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 "2014 AIME I Problems/Problem 7"

(→‎Solution)
(→‎Solution)
Line 18: Line 18:
  
 
Thus, we see that the maximum occurs when <math>\mathrm{cos}{x} = \dfrac{1}{10}</math>. Therefore, <math>\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\mathrm{tan^2}{\theta}</math> is <math>\dfrac{99}{99^2}</math>, or <math>\dfrac{1}{99}</math>, and our answer is <math>1 + 99 = \boxed{100}</math>.
 
Thus, we see that the maximum occurs when <math>\mathrm{cos}{x} = \dfrac{1}{10}</math>. Therefore, <math>\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\mathrm{tan^2}{\theta}</math> is <math>\dfrac{99}{99^2}</math>, or <math>\dfrac{1}{99}</math>, and our answer is <math>1 + 99 = \boxed{100}</math>.
 +
<asy>
 +
pair O = (0,0);
 +
pair A = (50,0);
 +
pair B = (49,sqrt(99));
 +
pair D = (sqrt(850),sqrt(850));
 +
draw(A--B--O--cycle);
 +
dotfactor = 3;
 +
dot("$A$",A,dir(135));
 +
dot("$B$",B,dir(215));
 +
dot("$C$",C,dir(305));
 +
dot("$D$",D,dir(45));
 +
pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));
 +
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
 +
dot("$H$",H,dir(90));
 +
dot("$F$",F,dir(270));
 +
draw(H--F);
 +
pair E = (0,(sqrt(850)-6)/2);
 +
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);
 +
dot("$E$",E,dir(180));
 +
dot("$G$",G,dir(0));
 +
draw(E--G);
 +
pair P = extension(H,F,E,G);
 +
dot("$P$",P,dir(60));
 +
label("$w$", intersectionpoint( A--P, E--H ));
 +
label("$x$", intersectionpoint( B--P, E--F ));
 +
label("$y$", intersectionpoint( C--P, G--F ));
 +
label("$z$", intersectionpoint( D--P, G--H ));</asy>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2014|n=I|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:25, 26 March 2014

Problem 7

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$. Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. (Note that $\arg(w)$, for $w \neq 0$, denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.

Solution

Let $w = \mathrm{cis}{(\alpha)}$ and $z = 10\mathrm{cis}{(\beta)}$. Then, $\dfrac{w - z}{z} = \dfrac{\mathrm{cis}{(\alpha)} - 10\mathrm{cis}{(\beta)}}{10\mathrm{cis}{\beta}}$.

Multiplying both the numerator and denominator of this fraction by $\mathrm{cis}{(-\beta)}$ gives us:

$\dfrac{w - z}{z} = \dfrac{1}{10}\mathrm{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\mathrm{cos}{(\alpha - \beta)} + \dfrac{1}{10}i\mathrm{sin}{(\alpha - \beta)} - 1$.

We know that $\mathrm{tan}{\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \alpha - \beta$. Then, we have that:

$\mathrm{tan}{\theta} = \dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10}.$

We need to find a maximum of this expression, so we take the derivative:

$\dfrac{d}{dx} \left (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10} \right) = \dfrac{1 - 10\mathrm{cos}{x}}{(\mathrm{cos}{x} - 10)^2}$

Thus, we see that the maximum occurs when $\mathrm{cos}{x} = \dfrac{1}{10}$. Therefore, $\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}$, and $\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}$. Thus, the maximum value of $\mathrm{tan^2}{\theta}$ is $\dfrac{99}{99^2}$, or $\dfrac{1}{99}$, and our answer is $1 + 99 = \boxed{100}$. [asy] pair O = (0,0); pair A = (50,0); pair B = (49,sqrt(99)); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

See also

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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=2014_AIME_I_Problems/Problem_7&oldid=61180"