Competition problem — Geometry — Maths Olympiad Prep

Library / /400 of 520

Geometry Difficulty 3.6 AMC 10/12 Find the answer

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 $\mathcal{R}$ ?

Pick one

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 $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$ .
~juicefruit

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.