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

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


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== Solution 1 ==
== Solution 1 ==
Noticing the symmetry, we begin with a substitution: <imath>w = z+2i</imath>. We know have the polynomial <imath>(w-i)(w)(w+i)+10</imath>. Expanding, we get
Noticing the symmetry, we begin with a substitution: <imath>w = z+2i</imath>. We now have the polynomial <imath>(w-i)(w)(w+i)+10</imath>. Expanding, we get
<cmath>w^3+w+10.</cmath>
<cmath>w^3+w+10.</cmath>
Using the Rational Root Theorem, we notice that <imath>-2</imath> is a root of this polynomial. Upon dividing the polynomial by <imath>w+2</imath>, we get that
Using the Rational Root Theorem, we notice that <imath>-2</imath> is a root of this polynomial. Upon dividing the polynomial by <imath>w+2</imath>, we get that
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~ScoutViolet
~ScoutViolet
==Solution 3 (Shoelace Theorem)==
Similarly to Solution 1, we find that <imath>w</imath> can be <imath>-2</imath>, <imath>1+2i</imath>, or <imath>1-2i</imath>. Therefore the 3 solutions of <imath>z</imath> are <imath>-2-2i</imath>, <imath>1</imath>, and <imath>1-4i</imath>.
Now, taking the real parts to be the x-coordinates and the imaginary parts to be the y-coordinates, we can use the Shoelace Theorem to find that area=<imath>\frac{1}{2}|(-2\cdot0+1\cdot(-4)+1\cdot(-2))-((-2)\cdot1+0\cdot1+(-4)\cdot(-2))|=6</imath>
~backtosq-1
~minor edit by 526


==Video Solution 1 by OmegaLearn==
==Video Solution 1 by OmegaLearn==
https://youtu.be/WIkOFDf3Fr4
https://youtu.be/WIkOFDf3Fr4


==Video Solution 2 by StressedPineapple==
https://youtube.com/watch?v=NWBPm3lThH4&t=127s
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==See Also==
==See Also==
{{AMC12 box|year=2025|ab=A|before=16|num-a=18}}
{{AMC12 box|year=2025|ab=A|num-b=16|num-a=18}}
* [[AMC 12]]
* [[AMC 12]]
* [[AMC 12 Problems and Solutions]]
* [[AMC 12 Problems and Solutions]]

Latest revision as of 11:37, 10 November 2025

Problem

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

Noticing the symmetry, we begin with a substitution: $w = z+2i$. We now have the polynomial $(w-i)(w)(w+i)+10$. Expanding, we get \[w^3+w+10.\] Using the Rational Root Theorem, we notice that $-2$ is a root of this polynomial. Upon dividing the polynomial by $w+2$, we get that \[w^3+w+10 = (w+2)(w^2-2w+5).\] Using the Quadratic Formula upon $w^2-2w+5$, we get that the other two roots are $1+2i$ and $1-2i$.

From here, notice that the area of the triangle formed by the roots of this polynomial will be equal to that of the original polynomial because the substitution only shifted the graph $2i$ up, not affecting the distances between each root.

Graphing the roots onto the complex plane, the vertical side of the triangle has length $4$, with the altitude to that side having length $3$. Therefore, the triangle has area $\frac{4 \cdot 3}{2} = \boxed{6}.$

~lprado

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 $\boxed{6}$.

~ScoutViolet

Solution 3 (Shoelace Theorem)

Similarly to Solution 1, we find that $w$ can be $-2$, $1+2i$, or $1-2i$. Therefore the 3 solutions of $z$ are $-2-2i$, $1$, and $1-4i$.

Now, taking the real parts to be the x-coordinates and the imaginary parts to be the y-coordinates, we can use the Shoelace Theorem to find that area=$\frac{1}{2}|(-2\cdot0+1\cdot(-4)+1\cdot(-2))-((-2)\cdot1+0\cdot1+(-4)\cdot(-2))|=6$

~backtosq-1 ~minor edit by 526

Video Solution 1 by OmegaLearn

https://youtu.be/WIkOFDf3Fr4

Video Solution 2 by StressedPineapple

https://youtube.com/watch?v=NWBPm3lThH4&t=127s

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

See Also

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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