Difference between revisions of "2019 CIME I Problems/Problem 12"

(problem and incomplete solution)
 
 
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=Solution 1=
 
=Solution 1=
Graph the equation <math>|\text{Re}(z)|=\text{Im}(z)+1</math> in the complex plane and you will find that the locus of all points <math>\frac{i}{z}</math> is the intersection of two circles and has area <math>75π+50</math>. The greatest integer less than or equal to <math>75π+50</math> is <math>\boxed{285}</math>.
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Graph the equation <math>|\text{Re}(z)|=\text{Im}(z)+1</math> in the complex plane and you will find that the locus of all points <math>\frac{i}{z}</math> is the intersection of two circles and has area <math>75\pi+50</math>. The greatest integer less than or equal to <math>75\pi+50</math> is <math>\boxed{285}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 17:37, 4 October 2020

Let $\mathcal{T}$ be the locus of all points $z$ in the complex plane satisfying $|\text{Re}(z)|=\text{Im}(z)+1$, and let $\mathcal{T}'$ be the locus of all points $z'=\frac{i}{z}$ where $z \in \mathcal{T}$ and $i=\sqrt{-1}$. If the area enclosed by $\mathcal{T}'$ is $\mathcal{A}$, compute $\lfloor 100\mathcal{A} \rfloor$.

Solution 1

Graph the equation $|\text{Re}(z)|=\text{Im}(z)+1$ in the complex plane and you will find that the locus of all points $\frac{i}{z}$ is the intersection of two circles and has area $75\pi+50$. The greatest integer less than or equal to $75\pi+50$ is $\boxed{285}$.

See also

2019 CIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

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