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Difference between revisions of "2024 AIME I Problems/Problem 7"

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Let <math>z=a+bi</math> such that <math>a^2+b^2=4^2=16</math>. The expression becomes:
  
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<cmath>(75+117i)(a+bi)+\dfrac{96+144i}{a+bi}.</cmath>
 +
 +
Call this complex number <math>w</math>. We simplify this expression.
 +
 +
\begin{align*}
 +
w&=(75+117i)(a+bi)+\dfrac{96+144i}{a+bi} \\
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&=(75a-117b)+(117a+75b)i+48\left(\dfrac{2+3i}{a+bi}\right) \\
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&=(75a-117b)+(116a+75b)i+48\left(\dfrac{(2+3i)(a-bi)}{(a+bi)(a-bi)}\right) \\
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&=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{a^2+b^2}\right) \\
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&=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{16}\right) \\
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&=(75a-117b)+(116a+75b)i+3\left(2a+3b+(3a-2b)i\right) \\
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&=(75a-117b)+(116a+75b)i+6a+9b+(9a-6b)i \\
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&=(81a-108b)+(125a+69b)i. \\
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\end{align*}
 +
 +
We want to maximize <math>\text{Re}(w)=81a-108b</math>. We can use elementary calculus for this, but to do so, we must put the expression in terms of one variable. Recall that <math>a^2+b^2=16</math>; thus, <math>b=\pm\sqrt{16-a^2}</math>. Notice that we have a <math>-108b</math> in the expression; to maximize the expression, we want <math>b</math> to be negative so that <math>-108b</math> is positive and thus contributes more to the expression. We thus let <math>b=-\sqrt{16-a^2}</math>. Let <math>f(a)=81a-108b</math>. We now know that <math>f(a)=81a+108\sqrt{16-a^2}</math>, and can proceed with normal calculus.
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 +
\begin{align*}
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f(a)&=81a+108\sqrt{16-a^2} \\
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&=27\left(3a+4\sqrt{16-a^2}\right) \\
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f'(a)&=27\left(3a+4\sqrt{16-a^2}\right)' \\
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&=27\left(3+4\left(\sqrt{16-a^2}\right)'\right) \\
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&=27\left(3+4\left(\dfrac{-2a}{2\sqrt{16-a^2}}\right)\right) \\
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&=27\left(3-4\left(\dfrac a{\sqrt{16-a^2}}\right)\right) \\
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&=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right). \\
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\end{align*}
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We want <math>f'(a)</math> to be <math>0</math> 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}} \\
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4a&=3\sqrt{16-a^2} \\
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16a^2&=9\left(16-a^2\right) \\
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16a^2&=144-9a^2 \\
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25a^2&=144 \\
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a^2&=\dfrac{144}{25} \\
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a&=\dfrac{12}5 \\
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&=2.4. \\
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\end{align*}
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We also find that <math>b=-\sqrt{16-2.4^2}=-\sqrt{16-5.76}=-\sqrt{10.24}=-3.2</math>.
 +
 +
Thus, the expression we wanted to maximize becomes <math>81\cdot2.4-108(-3.2)=81\cdot2.4+108\cdot3.2=\boxed{540}</math>.
 +
 +
~Technodoggo

Revision as of 13:39, 2 February 2024

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

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