Difference between revisions of "2008 AMC 12B Problems/Problem 19"
Mathmastersd (talk | contribs) m (→Solution) |
m (→Solution 1:) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==Problem== |
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ? | A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ? | ||
| Line 8: | Line 8: | ||
<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:== | ||
<math>f(1)=4+i+\alpha+\gamma</math> | <math>f(1)=4+i+\alpha+\gamma</math> | ||
| + | |||
<math>f(i)=-4-i+\alpha \cdot i +\gamma</math> | <math>f(i)=-4-i+\alpha \cdot i +\gamma</math> | ||
Latest revision as of 18:28, 26 September 2023
Contents
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 ![\[|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},\]](http://latex.artofproblemsolving.com/1/1/e/11ef7cae82a4f33f113baedb0472cdd897070286.png)
which reaches its minimum 
when 
by the Trivial Inequality. Thus, the answer is 
Solution 2:
Since and are both real we get,
![\[\alpha+\gamma=-i\]](http://latex.artofproblemsolving.com/0/b/9/0b9be1d6ebced83e6254403573be09115d96bae2.png)
![\[\alpha \cdot i+\gamma=i\]](http://latex.artofproblemsolving.com/c/0/7/c07af11d839913d5d4f792fe56235d930ae491b2.png)
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. 
