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Difference between revisions of "2024 AIME I Problems/Problem 7"
(Soln 3)
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~BH2019MV0
 
~BH2019MV0
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==Solution 3==
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Follow Solution 1 to get <math>81a-108b</math>. We can let <math>a=4\cos\theta</math> and <math>b=4\sin\theta</math> as <math>|z|=4</math>, and thus we have <math>324\cos\theta-432\sin\theta</math>. 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 <math>324\cos\theta+432\sin\theta</math> for obviously positive <math>\cos\theta</math> and <math>\sin\theta</math>.
 +
 +
 +
Using the previous fact, we can use the [[Cauchy-Schwarz Inequality]] to calculate the maximum. By the inequality, we have:
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<math>(324^2+432^2)(\cos^2\theta+\sin^2\theta)\ge(324\cos\theta+432\sin\theta)^2</math>
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<math>540^2\cdot1\ge(324\cos\theta+432\sin\theta)^2</math>
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<math>\boxed{540}\ge324\cos\theta+432\sin\theta</math>
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~eevee9406
  
 
==See also==
 
==See also==

Revision as of 13:42, 2 February 2024

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

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
All AIME Problems and Solutions

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