2024 AIME I Problems/Problem 7

Problem

Find the maximum real part of $(75+117i)z+\frac{96+144i}{z}$ , where $z$ is a complex number with $|z|=4$ . Here $i=\sqrt{-1}$ .

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 (Without Calculus)

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

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2024 AIME I Problems/Problem 7

Contents

Problem

Solution 1

Solution 2 (Without Calculus)

Solution 3

See also