Difference between revisions of "2014 AIME I Problems/Problem 7"
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== Problem 7 == | == Problem 7 == | ||
− | Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane | + | Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane) |
− | == Solution == | + | == Solution 1 == |
− | Let <math>w = \ | + | Let <math>w = \operatorname{cis}{(\alpha)}</math> and <math>z = 10\operatorname{cis}{(\beta)}</math>. Then, <math>\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\operatorname{cis}{\beta}}</math>. |
− | Multiplying both the numerator and denominator of this fraction by <math>\ | + | Multiplying both the numerator and denominator of this fraction by <math>\operatorname{cis}{(-\beta)}</math> gives us: |
− | <math>\dfrac{w - z}{z} = \dfrac{1}{10}\ | + | <math>\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</math>. |
+ | |||
+ | 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> | ||
+ | |||
+ | 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 [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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Without 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 equivalent 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> | ||
+ | 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: | ||
+ | <math>1^2=10^2+t^2-t*10*2\cos\theta</math> | ||
+ | rearranging: | ||
+ | <cmath>20t\cos\theta=t^2+99</cmath> | ||
+ | solving for <math>\cos\theta</math> we get: | ||
+ | |||
+ | <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> | ||
+ | , using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2\left(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}}\right)=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}</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>. | ||
+ | |||
+ | Note : You can also realize that the max <math>\theta</math> is when the line from <math>0</math> is tangent to the circle of radius <math>1</math> centered at <math>10.</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Note that <math>\frac{w-z}{z}=\frac{w}{z}-1</math>, and that <math>\left|\frac{w}{z}\right|=\frac{1}{10}</math>. Thus <math>\frac{w}{z}-1</math> is a complex number on the circle with radius <math>\frac{1}{10}</math> and centered at <math>-1</math> on the complex plane. Let <math>\omega</math> denote this circle. | ||
+ | |||
+ | Let <math>A</math> and <math>C</math> be the points that represent <math>\frac{w}{z}-1</math> and <math>-1</math> respectively on the complex plane. Let <math>O</math> be the origin. In order to maximize <math>\tan^2(\theta)</math>, we need to maximize <math>\angle{AOC}</math>. This angle is maximized when <math>AO</math> is tangent to <math>\omega</math>. Using the Pythagorean Theorem, we get | ||
+ | |||
+ | <cmath>AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}</cmath> | ||
+ | |||
+ | Thus | ||
+ | |||
+ | <cmath>\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}</cmath> | ||
+ | |||
+ | And the answer is <math>1+99=\boxed{100}</math>. | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | |||
+ | https://www.youtube.com/watch?v=yakhEuPy6Sg | ||
+ | |||
+ | ~sugar_rush | ||
== 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}} | ||
+ | |||
+ | [[Category:Complex numbers]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:25, 10 October 2023
Contents
Problem 7
Let and
be complex numbers such that
and
. Let
. The maximum possible value of
can be written as
, where
and
are relatively prime positive integers. Find
. (Note that
, for
, denotes the measure of the angle that the ray from
to
makes with the positive real axis in the complex plane)
Solution 1
Let and
. Then,
.
Multiplying both the numerator and denominator of this fraction by gives us:
.
We know that is equal to the imaginary part of the above expression divided by the real part. Let
. Then, we have that:
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,
Thus, we see that the maximum occurs when . Therefore,
, and
. Thus, the maximum value of
is
, or
, and our answer is
.
Solution 2
Without the loss of generality one can let lie on the positive x axis and since
is a measure of the angle if
then
and we can see that the question is equivalent to having a triangle
with sides
and
and trying to maximize the angle
using the Law of Cosines we get:
rearranging:
solving for
we get:
if we want to maximize
we need to minimize
, using AM-GM inequality we get that the minimum value for
hence using the identity
we get
and our answer is
.
Note : You can also realize that the max is when the line from
is tangent to the circle of radius
centered at
Solution 3
Note that , and that
. Thus
is a complex number on the circle with radius
and centered at
on the complex plane. Let
denote this circle.
Let and
be the points that represent
and
respectively on the complex plane. Let
be the origin. In order to maximize
, we need to maximize
. This angle is maximized when
is tangent to
. Using the Pythagorean Theorem, we get
Thus
And the answer is .
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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