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Geometry Difficulty 5.7 AIME, harder Prove it Mongolia

Let α,β,γ\alpha, \beta, \gamma be the angles opposite the sides a,b,ca, b, c of a triangle, respectively. Prove that if the lengths of a,b,ca, b, c form an arithmetic progression in this order, then the values cos⁡α,1−cos⁡β,cos⁡γ\cos \alpha, 1 - \cos \beta, \cos \gamma also form an arithmetic progression in that order.
(Otgonbayar Uuye)

Solution

By the Law of Sines, the values sin⁡α,sin⁡β,sin⁡γ\sin \alpha, \sin \beta, \sin \gamma form an arithmetic progression. Therefore
2cos⁡β2cos⁡γ−α2=2sin⁡α+γ2cos⁡γ−α2=sin⁡α+sin⁡γ=2sin⁡β=4sin⁡β2cos⁡β2. 2 \cos \frac{\beta}{2} \cos \frac{\gamma - \alpha}{2} = 2 \sin \frac{\alpha + \gamma}{2} \cos \frac{\gamma - \alpha}{2} = \sin \alpha + \sin \gamma = 2 \sin \beta = 4 \sin \frac{\beta}{2} \cos \frac{\beta}{2}.
Since cos⁡β2>0\cos \frac{\beta}{2} > 0, it follows cos⁡γ−α2=2sin⁡β2\cos \frac{\gamma - \alpha}{2} = 2 \sin \frac{\beta}{2}. Then
cos⁡α+cos⁡γ=2cos⁡α+γ2cos⁡γ−α2=4sin⁡2β2=2(1−cos⁡β). \cos \alpha + \cos \gamma = 2 \cos \frac{\alpha + \gamma}{2} \cos \frac{\gamma - \alpha}{2} = 4 \sin^2 \frac{\beta}{2} = 2(1 - \cos \beta).
Thus, cos⁡α,1−cos⁡β,cos⁡γ\cos \alpha, 1 - \cos \beta, \cos \gamma are in arithmetic progression.

Alternatively, by the Law of Cosines, we have
cos⁡α+cos⁡γ−2(1−cos⁡β)=b2+c2−a22bc+a2+b2−c22ab−2(1−a2+c2−b22ac)=(a+b−c)(b+c−a)2abc(a+c−2b)=0. \begin{aligned} \cos \alpha + \cos \gamma - 2(1 - \cos \beta) &= \frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + b^2 - c^2}{2ab} - 2\left(1 - \frac{a^2 + c^2 - b^2}{2ac}\right) \\ &= \frac{(a+b-c)(b+c-a)}{2abc}(a+c-2b) = 0. \end{aligned}

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