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

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.

$\begin{aligned} w & = (75 + 117 i) (a + b i) + \frac{96 + 144 i}{a + b i} \\ = (75 a - 117 b) + (117 a + 75 b) i + 48 (\frac{2 + 3 i}{a + b i}) \\ = (75 a - 117 b) + (116 a + 75 b) i + 48 (\frac{(2 + 3 i) (a - b i)}{(a + b i) (a - b i)}) \\ = (75 a - 117 b) + (116 a + 75 b) i + 48 (\frac{2 a + 3 b + (3 a - 2 b) i}{a^{2} + b^{2}}) \\ = (75 a - 117 b) + (116 a + 75 b) i + 48 (\frac{2 a + 3 b + (3 a - 2 b) i}{16}) \\ = (75 a - 117 b) + (116 a + 75 b) i + 3 (2 a + 3 b + (3 a - 2 b) i) \\ = (75 a - 117 b) + (116 a + 75 b) i + 6 a + 9 b + (9 a - 6 b) i \\ = (81 a - 108 b) + (125 a + 69 b) i . \end{aligned}$

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.

$\begin{aligned} f (a) & = 81 a + 108 \sqrt{16 - a^{2}} \\ = 27 (3 a + 4 \sqrt{16 - a^{2}}) \\ f^{'} (a) & = 27 {(3 a + 4 \sqrt{16 - a^{2}})}^{'} \\ = 27 (3 + 4 {(\sqrt{16 - a^{2}})}^{'}) \\ = 27 (3 + 4 (\frac{- 2 a}{2 \sqrt{16 - a^{2}}})) \\ = 27 (3 - 4 (\frac{a}{\sqrt{16 - a^{2}}})) \\ = 27 (3 - \frac{4 a}{\sqrt{16 - a^{2}}}) . \end{aligned}$

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

$\begin{aligned} 0 & = 27 (3 - \frac{4 a}{\sqrt{16 - a^{2}}}) \\ = 3 - \frac{4 a}{\sqrt{16 - a^{2}}} \\ 3 & = \frac{4 a}{\sqrt{16 - a^{2}}} \\ 4 a & = 3 \sqrt{16 - a^{2}} \\ 16 a^{2} & = 9 (16 - a^{2}) \\ 16 a^{2} & = 144 - 9 a^{2} \\ 25 a^{2} & = 144 \\ a^{2} & = \frac{144}{25} \\ a & = \frac{12}{5} \\ = 2.4 . \end{aligned}$

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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+Let <math>z=a+bi</math> such that <math>a^2+b^2=4^2=16</math>. The expression becomes:
+<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} \
+&=(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 <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.
+\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 <math>f'(a)</math> to be <math>0</math> to find the maximum.
+\begin{align*}
+&=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right) \
+&=3-\dfrac{4a}{\sqrt{16-a^2}} \
+&=\dfrac{4a}{\sqrt{16-a^2}} \
+a&=3\sqrt{16-a^2} \
+a^2&=9\left(16-a^2\right) \
+a^2&=144-9a^2 \
+a^2&=144 \
+a^2&=\dfrac{144}{25} \
+a&=\dfrac{12}5 \
+&=2.4. \
+\end{align*}
+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

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

Revision as of 13:39, 2 February 2024