Difference between revisions of "2018 AIME II Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>. | + | Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 16:34, 12 April 2018
Problem
Suppose that ,
, and
are complex numbers such that
,
, and
, where
. Then there are real numbers
and
such that
. Find
.
Solution 1
First we evaluate the magnitudes. ,
, and
. Therefore,
, or
. Divide to find that
,
, and
.
This allows us to see that the argument of
is
, and the argument of
is
. We need to convert the polar form to a standard form. Simple trig identities show
and
. More division is needed to find what
is.
Written by a1b2
Solution 2
Dividing the first equation by the second equation given, we find that . Substituting this into the third equation, we get
. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of
is the negative of that of
, and their magnitudes multiply to
. Thus we have
and
. To find
, we can use the previous substitution we made to find that
Therefore,
Solution by ktong
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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