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

Revision as of 14:40, 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.

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

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

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

Revision as of 14:38, 2 February 2024 (view source) Bh2019mv0 (talk \| contribs) (Second solution to AIME I Problem 7, uses distance to a line formula or something) ← Older edit		Revision as of 14:40, 2 February 2024 (view source) Puck 0 (talk \| contribs) Newer edit →
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		+	==Solution 1==
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	Let <math>z=a+bi</math> such that <math>a^2+b^2=4^2=16</math>. The expression becomes:		Let <math>z=a+bi</math> such that <math>a^2+b^2=4^2=16</math>. The expression becomes:

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

Revision as of 14:40, 2 February 2024

Solution 1

Solution 2 (Without Calculus)

See also