2024 AIME I Problems/Problem 7
Contents
Problem
Find the largest possible real part of where
is a complex number with
.
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 (Simple Analytic Geometry)
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
Solution 4 (Simple Quadratic Discriminant)
Similar to the solutions above, we find that , where
. To maximize this expression, we must maximize
. Let this value be
. Solving for
yields
. From the given information we also know that
. Substituting
in terms of
and
gives us
. Combining fractions, multiplying, and rearranging, gives
. This is useful because we want the maximum value of
such that this quadratic has real roots which is easy to find using the discriminant. For the roots to be real,
. Now all that is left to do is to solve this inequality. Simplifying this expression, we get
which means
and
. Therefore the maximum value of
is
and
~vsinghminhas
Video Solution in 3 minutes & Cauchy's Inequality by MegaMath
https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel
Solution 5 ("Completing the Triangle")
First, recognize the relationship between the reciprocal of a complex number with its conjugate
, namely:
Then, let and
.
Now, recognizing the 3 and 4 coefficients hinting at a 3-4-5 right triangle, we "complete the triangle" by rewriting our desired answer in terms of an angle of that triangle where
and
Since the simple trig ratio is bounded above by 1, our answer is \boxed{540}$
~ Cocoa @ https://www.corgillogical.com/
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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