2012 AIME II Problems/Problem 8
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 |
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