Difference between revisions of "2024 AIME I Problems/Problem 7"
(Second solution to AIME I Problem 7, uses distance to a line formula or something) |
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+ | ==Problem== | ||
+ | Find the largest possible real part of <cmath>(75+117i)z+\frac{96+144i}{z}</cmath>where <math>z</math> is a complex number with <math>|z|=4</math>. | ||
+ | ==Video Solution: Cauchy's Inequality== | ||
+ | https://www.youtube.com/watch?v=ejmrAJ9TpvM&ab_channel=MegaMathChannel | ||
+ | MegaMath | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
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: | ||
Line 49: | Line 57: | ||
~Technodoggo | ~Technodoggo | ||
− | ==Solution | + | ==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-4b)</math>. | |
− | Using <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> | + | |
+ | 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 <math>= 27 |z| |v| = 27(4)(5)=\boxed{540}</math>. | ||
+ | |||
+ | ~oinava | ||
+ | |||
+ | ==Solution 2b (Simple Analytic Geometry)== | ||
+ | Simplify rectangular form as in Solution 1 until we get <math>\text{Re}(w)=81x-108y = 27(3x-4y)</math>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | You can also prove this like so: | ||
+ | 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 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} | ||
Line 61: | Line 89: | ||
~BH2019MV0 | ~BH2019MV0 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Follow Solution 1 to get <math>81a-108b</math>. We can let <math>a=4\cos\theta</math> and <math>b=4\sin\theta</math> as <math>|z|=4</math>, and thus we have <math>324\cos\theta-432\sin\theta</math>. 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 <math>324\cos\theta+432\sin\theta</math> for obviously positive <math>\cos\theta</math> and <math>\sin\theta</math>. | ||
+ | |||
+ | |||
+ | Using the previous fact, we can use the [[Cauchy-Schwarz Inequality]] to calculate the maximum. By the inequality, we have: | ||
+ | |||
+ | <math>(324^2+432^2)(\cos^2\theta+\sin^2\theta)\ge(324\cos\theta+432\sin\theta)^2</math> | ||
+ | |||
+ | <math>540^2\cdot1\ge(324\cos\theta+432\sin\theta)^2</math> | ||
+ | |||
+ | <math>\boxed{540}\ge324\cos\theta+432\sin\theta</math> | ||
+ | |||
+ | ~eevee9406 | ||
+ | |||
+ | ==Solution 4 (Simple Quadratic Discriminant)== | ||
+ | Similar to the solutions above, we find that <math>Re((75+117i)z+\frac{96+144i}{z})=81a-108b=27(3a-4b)</math>, where <math>z=a+bi</math>. To maximize this expression, we must maximize <math>3a-4b</math>. Let this value be <math>x</math>. Solving for <math>a</math> yields <math>a=\frac{x+4b}{3}</math>. From the given information we also know that <math>a^2+b^2=16</math>. Substituting <math>a</math> in terms of <math>x</math> and <math>b</math> gives us <math>\frac{x^2+8bx+16b^2}{9}+b^2=16</math>. Combining fractions, multiplying, and rearranging, gives <math>25b^2+8xb+(x^2-144)=0</math>. This is useful because we want the maximum value of <math>x</math> such that this quadratic has real roots which is easy to find using the discriminant. For the roots to be real, <math>(8x)^2-4(25)(x^2-144) \ge 0</math>. Now all that is left to do is to solve this inequality. Simplifying this expression, we get <math>-36x^2+14400 \ge 0</math> which means <math>x^2 \le 400</math> and <math>x \le 20</math>. Therefore the maximum value of <math>x</math> is <math>20</math> and <math>27 \cdot 20 = \boxed{540}</math> | ||
+ | |||
+ | ~vsinghminhas | ||
+ | |||
+ | ==Solution 5 ("Completing the Triangle")== | ||
+ | |||
+ | First, recognize the relationship between the reciprocal of a complex number <math>z</math> with its conjugate <math>\overline{z}</math>, namely: | ||
+ | |||
+ | <cmath> \frac{1}{z} \cdot \frac{\overline{z}}{\overline{z}} = \frac{\overline{z}}{|z|^2} = \frac{\overline{z}}{16} </cmath> | ||
+ | |||
