2025 AMC 12A Problems/Problem 17: Difference between revisions
Mathkiddus (talk | contribs) |
Mathkiddus (talk | contribs) No edit summary |
||
| Line 1: | Line 1: | ||
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, where <imath>i = \sqrt{-1}</imath>. What is the area of the triangle formed by these three roots? | ||
<imath>\textbf{(A)}~6 \qquad \textbf{(B)}~8 \qquad \textbf{(C)}~10 \qquad \textbf{(D)}~12 \qquad \textbf{(E)}~14</imath> | |||
==Solution 1 (Symmetry)== | ==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, | Let <imath>w</imath> be a complex number such that <imath>w=z+2i,</imath> then we can change our polynomial to the following, | ||
Revision as of 17:07, 6 November 2025
The polynomial 
has three roots in the complex plane, where 
. What is the area of the triangle formed by these three roots?
Solution 1 (Symmetry)
Let 
be a complex number such that 
then we can change our polynomial to the following,
![\[(z+i)(z+2i)(z+3i) + 10 = (c-i)(c)(c+i)+10 = c^3 +c+10 = 0.\]](http://latex-new.aopstest.com/b/0/0/b00ec5ed4b90ce45d91e37a9cd7890bc4346aba8.png)
Notice 
is a root, and from this we get a quadratic and find the the possible values of 
Now we simply subtract 
from each root to get the roots to be 
and 
Moving back to the coordinate plane our points are 
and 
and using shoelace gives us an area of 
See Also
| 2025 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by 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 | |
These problems are copyrighted © by the Mathematical Association of America. 
