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2008 AIME II Problems/Problem 14: Difference between revisions

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Note:
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
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 <math>-15= \theta</math>. This yields <math>p = (1 + \sqrt{3})/2 > 4/3</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 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 two 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 \geq \frac{4}{3}</math>
<math>\boxed{007}</math>


== See also ==
== See also ==

Revision as of 04:52, 28 January 2019

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: \[\cos x\le k\sin(x+60)=k\cdot\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)\] \[\rightarrow 0\le \left(\frac{\sqrt{3}k}{2}-1\right)\cos x+\frac{k}{2}\sin x\rightarrow 0\le (\sqrt{3}k-2)\cos x+k\sin x\] \[\rightarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,\] 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 $\angle B = \angle D = 90$, 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 $4 + 3 = \boxed{007}$.

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 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 two 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 $\rho^2 \geq \frac{4}{3}$ $\boxed{007}$

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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