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Equality in regular (2^n-1)-gon

Source: Kvant Magazine No. 11-12 2021 M2680

March 9, 2023
geometryKvant

Problem Statement

Let n>1n>1 be a natural number and A0A1…A2n−2A_0A_1\ldots A_{2^n-2} be a regular polygon. Prove that 1A0A1=1A0A2+1A0A4+1A0A8+⋯+1A0A2n−1.\frac{1}{A_0A_1}=\frac{1}{A_0A_2}+\frac{1}{A_0A_4}+\frac{1}{A_0A_8}+\cdots+\frac{1}{A_0A_{2^{n-1}}}.Proposed by Le Hoang and Ngoc Thai (Vietnam)
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