Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Fixed link)
 
(32 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 +
__TOC__
 +
 
== Problem ==
 
== Problem ==
 
Let <math>a</math> and <math>b</math> be positive real numbers with <math>a\ge b</math>. Let <math>\rho</math> be the maximum possible value of <math>\frac {a}{b}</math> for which the system of equations
 
Let <math>a</math> and <math>b</math> be positive real numbers with <math>a\ge b</math>. Let <math>\rho</math> be the maximum possible value of <math>\frac {a}{b}</math> for which the system of equations
Line 6: Line 8:
 
has a solution in <math>(x,y)</math> satisfying <math>0\le x < a</math> and <math>0\le y < b</math>. Then <math>\rho^2</math> can be expressed as a fraction <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
 
has a solution in <math>(x,y)</math> satisfying <math>0\le x < a</math> and <math>0\le y < b</math>. Then <math>\rho^2</math> can be expressed as a fraction <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
−
__TOC__
+
== Solutions ==
−
== Solution ==
 
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Notice that the given equation implies
 
Notice that the given equation implies
Line 13: Line 14:
 
We have <math>2by \ge y^2</math>, so <math>2ax \le a^2 \implies x \le \frac {a}{2}</math>.
 
We have <math>2by \ge y^2</math>, so <math>2ax \le a^2 \implies x \le \frac {a}{2}</math>.
  
−
Then, notice <math>b^2 + x^2 = a^2 + y^2 \le a^2</math>, so <math>b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \ge \frac {4}{3}</math>.
+
Then, notice <math>b^2 + x^2 = a^2 + y^2 \ge a^2</math>, so <math>b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}</math>.
  
 
The solution <math>(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)</math> satisfies the equation, so <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3 + 4 = \boxed{007}</math>.
 
The solution <math>(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)</math> satisfies the equation, so <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3 + 4 = \boxed{007}</math>.
Line 22: Line 23:
 
Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>.
 
Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>.
  
−
A quick [[differentiation]] shows that <math>f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0</math>, so the maximum is at the endpoint <math>\theta = \frac {\pi}{6}</math>. We then get
+
Taking the [[derivative]] shows that <math>-f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0</math>, so the maximum is at the endpoint <math>\theta = 0</math>. We then get
−
<center><math>\rho = \frac {\cos{\frac {\pi}{6}}}{\sin{\frac {\pi}{2}}} = \frac {\sqrt {3}}{2}</math></center>
+
<center><math>\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}</math></center>
−
Then, <math>\rho^2 = \frac {3}{4}</math>, and the answer is <math>3+4=\boxed{007}</math>.
+
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}k}{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 ===
 
=== Solution 3 ===
 
Consider a [[cyclic quadrilateral]] <math>ABCD</math> with  
 
Consider a [[cyclic quadrilateral]] <math>ABCD</math> with  
−
<math>\angle B = \angle D = 90</math>, and <math>AB = y, BC = a, CD = b, AD = x</math>. Then
+
<math>\angle B = \angle D = 90^{\circ}</math>, and <math>AB = y, BC = a, CD = b, AD = x</math>. Then
 
<cmath>AC^2 = a^2 + y^2 = b^2 + x^2</cmath>
 
<cmath>AC^2 = a^2 + y^2 = b^2 + x^2</cmath>
 
From [[Ptolemy's Theorem]], <math>ax + by = AC(BD)</math>, so
 
From [[Ptolemy's Theorem]], <math>ax + by = AC(BD)</math>, so
Line 34: Line 46:
 
Simplifying, we have <math>BD = AC/2</math>.  
 
Simplifying, we have <math>BD = AC/2</math>.  
  
−
Note the [[circumcircle]] of <math>ABCD</math> has [[radius]] <math>r = AC/2</math>, so <math>BD = r</math> and has an arc of <math>60</math> degrees, so  
+
Note the [[circumcircle]] of <math>ABCD</math> has [[radius]] <math>r = AC/2</math>, so <math>BD = r</math> and has an arc of <math>60^{\circ}</math>, so  
−
<math>\angle C = 30</math>. Let <math>\angle BDC = \theta</math>.
+
<math>\angle C = 30^{\circ}</math>. Let <math>\angle BDC = \theta</math>.
 +
 
 +
<math>\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150^{\circ} - \theta)}</math>, where both <math>\theta</math> and <math>150^{\circ} - \theta</math> are <math>\leq 90^{\circ}</math> since triangle <math>BCD</math> must be [[acute triangle|acute]]. Since <math>\sin</math> is an increasing function over <math>(0, 90^{\circ})</math>, <math>\frac{\sin \theta}{\sin(150^{\circ} - \theta)}</math> is also increasing function over <math>(60^{\circ}, 90^{\circ})</math>.
 +
 
