Maths Olympiad Prep

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Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Romania

a) Let z1,z2,z3,z4z_1, z_2, z_3, z_4 be distinct complex numbers of zero sum, having equal absolute values. Prove that the points of affixes z1,z2,z3,z4z_1, z_2, z_3, z_4 are the vertices of a rectangle.

b) Let x,y,z,tx, y, z, t be real numbers such that sin⁡x+sin⁡y+sin⁡z+sin⁡t=0\sin x + \sin y + \sin z + \sin t = 0 and cos⁡x+cos⁡y+cos⁡z+cos⁡t=0\cos x + \cos y + \cos z + \cos t = 0. Prove that, for every integer nn,
sin⁡(2n+1)x+sin⁡(2n+1)y+sin⁡(2n+1)z+sin⁡(2n+1)t=0. \sin(2n + 1)x + \sin(2n + 1)y + \sin(2n + 1)z + \sin(2n + 1)t = 0.

Solution

a) The equality z1+z2+z3+z4=0z_1 + z_2 + z_3 + z_4 = 0 implies zˉ1+zˉ2+zˉ3+zˉ4=0\bar{z}_1 + \bar{z}_2 + \bar{z}_3 + \bar{z}_4 = 0 and furthermore 1z1+1z2+1z3+1z4=0\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4} = 0 (1), for ∣z1∣=∣z2∣=∣z3∣=∣z4∣≠0|z_1| = |z_2| = |z_3| = |z_4| \ne 0.
Suppose z1+z2=−z3−z4≠0z_1 + z_2 = -z_3 - z_4 \ne 0. The relation (1) gives z1z2=z3z4z_1 z_2 = z_3 z_4, so {z1,z2}={−z3,−z4}\{z_1, z_2\} = \{-z_3, -z_4\}. On the other hand, if z1+z2=0z_1 + z_2 = 0, then z3+z4=0z_3 + z_4 = 0. In both cases the numbers z1,z2,z3,z4z_1, z_2, z_3, z_4 form two pair of equal sum, hence the conclusion.

b) Let z1=cos⁡x+isin⁡xz_1 = \cos x + i \sin x, z2=cos⁡y+isin⁡yz_2 = \cos y + i \sin y, z3=cos⁡z+isin⁡zz_3 = \cos z + i \sin z and z4=cos⁡t+isin⁡tz_4 = \cos t + i \sin t to get z1+z2+z3+z4=0z_1 + z_2 + z_3 + z_4 = 0 and ∣z1∣=∣z2∣=∣z3∣=∣z4∣=1|z_1| = |z_2| = |z_3| = |z_4| = 1.
As before, the numbers z1,z2,z3,z4z_1, z_2, z_3, z_4 form two pairs of opposite numbers, so the same goes for numbers z12n+1,z22n+1,z32n+1,z42n+1z_1^{2n+1}, z_2^{2n+1}, z_3^{2n+1}, z_4^{2n+1}. Therefore z12n+1+z22n+1+z32n+1+z42n+1=0z_1^{2n+1} + z_2^{2n+1} + z_3^{2n+1} + z_4^{2n+1} = 0, implying the claim.

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