Complex Numbers
Equilateral triangle from squared modulus conditions
MJAT_TS6_P2
Grade 12
Question:
Let $A(z_1)$, $B(z_2)$, $C(z_3)$ be vertices of $\triangle ABC$ such that $|z_1-z_2|=3$, $z_1^2=z_2z_3$, $z_2^2=z_1z_3$ ($z_1z_2z_3\neq 0$). The area of $\triangle ABC$ equals:
A) $\dfrac{3}{4}\sqrt{3}$
B) $\dfrac{9}{4}\sqrt{3}$
C) $3\sqrt{3}$
D) $\dfrac{3}{4}$
Step-by-Step Solution
Key Concept: From $z_1^2=z_2z_3$ and $z_2^2=z_1z_3$: $z_3^2=z_1z_2$ follows. So $\sum z_k^2=\sum z_iz_j\Rightarrow\triangle ABC$ is equilateral with side $3$.
Area $=\dfrac{9\sqrt{3}}{4}$. Answer: **B**.
Correct Answer: B