Art of Problem Solving

2025 AMC 12A Problems/Problem 17: Difference between revisions

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The polynomial <imath>(z+i)(z+2i)(z+3i) + 10</imath> has three roots in the complex plane. What is the area of the triangle formed by these roots?
The polynomial <imath>(z+i)(z+2i)(z+3i) + 10</imath> has three roots in the complex plane. What is the area of the triangle formed by these roots?
==Solution 1==
SYMMETRY! Let <imath>w</imath> be a complex number such that <imath>w=z+2i,</imath> then we can change our polynomial to the following,
<cmath>(z+i)(z+2i)(z+3i) + 10 = (c-i)(c)(c+i)+10 = c^3 +c+10 = 0.</cmath>
Notice <imath>c = -2</imath> is a root, and from this we get a quadratic and find the the possible values of <imath>c = -2, 1\pm 2i.</imath> Now we simply subtract <imath>2i</imath> from each root to get the roots to be <imath>z = -2-2i, 1 + 0i,</imath> and <imath>1 - 4i.</imath> Moving back to the coordinate plane our points are <imath>(-2,-2), (1,-4),</imath> and <imath>(1,0),</imath> and using shoelace gives us an area of <imath>6, \boxed{A.}</imath>

Revision as of 16:59, 6 November 2025

The polynomial $(z+i)(z+2i)(z+3i) + 10$ has three roots in the complex plane. What is the area of the triangle formed by these roots?

Solution 1

SYMMETRY! Let $w$ be a complex number such that $w=z+2i,$ then we can change our polynomial to the following, \[(z+i)(z+2i)(z+3i) + 10 = (c-i)(c)(c+i)+10 = c^3 +c+10 = 0.\] Notice $c = -2$ is a root, and from this we get a quadratic and find the the possible values of $c = -2, 1\pm 2i.$ Now we simply subtract $2i$ from each root to get the roots to be $z = -2-2i, 1 + 0i,$ and $1 - 4i.$ Moving back to the coordinate plane our points are $(-2,-2), (1,-4),$ and $(1,0),$ and using shoelace gives us an area of $6, \boxed{A.}$