2022 AMC 12A Problem 13

Attempt Problem 13 of the 2022 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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13.

Let R\mathcal{R} be the region in the complex plane consisting of all complex numbers zz that can be written as the sum of complex numbers z1z_1 and z2,z_2, where z1z_1 lies on the segment with endpoints 33 and 4i,4i, and z2z_2 has magnitude at most 1.1. What integer is closest to the area of R?\mathcal{R}?
1313
1414
1515
1616
1717
Answer: A
Concepts:areadistance formula
Small hint:
R\mathcal{R} is the set of all points within distance 11 of the segment
Big hint:
The segment has length 5,5, so R\mathcal{R} is a 5×25\times2 rectangle capped by two half-disks
Solution:
Adding a disk of radius 11 to every point of the segment sweeps out all points within distance 11 of it. The segment from 33 to 4i4i has length 32+42=5.\sqrt{3^2+4^2}=5.
This “stadium” is a 5×25\times2 rectangle plus two half-disks of radius 1,1, with area 5⋅2+π(1)2=10+π≈13.14.5\cdot2+\pi(1)^2=10+\pi\approx13.14.
The closest integer is 13.13.
Thus, the correct answer is A.
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