 +
<math>\frac ab</math> maximizes at <math>\theta = 90^{\circ} \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 +
 +
3 = \boxed{007}</math>.
 +
 
 +
=== Note: ===
 +
None of the above solutions point out clearly the importance of the restriction that <math>a</math>, <math>b</math>, <math>x</math> and <math>y</math> 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 to 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 both 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 \leq \frac{4}{3}</math>
 +
<math>\boxed{007}</math>
 +
 
 +
=== Solution 5 ===
 +
The given system is equivalent to the points <math>(a,y)</math> and <math>(x,b)</math> forming an equilateral triangle with the origin. WLOG let this triangle have side length <math>1</math>, so <math>x=\sqrt{1-a^2}</math>. The condition <math>x<a</math> implies <math>(x,b)</math> lies to the left of <math>(a,y)</math>, so <math>(x,b)</math> is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that <math>b = \frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}</math>, so <math>\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}}</math>. Minimizing this is equivalent to minimizing the denominator, which happens when <math>\sqrt{1-a^2} = 0</math> and thus <math>a=1</math>, resulting in <math>\rho = \frac{2}{\sqrt{3}}</math>, so <math>\rho^2 = \frac{4}{3}</math> and the answer is <math>\boxed{007}</math>.
  
−
<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>.
+
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.
  
−
<math>\frac ab</math> maximizes at <math>\theta = 90 \Longrightarrow \frac ab</math> maximizes at <math>\frac 2{\sqrt {3}}</math>.
+
=== Solution 6 (Geometry and Trigonometry) ===
 +
Notice that by Pythagorean theorem, if we take a triangle with vertices <math>(0,0),</math> <math>(a,y),</math> and <math>(x,b)</math> forming an equilateral triangle. Now, take a rectangle with vertices <math>(0,0), (a,0), (0,b), (a,b).</math> Notice that <math>(a,y)</math> and <math>(x,b)</math> are on the sides. Let <math>\alpha</math> be the angle formed by the points <math>(0,b), (0,0), (x,b).</math> Then, we have that <cmath>\cos \alpha = \frac{b}{s},</cmath> where <math>s</math> is the side of the equilateral triangle. Also, we have that <math>30^{\circ}-\alpha</math> is the angle formed by the points <math>(a,0), (0,0), (a,y),</math> and so <cmath>\cos (30^{\circ}-\alpha) = \frac{a}{s}.</cmath> Thus, we have that
 +
<cmath>\frac{a}{b} = \frac{\cos (30^{\circ}-\alpha)}{\cos \alpha}.</cmath> We see that this expression is maximized when <math>\alpha</math> is maximized (at least when <math>\alpha</math> is in the interval <math>(0,90^{\circ}),</math> which it is). Then, <math>\alpha \ge 30^{\circ},</math> so ew have that the maximum of <math>\frac{a}{b}</math> is <cmath>\frac{\cos 0}{\cos 30^{\circ}} = \frac{2}{\sqrt{3}},</cmath> and so our answer is <math>4+3 = 7.</math>
  
 
== See also ==
 
== See also ==
Line 45: Line 78:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:56, 29 January 2024

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.

Solution 6 (Geometry and Trigonometry)

Notice that by Pythagorean theorem, if we take a triangle with vertices $(0,0),$ $(a,y),$ and $(x,b)$ forming an equilateral triangle. Now, take a rectangle with vertices $(0,0), (a,0), (0,b), (a,b).$ Notice that $(a,y)$ and $(x,b)$ are on the sides. Let $\alpha$ be the angle formed by the points $(0,b), (0,0), (x,b).$ Then, we have that \[\cos \alpha = \frac{b}{s},\] where $s$ is the side of the equilateral triangle. Also, we have that $30^{\circ}-\alpha$ is the angle formed by the points $(a,0), (0,0), (a,y),$ and so \[\cos (30^{\circ}-\alpha) = \frac{a}{s}.\] Thus, we have that \[\frac{a}{b} = \frac{\cos (30^{\circ}-\alpha)}{\cos \alpha}.\] We see that this expression is maximized when $\alpha$ is maximized (at least when $\alpha$ is in the interval $(0,90^{\circ}),$ which it is). Then, $\alpha \ge 30^{\circ},$ so ew have that the maximum of $\frac{a}{b}$ is \[\frac{\cos 0}{\cos 30^{\circ}} = \frac{2}{\sqrt{3}},\] and so our answer is $4+3 = 7.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_14&oldid=213739"