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Difference between revisions of "1992 AIME Problems/Problem 10"

(Undo revision 71028 by Starwars123 (talk))
(Undo revision 71027 by Starwars123 (talk))
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== Problem ==
 
== Problem ==
 
Consider the region <math>A</math> in the complex plane that consists of all points <math>z</math> such that both <math>\frac{z}{40}</math> and <math>\frac{40}{\overline{z}}</math> have real and imaginary parts between <math>0</math> and <math>1</math>, inclusive. What is the integer that is nearest the area of <math>A</math>?
 
Consider the region <math>A</math> in the complex plane that consists of all points <math>z</math> such that both <math>\frac{z}{40}</math> and <math>\frac{40}{\overline{z}}</math> have real and imaginary parts between <math>0</math> and <math>1</math>, inclusive. What is the integer that is nearest the area of <math>A</math>?
−
(If  <math>z=a+bi</math> with <math>a</math> and <math>b</math> real, then <math>z=a-bi</math> is the conjugate of <math>z</math>)
+
<math>If  </math>z=a+bi<math> with </math>a<math> and </math>b<math> real, then </math>z=a-bi<math> is the conjugate of </math>z<math>)
  
 
== Solution ==
 
== Solution ==
−
Let <math>z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i</math>. Since <math>0\leq \frac{a}{40},\frac{b}{40}\leq 1</math> we have the inequality <cmath>0\leq a,b \leq 40</cmath>which is a square of side length <math>40</math>.
+
Let </math>z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i<math>. Since </math>0\leq \frac{a}{40},\frac{b}{40}\leq 1<math> we have the inequality <cmath>0\leq a,b \leq 40</cmath>which is a square of side length </math>40<math>.
  
−
Also, <math>\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i</math> so we have <math>0\leq a,b \leq \frac{a^2+b^2}{40}</math>, which leads to:<cmath>(a-20)^2+b^2\geq 20^2</cmath>
+
Also, </math>\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i<math> so we have </math>0\leq a,b \leq \frac{a^2+b^2}{40}<math>, which leads to:<cmath>(a-20)^2+b^2\geq 20^2</cmath>
 
<cmath>a^2+(b-20)^2\geq 20^2</cmath>
 
<cmath>a^2+(b-20)^2\geq 20^2</cmath>
  
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<center>[[Image:AIME_1992_Solution_10.png]]</center>
 
<center>[[Image:AIME_1992_Solution_10.png]]</center>
  
−
We want the area outside the two circles but inside the square. Doing a little geometry, the area of the intersection of those three graphs is <math>40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68</math>
+
We want the area outside the two circles but inside the square. Doing a little geometry, the area of the intersection of those three graphs is </math>40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68<math>
  
−
<math>\boxed{572}</math>
+
</math>\boxed{572}$

Revision as of 13:38, 8 July 2015

Problem

Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$, inclusive. What is the integer that is nearest the area of $A$? $If$z=a+bi$with$a$and$b$real, then$z=a-bi$is the conjugate of$z$)

== Solution == Let$ (Error compiling LaTeX. Unknown error_msg)z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$. Since$0\leq \frac{a}{40},\frac{b}{40}\leq 1$we have the inequality <cmath>0\leq a,b \leq 40</cmath>which is a square of side length$40$.

Also,$ (Error compiling LaTeX. Unknown error_msg)\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$so we have$0\leq a,b \leq \frac{a^2+b^2}{40}$, which leads to:<cmath>(a-20)^2+b^2\geq 20^2</cmath> <cmath>a^2+(b-20)^2\geq 20^2</cmath>

We graph them:

<center>[[Image:AIME_1992_Solution_10.png]]</center>

We want the area outside the two circles but inside the square. Doing a little geometry, the area of the intersection of those three graphs is$ (Error compiling LaTeX. Unknown error_msg)40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68$$ (Error compiling LaTeX. Unknown error_msg)\boxed{572}$

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