A Chebyshev Criterion for Angle Multiples in a Triangle

The familiar condition A = 2B if and only if a^2 - b^2 = bc for a triangle is often treated as an isolated double-angle identity. We show that it is the first case of a uniform Chebyshev side equation. Let the angles of a nondegenerate triangle ABC be opposite the sides a, b, c, and set M = (a^2-b^2)/(bc) and u = (a^2+c^2-b^2)/(ac) = 2 cos B. For an integer k >= 2, if A = kB, then M = P_{k-1}(u), where P_0 = 0, P_1 = 1, and P_{n+1} = uP_n - P_{n-1} for n >= 1. Conversely, in the natural sector 0 < B < pi/(k+1), this same side equation is equivalent to A = kB or to the sine-reflected branch C = (k-3)B; for k = 2 and k = 3 the reflected branch is not a nondegenerate triangle. The cases k = 2 and k = 3 give, respectively, a^2-b^2 = bc and, in the sector 0 < B < pi/4, (a+b)(a-b)^2 = bc^2.

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Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2026-07-01
DOI
https://doi.org/10.5281/zenodo.21088833
Primary Topic
Mathematics and Applications
Type
article
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article

A Chebyshev Criterion for Angle Multiples in a Triangle

Zhongwei Liu
Zenodo (CERN European Organization for Nuclear Research)
Mathematics and Applications
article

A Chebyshev Criterion for Angle Multiples in a Triangle

Zhongwei Liu
article en

Abstract

The familiar condition A = 2B if and only if a^2 - b^2 = bc for a triangle is often treated as an isolated double-angle identity. We show that it is the first case of a uniform Chebyshev side equation. Let the angles of a nondegenerate triangle ABC be opposite the sides a, b, c, and set M = (a^2-b^2)/(bc) and u = (a^2+c^2-b^2)/(ac) = 2 cos B. For an integer k >= 2, if A = kB, then M = P_{k-1}(u), where P_0 = 0, P_1 = 1, and P_{n+1} = uP_n - P_{n-1} for n >= 1. Conversely, in the natural sector 0 < B < pi/(k+1), this same side equation is equivalent to A = kB or to the sine-reflected branch C = (k-3)B; for k = 2 and k = 3 the reflected branch is not a nondegenerate triangle. The cases k = 2 and k = 3 give, respectively, a^2-b^2 = bc and, in the sector 0 < B < pi/4, (a+b)(a-b)^2 = bc^2.

Zenodo (CERN European Organization for Nuclear Research)
Educational Department of Liaoning Province (CN)
Openalex Percentile: Top 4%
Mathematics and Applications
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