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Difference between revisions of "2024 AIME I Problems/Problem 7"
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- spectraldragon8
 
- spectraldragon8
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==Solution 7 (Geometry)==
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Follow solution 2 to get that we want to find the line <math>81a-108b=k</math> tangent to circle <math>a^2+b^2=16</math>. The line turns into <math>a=\frac{k}{81}+\frac{4b}{3}</math>
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Connect the center of the circle to the tangency point and the y-intercept of the line. Let the tangency point be <math>A</math>, the y-intercept be <math>C</math>, and the center be <math>B</math>. Drop the perpendicular from <math>A</math> to <math>BC</math> and call it <math>D</math>. Let <math>AD=3x</math>, <math>DC=4x</math>. Then, <math>BD=\sqrt{AB^2-AD^2}=\sqrt{16-9x^2}</math>. By similar triangles, get that <math>\frac{BD}{AD}=\frac{AD}{DC}</math>, so <math>\frac{\sqrt{16-9x^2}}{3x}=\frac{3x}{4x}</math>. Solve this to get that <math>x=\frac{16}{15}</math>, so <math>BC=\frac{20}{3}</math> and <math>\frac{k}{81}=\frac{20}{3}</math>, so <math>k=\boxed{540}</math>
 
==Video Solution by MOP 2024==
 
==Video Solution by MOP 2024==
 
https://www.youtube.com/watch?v=nH7dUh0HghA
 
https://www.youtube.com/watch?v=nH7dUh0HghA

Revision as of 19:54, 2 February 2024

Problem

Find the largest possible real part of \[(75+117i)z+\frac{96+144i}{z}\]where $z$ is a complex number with $|z|=4$.

Solution 1

Let $z=a+bi$ such that $a^2+b^2=4^2=16$. The expression becomes:

\[(75+117i)(a+bi)+\dfrac{96+144i}{a+bi}.\]

Call this complex number $w$. We simplify this expression.

\begin{align*} w&=(75+117i)(a+bi)+\dfrac{96+144i}{a+bi} \\ &=(75a-117b)+(117a+75b)i+48\left(\dfrac{2+3i}{a+bi}\right) \\ &=(75a-117b)+(116a+75b)i+48\left(\dfrac{(2+3i)(a-bi)}{(a+bi)(a-bi)}\right) \\ &=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{a^2+b^2}\right) \\ &=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{16}\right) \\ &=(75a-117b)+(116a+75b)i+3\left(2a+3b+(3a-2b)i\right) \\ &=(75a-117b)+(116a+75b)i+6a+9b+(9a-6b)i \\ &=(81a-108b)+(125a+69b)i. \\ \end{align*}

We want to maximize $\text{Re}(w)=81a-108b$. We can use elementary calculus for this, but to do so, we must put the expression in terms of one variable. Recall that $a^2+b^2=16$; thus, $b=\pm\sqrt{16-a^2}$. Notice that we have a $-108b$ in the expression; to maximize the expression, we want $b$ to be negative so that $-108b$ is positive and thus contributes more to the expression. We thus let $b=-\sqrt{16-a^2}$. Let $f(a)=81a-108b$. We now know that $f(a)=81a+108\sqrt{16-a^2}$, and can proceed with normal calculus.

\begin{align*} f(a)&=81a+108\sqrt{16-a^2} \\ &=27\left(3a+4\sqrt{16-a^2}\right) \\ f'(a)&=27\left(3a+4\sqrt{16-a^2}\right)' \\ &=27\left(3+4\left(\sqrt{16-a^2}\right)'\right) \\ &=27\left(3+4\left(\dfrac{-2a}{2\sqrt{16-a^2}}\right)\right) \\ &=27\left(3-4\left(\dfrac a{\sqrt{16-a^2}}\right)\right) \\ &=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right). \\ \end{align*}

We want $f'(a)$ to be $0$ to find the maximum.

\begin{align*} 0&=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right) \\ &=3-\dfrac{4a}{\sqrt{16-a^2}} \\ 3&=\dfrac{4a}{\sqrt{16-a^2}} \\ 4a&=3\sqrt{16-a^2} \\ 16a^2&=9\left(16-a^2\right) \\ 16a^2&=144-9a^2 \\ 25a^2&=144 \\ a^2&=\dfrac{144}{25} \\ a&=\dfrac{12}5 \\ &=2.4. \\ \end{align*}

We also find that $b=-\sqrt{16-2.4^2}=-\sqrt{16-5.76}=-\sqrt{10.24}=-3.2$.

Thus, the expression we wanted to maximize becomes $81\cdot2.4-108(-3.2)=81\cdot2.4+108\cdot3.2=\boxed{540}$.

~Technodoggo

Solution 2 (Simple Analytic Geometry)

Same steps as solution one until we get $\text{Re}(w)=81a-108b$. We also know $|z|=4$ or $a^2+b^2=16$. We want to find the line $81a-108b=k$ tangent to circle $a^2+b^2=16$. Using $\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=r$ we can substitute and get $\frac{|81(0)-108(0)-k|}{\sqrt{81^2+108^2}}=4$ \begin{align*} \frac{k}{\sqrt{18225}}&=4 \\\frac{k}{135}&=4 \\k&=\boxed{540} \end{align*}

~BH2019MV0

Solution 3

Follow Solution 1 to get $81a-108b$. We can let $a=4\cos\theta$ and $b=4\sin\theta$ as $|z|=4$, and thus we have $324\cos\theta-432\sin\theta$. Furthermore, we can ignore the negative sign in front of the second term as we are dealing with sine and cosine, so we finally wish to maximize $324\cos\theta+432\sin\theta$ for obviously positive $\cos\theta$ and $\sin\theta$.


