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Difference between revisions of "2022 AMC 12A Problems/Problem 13"

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If <math>z</math> is a complex number and <math>z = a + bi</math>, then the magnitude (length) of <math>z</math> is <math>\sqrt{a^2 + b^2}</math>. Therefore, <math>z_1</math> has a magnitude of 5. If <math>z_2</math> has a magnitude of at most one, that means for each point on the segment given by <math>z_1</math>, the bounds of the region <math>\mathcal{R}</math> could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of <math>\pi \approx 3</math>.
 
If <math>z</math> is a complex number and <math>z = a + bi</math>, then the magnitude (length) of <math>z</math> is <math>\sqrt{a^2 + b^2}</math>. Therefore, <math>z_1</math> has a magnitude of 5. If <math>z_2</math> has a magnitude of at most one, that means for each point on the segment given by <math>z_1</math>, the bounds of the region <math>\mathcal{R}</math> could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of <math>\pi \approx 3</math>.
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Therefore, the total area is <math>5(2) + \pi \approx 10 + 3 = 13</math> <math>\boxed{\textbf{(A)}}</math>.
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Therefore, the total area is <math>5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}</math>.
  
 
~juicefruit
 
~juicefruit
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~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 +
==Video Solution 1 (Simple and Fun!!!)==
 +
https://youtu.be/7yAh4MtJ8a8?si=5AxafzIVqF71CtGL&t=2712
 +
 +
~Math-X
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2022|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2022|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:41, 25 October 2023

Problem

Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$, where $z_1$ lies on the segment with endpoints $3$ and $4i$, and $z_2$ has magnitude at most $1$. What integer is closest to the area of $\mathcal{R}$?

$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$

Solution

[asy] size(250); import TrigMacros; rr_cartesian_axes(-2,6,-2,6,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-1,5,Ticks(f, 1.0)); yaxis(-1,5,Ticks(f, 1.0)); dot((3,0)); dot((0,4)); draw((0,4)--(3,0), blue); draw((0.8, 4.6)..(-.6,4.8)..(-.8, 3.4),red); draw((-.8, 3.4)--(2.2, -0.6), red); draw((2.2, -0.6)..(3.6,-0.8)..(3.8,0.6), red); draw((0.8, 4.6)--(3.8,0.6),red); draw((0.8, 4.6)--(-.8, 3.4),red+dashed); draw((2.2, -0.6)--(3.8,0.6),red+ dashed); draw((3,0)--(3,-1),Arrow); label("1",(3,0)--(3,-1),E); draw((0,4)--(-.6,4.8),Arrow); label("1",(0,4)--(-.6,4.8),SW); draw((1.5,2)--(2.3,2.6),Arrow); label("1",(1.5,2)--(2.3,2.6),SE); [/asy]

If $z$ is a complex number and $z = a + bi$, then the magnitude (length) of $z$ is $\sqrt{a^2 + b^2}$. Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$, the bounds of the region $\mathcal{R}$ could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of $\pi \approx 3$. Therefore, the total area is $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$.

~juicefruit

Video Solution 1 (Quick and Simple)

https://youtu.be/z-Ay2nNejnY

~Education, the Study of Everything

Video Solution 1 (Simple and Fun!!!)

https://youtu.be/7yAh4MtJ8a8?si=5AxafzIVqF71CtGL&t=2712

~Math-X

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

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