Difference between revisions of "2002 AIME I Problems/Problem 12"
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== Problem == | == Problem == | ||
| + | Let <math>F(z)=\dfrac{z+i}{z-i}</math> for all complex numbers <math>z\neq i</math>, and let <math>z_n=F(z_{n-1})</math> for all positive integers <math>n</math>. Given that <math>z_0=\dfrac{1}{137}+i</math> and <math>z_{2002}=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers, find <math>a+b</math>. | ||
== Solution == | == Solution == | ||
| + | Iterating <math>F</math> we get: | ||
| + | |||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | F(z) &= \frac{z+i}{z-i}\\ | ||
| + | F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ | ||
| + | &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ | ||
| + | F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | |||
| + | From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus | ||
| + | <math>z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.</math> | ||
| + | |||
| + | Thus <math>a+b = 1+274 = \boxed{275}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2002|n=I|num-b=11|num-a=13}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 21:39, 21 November 2018
Problem
Let 
for all complex numbers 
, and let 
for all positive integers 
. Given that 
and 
, where 
and 
are real numbers, find 
.
Solution
Iterating 
we get:
From this, it follows that 
, for all 
. Thus
Thus 
.
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
