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Geometry Difficulty 4.6 AIME Prove it North Macedonia

Prove that tg⁡αtg⁡β=a2+c2−b2b2+c2−a2\frac{\operatorname{tg}\alpha}{\operatorname{tg}\beta} = \frac{a^2+c^2-b^2}{b^2+c^2-a^2} for every triangle ABCABC (a,b,ca, b, c and α,β\alpha, \beta are the sides and the angles in the triangle, respectively).

Solution

From the cosine theorem, we have
b2=a2+c2−2accos⁡β,a2=b2+c2−2bccos⁡α, b^2 = a^2 + c^2 - 2ac \cos \beta, \quad a^2 = b^2 + c^2 - 2bc \cos \alpha,
Therefore a2+c2−b2=2accos⁡βa^2 + c^2 - b^2 = 2ac \cos \beta, b2+c2−a2=2bccos⁡αb^2 + c^2 - a^2 = 2bc \cos \alpha.
So a2+c2−b2b2+c2−a2=2accos⁡β2bccos⁡α=acos⁡βbcos⁡α=(∗)\frac{a^2+c^2-b^2}{b^2+c^2-a^2} = \frac{2ac \cos \beta}{2bc \cos \alpha} = \frac{a \cos \beta}{b \cos \alpha} = (*).
From the law of sine, we have
asin⁡α=bsin⁡β, i.e. ab=sin⁡αsin⁡β. Now (∗)=sin⁡αcos⁡βsin⁡βcos⁡α=tg⁡αtg⁡β. \frac{a}{\sin \alpha} = \frac{b}{\sin \beta}, \text{ i.e. } \frac{a}{b} = \frac{\sin \alpha}{\sin \beta}. \text{ Now } (*) = \frac{\sin \alpha \cos \beta}{\sin \beta \cos \alpha} = \frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.