Complex Numbers
Triangle with complex vertices; angle bisector
MMTS_Full_Test_20
Grade 12
Question:
In $\triangle ABC$: $A=2025\omega+2024i$, $B=2024i\omega^2+2025$, $C=2024\omega^2-2025i$ where $\omega$ is non-real cube root of unity with positive imaginary part. The internal bisector of $\angle ACB$ meets $AB$ at $D$ and $\angle BDC=\frac{k\pi}{24}$. Then $k$ is
(A) 6
(B) 12
(C) 14
(D) 16
Step-by-Step Solution
Key Concept: Substitute $\omega=e^{i\cdot2\pi/3}$. Compute $A,B,C$ explicitly and find the angles using argument formulas.
$\angle BDC=7\pi/12=14\pi/24\Rightarrow k=14$.
Correct Answer: (C) 14