2025 AMC 12A Problems/Problem 17: Difference between revisions
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<cmath>(z+i)(z+2i)(z+3i) + 10 = (c-i)(c)(c+i)+10 = c^3 +c+10 = 0.</cmath> | <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> | 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> | 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> \boxed{6,A.}</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> \boxed{6,\textbf{A.}}</imath> | |||
~mathkiddus | |||
~[[User:Mathkiddus|mathkiddus]] | |||
Revision as of 17:03, 6 November 2025
The polynomial 
has three roots in the complex plane. What is the area of the triangle formed by these 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 
