Difference between revisions of "2024 AIME I Problems/Problem 7"
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==Solution 2a (Cauchy-Schwartz and vector algebra)== | ==Solution 2a (Cauchy-Schwartz and vector algebra)== | ||
− | Simplify rectangular form as in Solution 1 until we get <math>\text{Re}(w)=81a-108b = 27(3a | + | Simplify rectangular form as in Solution 1 until we get <math>\text{Re}(w)=81a-108b = 27(3a-4b)</math>. |
− | + | By Cauchy-Schwartz, to maximize <math>\text{Re}(w)</math>, the vector <math>z=[a,b]</math> ( <math>|z| =4</math>) is <math>\frac{4}{|[3,-4]|}[3,-4]</math>. | |
− | + | We don't need to bash the arithmetic next, because the unit vector <math>u</math> that maximizes <math>v \cdot u</math> is <math>u=v/|v|</math>, so <math>v \cdot u= v\cdot v = |v|^2/|v| = |v|</math>, which here is just <math>\sqrt{3^2+(-4)^2} =5</math>. | |
− | Combining what remains, we get answer < | + | Combining what remains, we get answer <math>= 27 |z| |v| = 27(4)(5)=\boxed{540}</math>. |
~oinava | ~oinava | ||
==Solution 2b (Simple Analytic Geometry)== | ==Solution 2b (Simple Analytic Geometry)== | ||
− | Simplify rectangular form as in Solution 1 until we get < | + | Simplify rectangular form as in Solution 1 until we get <math>\text{Re}(w)=81x-108y = 27(3x-4y)</math>. |
− | We also know < | + | We also know <math>|z|=4</math> or <math>x^2+y^2=16</math>. |
By AM-GM or Cauchy-Schwartz, b = 4a/3, so | By AM-GM or Cauchy-Schwartz, b = 4a/3, so | ||
You can also prove this like so: | You can also prove this like so: | ||
− | We want to find the line < | + | We want to find the line <math>81x-108y=k</math> tangent to circle <math>x^2+y^2=16</math>, which is perpendicular to the line connecting tangent point to circle's center <math>(0,0)</math>. |
− | Using < | + | |
+ | Using the formula for (perpendicular) distance from a point to a line: <math>\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=r</math> we can substitute and get | ||
+ | <math>\frac{|81(0)-108(0)-k|}{\sqrt{81^2+108^2}}=4</math> | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{k}{\sqrt{ | + | \frac{k}{27\sqrt{3^2+4^2}}&=4 |
\\\frac{k}{135}&=4 | \\\frac{k}{135}&=4 | ||
\\k&=\boxed{540} | \\k&=\boxed{540} |
Latest revision as of 00:24, 19 July 2024
Contents
- 1 Problem
- 2 Video Solution: Cauchy's Inequality
- 3 Solution 1
- 4 Solution 2a (Cauchy-Schwartz and vector algebra)
- 5 Solution 2b (Simple Analytic Geometry)
- 6 Solution 3
- 7 Solution 4 (Simple Quadratic Discriminant)
- 8 Solution 5 ("Completing the Triangle")
- 9 Solution 6 (Cauchy-Schwarz Inequality ) (Fastest)
- 10 Solution 7 (Geometry)
- 11 Solution 8 (Euler's Formula and Trig Optimization)
- 12 Solution 9 (Calc semi-bash)
- 13 Video Solution by MOP 2024
- 14 Video Solution by OmegaLearn.org
- 15 Video Solution
- 16 See also
Problem
Find the largest possible real part of where
is a complex number with
.
Video Solution: Cauchy's Inequality
https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel MegaMath
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 2a (Cauchy-Schwartz and vector algebra)
Simplify rectangular form as in Solution 1 until we get .
By Cauchy-Schwartz, to maximize , the vector
(
) is
.
We don't need to bash the arithmetic next, because the unit vector that maximizes
is
, so
, which here is just
.
Combining what remains, we get answer .
~oinava
Solution 2b (Simple Analytic Geometry)
Simplify rectangular form as in Solution 1 until we get .
We also know or
.
By AM-GM or Cauchy-Schwartz, b = 4a/3, so
You can also prove this like so:
We want to find the line tangent to circle
, which is perpendicular to the line connecting tangent point to circle's center
.
Using the formula for (perpendicular) distance from a point to a line: 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
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
~ Cocoa @ https://www.corgillogical.com/ (yes i am a corgi that does math)
Solution 6 (Cauchy-Schwarz Inequality ) (Fastest)
Follow as solution 1 would to obtain
By the Cauchy-Schwarz Inequality, we have
so
and we obtain that
Solution 7 (Geometry)
Follow solution 2 to get that we want to find the line tangent to circle
. The line turns into
Connect the center of the circle to the tangency point and the y-intercept of the line. Let the tangency point be
, the y-intercept be
, and the center be
. Drop the perpendicular from
to
and call it
. Let
,
. Then,
. By similar triangles, get that
, so
. Solve this to get that
, so
and
, so
~ryanbear
Solution 8 (Euler's Formula and Trig Optimization)
Because , we can let
. Then, substituting
, we get that the complex number is
\begin{align*} w&=4e^{i\theta}(75+117e^{\frac{i\pi}{2}})+\dfrac{1}{4}e^{-i\theta}(96+144e^{\frac{i\pi}{2}})\\ &=300e^{i\theta}+468e^{i(\frac{\pi}{2}+\theta)}+24e^{-i\theta}+36e^{i(\frac{\pi}{2}-\theta)}\\ \end{align*}
We know that the from Euler's formula, so applying this and then applying trig identities yields
\begin{align*} \text{Re}(w)&=300\cos{(\theta)}+468\cos{(\dfrac{\pi}{2}+\theta)}+24\cos{(-\theta)}+36\cos{(\dfrac{\pi}{2}-\theta)}\\ &=300\cos{(\theta)}-468sin{(\theta)}+24\cos{(\theta)}+36\sin{(\theta)}\\ &=324\cos{(\theta)}-432\sin{(\theta)}\\ \implies \dfrac{1}{108}\text{Re}(w)&=3\cos{(\theta)}-4\sin{(\theta)}\\ \end{align*}
We can see that the right-hand side looks an awful lot like the sum of angles formula for cosine, but 3 and 4 don't satisfy the pythagorean identity. To make them do so, we can divide everything by and set
and
. Now we have that
Obviously the maximum value of the right hand side is 1, so the maximum value of the real part is
.
~Mooshiros
Solution 9 (Calc semi-bash)
Let denote value of the above expression such that
is maximized. We write
and multiply the second term in the expression by
turning the expression into
Now, we write
. Since
is even and
is odd,
We want to maximize this expression, so we take its derivative and set it equal to
(and quickly check the second derivative for inflection points):
so
which is reminiscent of a
right triangle in the fourth quadrant (side lengths of
). Since
we quickly see that
and
Therefore,
-Benedict T (countmath1)
Video Solution by MOP 2024
https://www.youtube.com/watch?v=nH7dUh0HghA
~r00tsOfUnity
Video Solution by OmegaLearn.org
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.