2024 AIME I Problems/Problem 7

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Problem

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

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

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

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.

w=(75+117i)(a+bi)+96+144ia+bi=(75a117b)+(117a+75b)i+48(2+3ia+bi)=(75a117b)+(116a+75b)i+48((2+3i)(abi)(a+bi)(abi))=(75a117b)+(116a+75b)i+48(2a+3b+(3a2b)ia2+b2)=(75a117b)+(116a+75b)i+48(2a+3b+(3a2b)i16)=(75a117b)+(116a+75b)i+3(2a+3b+(3a2b)i)=(75a117b)+(116a+75b)i+6a+9b+(9a6b)i=(81a108b)+(125a+69b)i.

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.

f(a)=81a+10816a2=27(3a+416a2)f(a)=27(3a+416a2)=27(3+4(16a2))=27(3+4(2a216a2))=27(34(a16a2))=27(34a16a2).

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

0=27(34a16a2)=34a16a23=4a16a24a=316a216a2=9(16a2)16a2=1449a225a2=144a2=14425a=125=2.4.

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

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

See also

2024 AIME I (ProblemsAnswer KeyResources)
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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