Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \Im(f(1)) & = i+i\Im(\alpha)+i\Im(\gamma) \ | + | \text{Im}(f(1)) & = i+i\text{Im}(\alpha)+i\text{Im}(\gamma) \ |
− | \Im(f(i)) & = -i+i\Re(\alpha)+i\Im(\gamma) | + | \text{Im}(f(i)) & = -i+i\text{Re}(\alpha)+i\text{Im}(\gamma) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Let <math>p=\Im(\gamma)</math> and <math>q=\Re{(\gamma)},</math> then we know <math>\Im(\alpha)=-p-1</math> and <math>\Re(\alpha)=1-p.</math> Therefore <cmath>|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},</cmath> which reaches its minimum <math>\sqrt 2</math> when <math>p=q=0</math> by the Trivial Inequality. Thus, the answer is <math>\boxed B.</math> | + | Let <math>p=\text{Im}(\gamma)</math> and <math>q=\text{Re}{(\gamma)},</math> then we know <math>\text{Im}(\alpha)=-p-1</math> and <math>\text{Re}(\alpha)=1-p.</math> Therefore <cmath>|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},</cmath> which reaches its minimum <math>\sqrt 2</math> when <math>p=q=0</math> by the Trivial Inequality. Thus, the answer is <math>\boxed B.</math> |
==Solution 2:== | ==Solution 2:== |
Latest revision as of 18:28, 26 September 2023
Contents
[hide]Problem
A function is defined by for all complex numbers , where and are complex numbers and . Suppose that and are both real. What is the smallest possible value of ?
Solution 1:
We need only concern ourselves with the imaginary portions of and (both of which must be 0). These are:
Let and then we know and Therefore which reaches its minimum when by the Trivial Inequality. Thus, the answer is
Solution 2:
Since and are both real we get,
Solving, we get , can be anything, to minimize the value we set , so then the answer is . Thus, the answer is
By: Quaratinium
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.