+ | Then, let <math>z = 4(\cos\theta + i\sin\theta)</math> and <math>\overline{z} = 4(\cos\theta - i\sin\theta)</math>. | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | Re \left ((75+117i)z+\frac{96+144i}{z} \right) &= Re\left ( (75+117i)z + (6+9i)\overline{z} \right ) \\ | ||
+ | &= 4 \cdot Re\left ( (75+117i)(\cos\theta + i\sin\theta) + (6+9i)(\cos\theta - i\sin\theta) \right ) \\ | ||
+ | &= 4 \cdot (75\cos\theta - 117\sin\theta + 6\cos\theta + 9\sin\theta) \\ | ||
+ | &= 4 \cdot (81\cos\theta - 108\sin\theta) \\ | ||
+ | &= 4\cdot 27 \cdot (3\cos\theta - 4\sin\theta) | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | 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 <math>\phi</math> where <math>\cos\phi = \frac{3}{5}</math> and <math>\sin\phi = \frac{4}{5}</math> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 4\cdot 27 \cdot(3\cos\theta - 4\sin\theta) &= 4\cdot 27 \cdot 5 \cdot (\frac{3}{5}\cos\theta - \frac{4}{5}\sin\theta) \\ | ||
+ | &= 540 \cdot (\cos\phi\cos\theta - \sin\phi\sin\theta) \\ | ||
+ | &= 540 \cos(\theta + \phi) | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Since the simple trig ratio is bounded above by 1, our answer is <math>\boxed{540}</math> | ||
+ | |||
+ | ~ 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 <math>81a + 108\sqrt{16-a^2}.</math> | ||
+ | |||
+ | By the Cauchy-Schwarz Inequality, we have | ||
+ | |||
+ | <cmath>(a^2 + (\sqrt{16-a^2})^2)(81^2 + 108^2) \geq (81a + 108\sqrt{16-a^2})^2,</cmath> | ||
+ | |||
+ | so | ||
+ | |||
+ | <cmath>4^2 \cdot 9^2 \cdot 15^2 \geq (81a + 108\sqrt{16-a^2})^2</cmath> | ||
+ | |||
+ | and we obtain that <math>81a + 108\sqrt{16-a^2} \leq 4 \cdot 9 \cdot 15 = \boxed{540}.</math> | ||
+ | |||
+ | |||
+ | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] | ||
+ | |||
+ | ==Solution 7 (Geometry)== | ||
+ | Follow solution 2 to get that we want to find the line <math>81a-108b=k</math> tangent to circle <math>a^2+b^2=16</math>. The line turns into <math>a=\frac{k}{81}+\frac{4b}{3}</math> | ||
+ | Connect the center of the circle to the tangency point and the y-intercept of the line. Let the tangency point be <math>A</math>, the y-intercept be <math>C</math>, and the center be <math>B</math>. Drop the perpendicular from <math>A</math> to <math>BC</math> and call it <math>D</math>. Let <math>AD=3x</math>, <math>DC=4x</math>. Then, <math>BD=\sqrt{AB^2-AD^2}=\sqrt{16-9x^2}</math>. By similar triangles, get that <math>\frac{BD}{AD}=\frac{AD}{DC}</math>, so <math>\frac{\sqrt{16-9x^2}}{3x}=\frac{3x}{4x}</math>. Solve this to get that <math>x=\frac{16}{15}</math>, so <math>BC=\frac{20}{3}</math> and <math>\frac{k}{81}=\frac{20}{3}</math>, so <math>k=\boxed{540}</math> | ||
+ | ~ryanbear | ||
+ | |||
+ | ==Solution 8 (Euler's Formula and Trig Optimization)== | ||
+ | Because <math>|z|=4</math>, we can let <math>z=4e^{i\theta}</math>. Then, substituting <math>i=e^{\frac{i\pi}{2}}</math>, 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 <math>\text{Re}(e^{i\alpha})=\cos(\alpha)</math> 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 <math>\sqrt{3^2+4^2}=5</math> and set <math>\cos{(\alpha)}::=\frac{3}{5}</math> and <math>\sin{(\alpha)}::=\frac{4}{5}</math>. Now we have that | ||
+ | <cmath>\dfrac{1}{540}\text{Re}(w)=\cos{(\theta+\alpha)}</cmath> | ||
+ | Obviously the maximum value of the right hand side is 1, so the maximum value of the real part is <math>\boxed{540}</math>. | ||
+ | |||
+ | ~Mooshiros | ||
+ | |||
+ | ==Solution 9 (Calc semi-bash)== | ||
+ | Let <math>c</math> denote value of the above expression such that <math>\mathsf{Re} (c)</math> is maximized. We write <math>z=4e^{i\theta}</math> and multiply the second term in the expression by <math>\overline{z} = 4e^{-i\theta},</math> turning the expression into | ||
