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

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~[[User:Mathkiddus|mathkiddus]]
~[[User:Mathkiddus|mathkiddus]]
==Solution 2 (bash)==
Expand the left hand side, and we get <imath>x^3-11x+10+(6x^2-6)i</imath>. We immediately see that <imath>x=1</imath> is a root, so factor this out, and we get <imath>x^2+(6i+1)x+(6i-10)</imath>. We put this into the quadratic formula, and we get the other two roots are <imath>\frac{-6i-1 \pm \sqrt{5-12i}}{2}</imath>. Note that <imath>5-12i = (3-2i)^2</imath>, hence we get <imath>\frac{2-8i}{2} = 1-4i</imath> and <imath>\frac{-4i-4}{2} = -2-2i</imath> are the other two roots. We convert into coordinates to get <imath>(1,-4)</imath>, <imath>(1,0)</imath>, and <imath>(-2,-2)</imath>. Note that one of these lines is vertical (<imath>(1,-4)</imath> to <imath>(1,0)</imath>), so the area is the base (<imath>4</imath>) times the height (<imath>1-(-2)=3</imath>) over <imath>2</imath>, aka <imath>\boxed{6}</imath>.


==See Also==
==See Also==

Revision as of 19:45, 6 November 2025

The polynomial $(z + i)(z + 2i)(z + 3i) + 10$ has three roots in the complex plane, where $i = \sqrt{-1}$. What is the area of the triangle formed by these three roots?

$\textbf{(A)}~6 \qquad \textbf{(B)}~8 \qquad \textbf{(C)}~10 \qquad \textbf{(D)}~12 \qquad \textbf{(E)}~14$

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 $\boxed{6,\textbf{A.}}$


~mathkiddus

Solution 2 (bash)

Expand the left hand side, and we get $x^3-11x+10+(6x^2-6)i$. We immediately see that $x=1$ is a root, so factor this out, and we get $x^2+(6i+1)x+(6i-10)$. We put this into the quadratic formula, and we get the other two roots are $\frac{-6i-1 \pm \sqrt{5-12i}}{2}$. Note that $5-12i = (3-2i)^2$, hence we get $\frac{2-8i}{2} = 1-4i$ and $\frac{-4i-4}{2} = -2-2i$ are the other two roots. We convert into coordinates to get $(1,-4)$, $(1,0)$, and $(-2,-2)$. Note that one of these lines is vertical ($(1,-4)$ to $(1,0)$), so the area is the base ($4$) times the height ($1-(-2)=3$) over $2$, aka $\boxed{6}$.

See Also

2025 AMC 12A (Problems • Answer Key • Resources)
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Problem 18
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