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Difference between revisions of "2014 AIME I Problems/Problem 7"

(→‎Solution)
(→‎Solution 2 (No calculus))
Line 46: Line 46:
 
<cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath>
 
<cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath>
 
if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math>
 
if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math>
−
, using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t^2}{20t}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}</math>
+
, using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}</math>
 
hence using the identity <math>\tan^2\theta=\sec^2\theta-1</math>
 
hence using the identity <math>\tan^2\theta=\sec^2\theta-1</math>
 
we get <math>\tan^2\theta=\frac{1}{99}</math>and our answer is <math>1 + 99 = \boxed{100}</math>.
 
we get <math>\tan^2\theta=\frac{1}{99}</math>and our answer is <math>1 + 99 = \boxed{100}</math>.

Revision as of 20:02, 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 $1 + 99 = \boxed{100}$.

Solution 2 (No calculus)

with out the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivelent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ [asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(45)); dot("$B$",B,dir(45)); dot("$O$",O,dir(135)); dot("$ \theta$",O,dir(30)); label("$1$", ( A--B )); label("$10$",(O--A)); label("$t$",(O--B)); [/asy]

using the law of cosines we get: $1^2=10^2+t^2-t*10*2\cos\theta$ rearranging: \[20t\cos\theta=t^2+99\] solving for $\cos\theta$ we get:

\[\frac{99}{20t}+\frac{t}{20}=\cos\theta\] if we want to maximize $\theta$ we need to minimize $\cos\theta$ , using AM-GM inequality we get that the minimum value for $\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$ hence using the identity $\tan^2\theta=\sec^2\theta-1$ we get $\tan^2\theta=\frac{1}{99}$and our answer is $1 + 99 = \boxed{100}$.

See also

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions

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