Difference between revisions of "2012 AIME II Problems/Problem 8"
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== Problem 8 == | == Problem 8 == | ||
The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath> | The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath> | ||
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== Solution == | == Solution == | ||
− | Multiplying the two equations together gives us <cmath>zw + 32i - \frac{240}{zw} = -30 + 46i</cmath> and multiplying by <math>zw</math> then gives us a quadratic in <math>zw</math>: <cmath>(zw)^2 + (30-14i)zw - 240 =0.</cmath> Using the quadratic formula, we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(15-7i)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The smallest possible value of <math>\vert zw\vert^2</math> is then obviously <math>6^2 + 2^2 = \boxed{040.}</math> | + | Multiplying the two equations together gives us <cmath>zw + 32i - \frac{240}{zw} = -30 + 46i</cmath> and multiplying by <math>zw</math> then gives us a quadratic in <math>zw</math>: <cmath>(zw)^2 + (30-14i)zw - 240 =0.</cmath> Using the quadratic formula, we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(15-7i)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The smallest possible value of <math>\vert zw\vert^2</math> is then obviously <math>6^2 + 2^2 = \boxed{040}</math>. |
+ | ===Note=== | ||
+ | A key thing to note here is that <math>|zw|^2=|z|^2\cdot|w|^2,</math> which can be proved as follows: | ||
+ | |||
+ | Proof: Using the values for <math>z</math> and <math>w</math> that we used above, we get: | ||
+ | |||
+ | \begin{align*} | ||
+ | |zw|^2&=|(ac-bd)+i(bc+ad)|^2\\ | ||
+ | &=(ac-bd)^2+(bc+ad)^2\\ | ||
+ | &=a^2c^2+b^2d^2+b^2c^2+a^2d^2-2abcd+2abc\\ | ||
+ | &=a^2c^2+b^2d^2+b^2c^2+a^2d^2 | ||
+ | \end{align*} | ||
+ | Also, <math>|z|^2=a^2+b^2</math> and <math>|w|^2=c^2+d^2</math>. Therefore: | ||
+ | <cmath>|z|^2\cdot|w|^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2</cmath> | ||
+ | and our proof is complete. | ||
+ | |||
+ | Now, also note that we found <math>\sqrt{416-210i}</math> by letting <math>416-210i=(a-bi)^2</math> and solving for <math>a</math> and <math>b</math> by considering real and imaginary parts. Then, we substitute that into <math>a-bi</math> which is the value of <math>\sqrt{416-210i}</math> and continue from there. | ||
+ | |||
+ | '''mathboy282''' | ||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=7|num-a=9}} | {{AIME box|year=2012|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:54, 14 January 2024
Contents
Problem 8
The complex numbers and
satisfy the system
Find the smallest possible value of
.
Solution
Multiplying the two equations together gives us and multiplying by
then gives us a quadratic in
:
Using the quadratic formula, we find the two possible values of
to be
=
The smallest possible value of
is then obviously
.
Note
A key thing to note here is that which can be proved as follows:
Proof: Using the values for and
that we used above, we get:
\begin{align*}
|zw|^2&=|(ac-bd)+i(bc+ad)|^2\\
&=(ac-bd)^2+(bc+ad)^2\\
&=a^2c^2+b^2d^2+b^2c^2+a^2d^2-2abcd+2abc\\
&=a^2c^2+b^2d^2+b^2c^2+a^2d^2
\end{align*}
Also, and
. Therefore:
and our proof is complete.
Now, also note that we found by letting
and solving for
and
by considering real and imaginary parts. Then, we substitute that into
which is the value of
and continue from there.
mathboy282
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.