Difference between revisions of "Mock AIME I 2012 Problems/Problem 8"
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-Solution by '''thecmd999''' | -Solution by '''thecmd999''' | ||
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+ | ==Solution 3 (Very nice)== | ||
+ | Note that the "<math>+1</math>" will make <math>z</math> maximized when it makes the RHS "minimized", or when it is pointing in the opposite direction of <math>z</math>. Therefore, <math>z</math> is pointing to the left, so <math>z=z^2-1</math>, <math>z^2-z-1=0</math>, and <math>z=\frac{\sqrt5+1}{2}</math>, <math>r + s + t = \boxed{054}</math>. |
Revision as of 18:11, 7 May 2020
Problem
Suppose that the complex number satisfies
. If
is the maximum possible value of
,
can be expressed in the form
. Find
.
Solution 1
We begin by dividing both sides by to obtain
. Now, consider that we may write
with
a positive real number so that
and
some real number. Then,
Since we need
, we must have
, or equivalently,
. By the quadratic equation, this has roots
and to maximize
, we take the larger root
which is clearly maximized when
is minimized. Since
, the maximum value of
will occur where
, so the maximum value of
occurs where
and finally we find that the maximum value of
is
Taking the fourth power, the desired answer is
.
Solution 2
Let
.
Note that
.
To compute , connect
to the origin and construct an altitude to the x-axis. Extend this line one unit above
and connect that point to the origin to create
.
Now, we use the Law of Cosines on the triangle with vertices at the origin, , and
, giving
(or you could use the Pythagorean Theorem). Either way, we proceed from here as we did with Solution 1 to get .
-Solution by thecmd999
Solution 3 (Very nice)
Note that the "" will make
maximized when it makes the RHS "minimized", or when it is pointing in the opposite direction of
. Therefore,
is pointing to the left, so
,
, and
,
.