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Geometry Difficulty 4.5 AIME Prove it Estonia

Given a triangle ABCABC where ∣BC∣=a|BC| = a, ∣CA∣=b|CA| = b and ∣AB∣=c|AB| = c, prove that the equality
1a+b+1b+c=3a+b+c \frac{1}{a+b} + \frac{1}{b+c} = \frac{3}{a+b+c}
holds if and only if ∠ABC=60∘\angle ABC = 60^\circ.

Solution

By finding the common denominator on the left hand side, transform the equation to (a+2b+c)(a+b+c)=3(a+b)(b+c)(a+2b+c)(a+b+c) = 3(a+b)(b+c). Expanding the brackets and simplifying gives b2=a2+c2−acb^2 = a^2 + c^2 - ac. Comparing the latter with the cosine law b2=a2+c2−2accos⁡βb^2 = a^2 + c^2 - 2ac \cos \beta, we see that the equality holds if and only if cos⁡β=12\cos \beta = \frac{1}{2}, i.e., β=60∘\beta = 60^\circ.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.