A Cyclic Absolute Chain

★ 9/10Absolute ValuealgebrasequencesCasework

Problem

Real numbers a1,a2,…,a2017a_1, a_2, \ldots, a_{2017} satisfy a1+a2+⋯+a2017=0a_1 + a_2 + \cdots + a_{2017} = 0 and

∣a1−2a2∣=∣a2−2a3∣=⋯=∣a2016−2a2017∣=∣a2017−2a1∣.|a_1 - 2a_2| = |a_2 - 2a_3| = \cdots = |a_{2016} - 2a_{2017}| = |a_{2017} - 2a_1|.

Prove that a1=a2=⋯=a2017=0a_1 = a_2 = \cdots = a_{2017} = 0.

Answer

Solution

Difficulty9/10
TopicsAbsolute Value, algebra, sequences, Casework

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