Difference between revisions of "2022 AMC 12A Problems/Problem 13"
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<math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> | <math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> | ||
| + | |||
| + | ==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 <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>. | ||
| + | Therefore, the total area is <math>5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}</math>. | ||
| + | |||
| + | ~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 == | ||
| + | {{AMC12 box|year=2022|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:41, 25 October 2023
Contents
Problem
Let 
be the region in the complex plane consisting of all complex numbers 
that can be written as the sum of complex numbers 
and 
, where lies on the segment with endpoints 
and 
, and has magnitude at most 
. What integer is closest to the area of ?
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]](http://latex.artofproblemsolving.com/b/5/8/b5865dd020cde786c1c7723b1e0e1b00df3a402f.png)
If is a complex number and 
, then the magnitude (length) of is 
. Therefore, has a magnitude of 5. If has a magnitude of at most one, that means for each point on the segment given by , the bounds of the region 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 
.
Therefore, the total area is 
.
~juicefruit
Video Solution 1 (Quick and Simple)
~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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
