Difference between revisions of "2022 AMC 12A Problems/Problem 13"
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==Solution== | ==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>. | 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 = 13</math> (A | + | 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 15:41, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.