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

(→‎Solution)
(→‎Solution)
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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>.
 +
 +
== Solution 2 (No calculus) ==
 +
with out the loss of generality one can let <math>z</math> lie on the positive x axis and since <math>arg(\theta)</math> is a measure of the angle if <math>z=10</math> then <math>arg(\dfrac{w-z}{z})=arg(w-z)</math> and we can see that the question is equivelent to having a triangle <math>OAB</math> with sides <math>OA =10</math> <math>AB=1</math> and <math>OB=t</math> and trying to maximize the angle <math>BOA</math>
 
<asy>
 
<asy>
 
pair O = (0,0);
 
pair O = (0,0);
−
pair A = (50,0);
+
pair A = (100,0);
−
pair B = (49,sqrt(99));
+
pair B = (80,30);
 
pair D = (sqrt(850),sqrt(850));
 
pair D = (sqrt(850),sqrt(850));
 
draw(A--B--O--cycle);
 
draw(A--B--O--cycle);
 
dotfactor = 3;
 
dotfactor = 3;
−
dot("$A$",A,dir(135));
+
dot("$A$",A,dir(45));
−
dot("$B$",B,dir(215));
+
dot("$B$",B,dir(45));
−
dot("$C$",C,dir(305));
+
dot("$O$",O,dir(135));
−
dot("$D$",D,dir(45));
+
dot("$ \theta$",O,dir(30));
−
pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));
+
 
−
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
+
label("$1$", ( A--B ));
−
dot("$H$",H,dir(90));
+
label("$10$",(O--A));
−
dot("$F$",F,dir(270));
+
label("$t$",(O--B));
−
draw(H--F);
+
</asy>
−
pair E = (0,(sqrt(850)-6)/2);
+
 
−
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);
+
using the law of cosines we get:
−
dot("$E$",E,dir(180));
+
<math>1^2=10^2+t^2-t*10*2\cos\theta</math>
−
dot("$G$",G,dir(0));
+
rearranging:
−
draw(E--G);
+
<cmath>20t\cos\theta=t^2+99</cmath>
−
pair P = extension(H,F,E,G);
+
solving for <math>\cos\theta</math> we get:
−
dot("$P$",P,dir(60));
+
 
−
label("$w$", intersectionpoint( A--P, E--H ));
+
<cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath>
−
label("$x$", intersectionpoint( B--P, E--F ));
+
if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math>
−
label("$y$", intersectionpoint( C--P, G--F ));
+
, 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>
−
label("$z$", intersectionpoint( D--P, G--H ));</asy>
+
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>.
  
 
== 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 20:01, 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 $\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t^2}{20t}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$ 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)
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