Maths Olympiad Prep

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Problem 601

AMC 10/12, early questions
Geometry Difficulty 3.6 Find the answer China Mathematical Competition · China

It is given that complex numbers z1z_1 and z2z_2 satisfy ∣z1∣=2|z_1| = 2 and ∣z2∣=3|z_2| = 3. If the included angle of their corresponding vectors is 60∘60^\circ, then ∣z1+z2z1−z2∣=‾\left|\frac{z_1 + z_2}{z_1 - z_2}\right| = \underline{\hspace{2cm}}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

By the cosine rule, we obtain
∣z1+z2∣=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣cos⁡120∘=19, |z_1 + z_2| = \sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos 120^\circ} = \sqrt{19},
and
∣z1−z2∣=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣cos⁡60∘=7. |z_1 - z_2| = \sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos 60^\circ} = \sqrt{7}.
Therefore,
∣z1+z2z1−z2∣=197=1337. \left| \frac{z_1 + z_2}{z_1 - z_2} \right| = \frac{\sqrt{19}}{\sqrt{7}} = \frac{\sqrt{133}}{7}.

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