Using the previous fact, we can use the Cauchy-Schwarz Inequality to calculate the maximum. By the inequality, we have:

$(324^2+432^2)(\cos^2\theta+\sin^2\theta)\ge(324\cos\theta+432\sin\theta)^2$

$540^2\cdot1\ge(324\cos\theta+432\sin\theta)^2$

$\boxed{540}\ge324\cos\theta+432\sin\theta$

~eevee9406

Solution 4 (Simple Quadratic Discriminant)

Similar to the solutions above, we find that $Re((75+117i)z+\frac{96+144i}{z})=81a-108b=27(3a-4b)$, where $z=a+bi$. To maximize this expression, we must maximize $3a-4b$. Let this value be $x$. Solving for $a$ yields $a=\frac{x+4b}{3}$. From the given information we also know that $a^2+b^2=16$. Substituting $a$ in terms of $x$ and $b$ gives us $\frac{x^2+8bx+16b^2}{9}+b^2=16$. Combining fractions, multiplying, and rearranging, gives $25b^2+8xb+(x^2-144)=0$. This is useful because we want the maximum value of $x$ such that this quadratic has real roots which is easy to find using the discriminant. For the roots to be real, $(8x)^2-4(25)(x^2-144) \ge 0$. Now all that is left to do is to solve this inequality. Simplifying this expression, we get $-36x^2+144000 \ge 0$ which means $x^2 \le 400$ and $x \le 20$. Therefore the maximum value of $x$ is $20$ and $27 \cdot 20 = \boxed{540}$

~vsinghminhas

Solution 5 ("Completing the Triangle")

First, recognize the relationship between the reciprocal of a complex number $z$ with its conjugate $\overline{z}$, namely:

\[\frac{1}{z} \cdot \frac{\overline{z}}{\overline{z}} = \frac{\overline{z}}{|z|^2} = \frac{\overline{z}}{16}\]

Then, let $z = 4(\cos\theta + i\sin\theta)$ and $\overline{z} = 4(\cos\theta - i\sin\theta)$.

\begin{align*} Re \left ((75+117i)z+\frac{96+144i}{z} \right) &= Re\left ( (75+117i)z + (6+9i)\overline{z} \right ) \\ &= 4 \cdot Re\left ( (75+117i)(\cos\theta + i\sin\theta) + (6+9i)(\cos\theta - i\sin\theta) \right ) \\ &= 4 \cdot (75\cos\theta - 117\sin\theta + 6\cos\theta + 9\sin\theta) \\ &= 4 \cdot (81\cos\theta - 108\sin\theta) \\ &= 4\cdot 27 \cdot (3\cos\theta - 4\sin\theta) \end{align*}

Now, recognizing the 3 and 4 coefficients hinting at a 3-4-5 right triangle, we "complete the triangle" by rewriting our desired answer in terms of an angle of that triangle $\phi$ where $\cos\phi = \frac{3}{5}$ and $\sin\phi = \frac{4}{5}$

\begin{align*} 4\cdot 27 \cdot(3\cos\theta - 4\sin\theta) &= 4\cdot 27 \cdot 5 \cdot (\frac{3}{5}\cos\theta - \frac{4}{5}\sin\theta) \\ &= 540 \cdot (\cos\phi\cos\theta - \sin\phi\sin\theta) \\ &= 540 \cos(\theta + \phi) \end{align*}

Since the simple trig ratio is bounded above by 1, our answer is $\boxed{540}$

~ Cocoa @ https://www.corgillogical.com/ (yes i am a corgi that does math)


Solution 6 (Cauchy-Schwarz Inequality ) (Fastest)

Follow as solution 1 would to obtain $81a + 108\sqrt{16-a^2}.$

By the Cauchy-Schwarz Inequality, we have

\[(a^2 + (\sqrt{16-a^2})^2)(81^2 + 108^2) \geq (81a + 108\sqrt{16-a^2})^2,\]

so

\[4^2 \cdot 9^2 \cdot 15^2 \geq (81a + 108\sqrt{16-a^2})^2\]

and we obtain that $81a + 108\sqrt{16-a^2} \leq 4 \cdot 9 \cdot 15 = \boxed{540}.$


- spectraldragon8

Solution 7 (Geometry)

Follow solution 2 to get that we want to find the line $81a-108b=k$ tangent to circle $a^2+b^2=16$. The line turns into $a=\frac{k}{81}+\frac{4b}{3}$ Connect the center of the circle to the tangency point and the y-intercept of the line. Let the tangency point be $A$, the y-intercept be $C$, and the center be $B$. Drop the perpendicular from $A$ to $BC$ and call it $D$. Let $AD=3x$, $DC=4x$. Then, $BD=\sqrt{AB^2-AD^2}=\sqrt{16-9x^2}$. By similar triangles, get that $\frac{BD}{AD}=\frac{AD}{DC}$, so $\frac{\sqrt{16-9x^2}}{3x}=\frac{3x}{4x}$. Solve this to get that $x=\frac{16}{15}$, so $BC=\frac{20}{3}$ and $\frac{k}{81}=\frac{20}{3}$, so $k=\boxed{540}$

Video Solution by MOP 2024

https://www.youtube.com/watch?v=nH7dUh0HghA

~r00tsOfUnity

Video Solution in 3 minutes & Cauchy's Inequality by MegaMath

https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel

See also

2024 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
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