Difference between revisions of "2019 CIME I Problems/Problem 12"
Icematrix2 (talk | contribs) (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> | + | 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 be the locus of all points
in the complex plane satisfying
, and let
be the locus of all points
where
and
. If the area enclosed by
is
, compute
.
Solution 1
Graph the equation in the complex plane and you will find that the locus of all points
is the intersection of two circles and has area
. The greatest integer less than or equal to
is
.
See also
2019 CIME I (Problems • Answer Key • Resources) | ||
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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