$ . The vertex of $P$ is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the $H$ , which is $\sqrt{a^2 - b^2}$ . So for the intersection to exist with $x<a$ and $y \geq 0$ , $P$ needs to cross x-axis between $\sqrt{a^2 - b^2}$ , and $a$ , meaning, $(\sqrt{a^2 - b^2}-a)^2 + b^2-a^2 \geq 0$ Divide both side by $b^2$ , $(\sqrt{\rho^2 - 1}-\rho)^2 + 1-\rho^2 \geq 0$ which can be easily solved by moving $1-\rho^2$ to RHS and taking square roots. Final answer $\rho^2 \leq \frac{4}{3}$ $\boxed{007}$

Solution 5

The given system is equivalent to the points $(a,y)$ and $(x,b)$ forming an equilateral triangle with the origin. WLOG let this triangle have side length $1$ , so $x=\sqrt{1-a^2}$ . The condition $x<a$ implies $(x,b)$ lies to the left of $(a,y)$ , so $(x,b)$ is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that $b = \frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}$ , so $\frac{a}{b} = \frac{a}{\frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}} = \frac{1}{\frac{\sqrt{3}}{2} + \frac{1}{2a}\sqrt{1-a^2}}$ . Minimizing this is equivalent to minimizing the denominator, which happens when $\sqrt{1-a^2} = 0$ and thus $a=1$ , resulting in $\rho = \frac{2}{\sqrt{3}}$ , so $\rho^2 = \frac{4}{3}$ and the answer is $\boxed{007}$ .

As a remark, expressing the condition that the triangle is equilateral purely algebraically instead of using trig eliminates the need for calculus or analyzing the behavior of sine.

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