Maths Olympiad Prep

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, 2024

Geometry Difficulty 5.2 AIME, harder Find the answer United States

Suppose zz is a complex number with positive imaginary part, with real part greater than 11, and with z=2|z| = 2. In the complex plane, the four values 00, zz, z2z^2, and z3z^3 are the vertices of a quadrilateral with area 1515. What is the imaginary part of zz?

Pick one

Solution

Let θ\theta be the argument of zz. Because z=2|z| = 2 and the real part of zz is greater than 11, it follows that θ\theta is less than 6060^\circ. This ensures that the imaginary parts of z2z^2 and z3z^3 are positive and all the vertices of the quadrilateral other than 00 lie in the upper half-plane. Thus the area of the quadrilateral is the sum of the areas of the triangle with vertices 00, zz, and z2z^2 and the triangle with vertices 00, z2z^2, and z3z^3. See the figure.

Figure 1

Because the area of a triangle with side lengths aa and bb with included angle α\alpha is 12absinα\frac{1}{2}ab \sin \alpha, the area of the quadrilateral must be
12(zz2sinθ+z2z3sinθ)=20sinθ=15. \frac{1}{2} (|z| \cdot |z|^2 \cdot \sin \theta + |z|^2 \cdot |z|^3 \cdot \sin \theta) = 20 \sin \theta = 15.
It follows that sinθ=34\sin \theta = \frac{3}{4}, and the imaginary part of zz is 2sinθ=322 \cdot \sin \theta = \frac{3}{2}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.