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

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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 = 13</math> (A)
  
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Revision as of 16:41, 12 November 2022

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

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 $5(2) + \pi \approx 10 + 3 = 13$ (A)

~juicefruit

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