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

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We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divided by the real part. Let <math>x = \alpha - \beta</math>. Then, we have that:
 
We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divided by the real part. Let <math>x = \alpha - \beta</math>. Then, we have that:
  
−
<math>\tan{\theta} = \dfrac{\sin{x}}{cos{x} - 10}.</math>
+
<math>\tan{\theta} = \dfrac{\sin{x}}{\cos{x} - 10}.</math>
  
 
We need to find a maximum of this expression, so we take the derivative:
 
We need to find a maximum of this expression, so we take the derivative:
  
−
<math>\dfrac{d}{dx} \left (\dfrac{\sin{x}}{\cos{x} - 10} \right) = \dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}</math>
+
Note (not from author): To take the derivative, we need to use the [https://en.wikipedia.org/wiki/Quotient_rule Quotient Rule]. In this case, <cmath>\frac{d}{dx}\left(\frac{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}</cmath>
  
 
Thus, we see that the maximum occurs when <math>\cos{x} = \dfrac{1}{10}</math>. Therefore, <math>\sin{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\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>\cos{x} = \dfrac{1}{10}</math>. Therefore, <math>\sin{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\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>.

Revision as of 12:15, 21 August 2021

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 1

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

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

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

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

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

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

Note (not from author): To take the derivative, we need to use the Quotient Rule. In this case, \[\frac{d}{dx}\left(\frac{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}\]

Thus, we see that the maximum occurs when $\cos{x} = \dfrac{1}{10}$. Therefore, $\sin{x} = \pm\dfrac{\sqrt{99}}{10}$, and $\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}$. Thus, the maximum value of $\tan^2{\theta}$ is $\dfrac{99}{99^2}$, or $\dfrac{1}{99}$, and our answer is $1 + 99 = \boxed{100}$.

Solution 2

Without 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 equivalent 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,(7,1.2)); 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 $1 + 99 = \boxed{100}$.

Solution 3

Note that $\frac{w-z}{z}=\frac{w}{z}-1$, and that $\left|\frac{w}{z}\right|=\frac{1}{10}$. Thus $\frac{w}{z}-1$ is a complex number on the circle with radius $\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\omega$ denote this circle.

Let $A$ and $C$ be the points that represent $\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In order to maximize $\tan^2(\theta)$, we need to maximize $\angle{AOC}$. This angle is maximized when $AO$ is tangent to $\omega$. Using the Pythagorean Theorem, we get

\[AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}\]

Thus

\[\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}\]

And the answer is $1+99=\boxed{100}$.

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=yakhEuPy6Sg

~sugar_rush

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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