AMC 12B 2025 (Problem 22)What is the greatest possible area of the triangle in the complex plane with vertices 2z2z2z, (1+i)z(1+i)z(1+i)z, and (1−i)z(1-i)z(1−i)z, where zzz is a complex number satisfying ∣4z−2∣=1|4z-2|=1∣4z−2∣=1?(A) 14\text{(A)}\;\frac14(A)41(B) 12\text{(B)}\;\frac12(B)21(C) 916\text{(C)}\;\frac{9}{16}(C)169(D) 34\text{(D)}\;\frac34(D)43(E) 1\text{(E)}\;1(E)1Related TopicsCoreToolkit 120 — Complex Plane TransformationsMajorToolkit 119 — Circle Distance InequalitiesHints (5)Hint 1Hint 2∣z∣=r|z|=r∣z∣=r1+i=2ei45∘1+i=\sqrt2e^{i45^\circ}1+i=2ei45∘1−i=2e−i45∘1-i=\sqrt2e^{-i45^\circ}1−i=2e−i45∘Hint 3z=reiθz=re^{i\theta}z=reiθz(1+i)=r2ei(θ+45∘)z(1+i)=r\sqrt2e^{i(\theta+45^\circ)}z(1+i)=r2ei(θ+45∘)z(1−i)=r2ei(θ−45∘)z(1-i)=r\sqrt2e^{i(\theta-45^\circ)}z(1−i)=r2ei(θ−45∘)2z=2reiθ2z=2re^{i\theta}2z=2reiθHint 4Area2z, z(1+i), z(1−i)=r2\operatorname{Area}_{2z,\,z(1+i),\,z(1-i)}=r^2Area2z,z(1+i),z(1−i)=r2Hint 5By Toolkit 119 — Circle Distance Inequalities,max{∣z∣=r}=34\max\{|z|=r\}=\frac34max{∣z∣=r}=43Ans=max{r2}=(34)2=916\text{Ans}=\max\{r^2\}=\left(\frac34\right)^2=\frac{9}{16}Ans=max{r2}=(43)2=169Final Answer(C) 916\frac{9}{16}169
