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

Line 1: Line 1:
 +
==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:  
  

Revision as of 13:40, 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.

w=(75+117i)(a+bi)+96+144ia+bi=(75a117b)+(117a+75b)i+48(2+3ia+bi)=(75a117b)+(116a+75b)i+48((2+3i)(abi)(a+bi)(abi))=(75a117b)+(116a+75b)i+48(2a+3b+(3a2b)ia2+b2)=(75a117b)+(116a+75b)i+48(2a+3b+(3a2b)i16)=(75a117b)+(116a+75b)i+3(2a+3b+(3a2b)i)=(75a117b)+(116a+75b)i+6a+9b+(9a6b)i=(81a108b)+(125a+69b)i.

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.

f(a)=81a+10816a2=27(3a+416a2)f(a)=27(3a+416a2)=27(3+4(16a2))=27(3+4(2a216a2))=27(34(a16a2))=27(34a16a2).

We want $f'(a)$ to be $0$ to find the maximum.

0=27(34a16a2)=34a16a23=4a16a24a=316a216a2=9(16a2)16a2=1449a225a2=144a2=14425a=125=2.4.

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

See also

2024 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png