+ | <cmath>4e^{i\theta}(75+117i) + \frac{(96 + 144i)\cdot 4e^{-i\theta}}{4e^{i\theta}\cdot 4e^{-i\theta}} = 300e^{i\theta} + 468ie^{i\theta} + (24+ 36i)e^{-i\theta}.</cmath> | ||
+ | Now, we write <math>e^{i\theta} = \cos\theta + i\sin\theta</math>. Since <math>\cos</math> is even and <math>\sin</math> is odd, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | &300(\cos\theta + i\sin\theta) +468i + (24+36i)(\cos\theta -i\sin\theta) \\ | ||
+ | \iff & \mathsf{Re}(c) = 324\cos\theta -468\sin\theta | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | We want to maximize this expression, so we take its derivative and set it equal to <math>0</math> (and quickly check the second derivative for inflection points): | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | &\mathsf{Re}(c) = 108\left(3\cos\theta - 4\sin\theta\right)\\ | ||
+ | \frac{d}{d\theta} &\mathsf{Re}(c) = -324\sin\theta -468\cos\theta = 0, | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | so <math>\tan\theta = -\dfrac{468}{324} = -\dfrac{4}{3},</math> which is reminiscent of a <math>3-4-5</math> right triangle in the fourth quadrant (side lengths of <math>3, -4, 5</math>). Since <math>\tan\theta = -\frac{4}{3},</math> we quickly see that <math>\sin\theta = -\dfrac{4}{5}</math> and <math>\cos\theta = \dfrac{3}{5}.</math> Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \mathsf{Re}(c) &= 108\left(3\cos\theta - 4\sin\theta \right) = 108\left(\frac{9}{5} + \frac{16}{5} \right) = 108\cdot 5 = \boxed{\textbf{(540)}} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Solution 10 (Lagrange multipliers)== | ||
+ | With <math>z = a + bi</math> such that <math>a^2 + b^2 = 16,</math> we have | ||
+ | <cmath> | ||
+ | \begin{align*} (75 +117i)(a + bi) + \frac{48}{a + bi} (2 + 3i) &= 75a + 75bi + 117ai - 117b + \frac{48}{a + bi}(2 + 3i) \\ &= 75a - 117b + (117a + 75b)i + 48 (2 + 3i) \cdot \frac{a - bi}{16} \\ &= 75a - 117b + (117a + 75b)i + 3 (2 + 3i)(a - bi) \end{align*} | ||
+ | </cmath> | ||
+ | where we use <math>z^{-1} = \frac{\bar z}{|z|^2}.</math> With <math>3 (2 + 3i)(a - bi) = 3 [2a - 2bi + 3ai + 3b] = 6a +9b +9ai-6bi,</math> the expression becomes <math>81a-108b+ (126a + 69b)i</math> and we would like to maximize <math>81a - 108b = 9(9a - 12b) = 27(3a - 4b)</math> with <math>a^2 + b^2 = 16.</math> | ||
+ | With <math>f(a, b) = 3a - 4b</math> and <math>g(a, b) = a^2 + b^2 = 16,</math> we have | ||
+ | <cmath> | ||
+ | \begin{align*} 3 = 2\lambda a, \quad -4 = 2\lambda b \implies -\frac{3}{4} = \frac ab \implies -3b = 4a \implies b = -\frac 43 a\end{align*} | ||
+ | </cmath> | ||
+ | so | ||
+ | <cmath> | ||
+ | a^2 + \frac{16}{9}a^2 = \frac{25}{9}a^2 = 16 \implies \frac{5}{3}a = 4 \implies a = \frac {12}5, b = -\frac{16}{5} | ||
+ | </cmath> | ||
+ | and we have <math>3a - 4b = \frac{36}{5} + \frac{64}{5} = 20,</math> so the maximum is <math>27 \cdot 20 = \boxed{540}.</math> | ||
+ | -centslordm | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://www.youtube.com/watch?v=nH7dUh0HghA | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==Video Solution by OmegaLearn.org== | ||
+ | https://youtu.be/fErzDCu_utY | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/fC6CPlQIRB8 | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
{{AIME box|year=2024|n=I|num-b=6|num-a=8}} | {{AIME box|year=2024|n=I|num-b=6|num-a=8}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:57, 6 October 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 Solution 10 (Lagrange multipliers)
- 14 Video Solution by MOP 2024
- 15 Video Solution by OmegaLearn.org
- 16 Video Solution
- 17 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)
Solution 10 (Lagrange multipliers)
With such that we have where we use With the expression becomes and we would like to maximize with With and we have so and we have so the maximum is -centslordm
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 |
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