2024 AIME I Problems/Problem 7
Contents
Problem
Find the maximum real part of , where is a complex number with . Here .
Video Solution in 3 minutes & Cauchy's Inequality by MegaMath
https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel
Solution 1
Let such that . The expression becomes:
Call this complex number . 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 . We can use elementary calculus for this, but to do so, we must put the expression in terms of one variable. Recall that ; thus, . Notice that we have a in the expression; to maximize the expression, we want to be negative so that is positive and thus contributes more to the expression. We thus let . Let . We now know that , 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 to be 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 .
Thus, the expression we wanted to maximize becomes .
~Technodoggo
Solution 2 (Without Calculus)
Same steps as solution one until we get . We also know or . We want to find the line tangent to circle . Using we can substitute and get
~BH2019MV0
Solution 3
Follow Solution 1 to get . We can let and as , and thus we have . 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 for obviously positive and .
Using the previous fact, we can use the Cauchy-Schwarz Inequality to calculate the maximum. By the inequality, we have:
~eevee9406
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AIME Problems and Solutions |
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