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2024 AMC 12B Problem 12

Problem 12 of 25IntermediateAlgebraGeometry

Suppose zz is a complex number with positive imaginary part, with real part greater than 1,1, and with ∣z∣=2.|z| = 2. In the complex plane, the four points 0,0, z,z, z2,z^2, and z3z^3 are the vertices of a quadrilateral with area 15.15. What is the imaginary part of z?z?

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Solution

For vertices 0,z,z2,z30, z, z^2, z^3 the shoelace formula gives area 12 ∣Im⁡(zˉz2+z2‾ z3)∣=12 ∣Im⁡((∣z∣2+∣z∣4)z)∣=12(∣z∣2+∣z∣4)Im⁡(z). \begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z). \end{aligned} With ∣z∣=2,|z| = 2, this is 12(4+16)Im⁡(z)=10Im⁡(z).\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z). Setting 10Im⁡(z)=1510\operatorname{Im}(z) = 15 gives Im⁡(z)=32.\operatorname{Im}(z) = \dfrac32. (Then Re⁡(z)=4−94=72>1,\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1, as required.) Thus, the correct answer is D.
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