This question was uploaded on 29/12/21 on social media accounts.
Solution:
Given equation:
\[\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac a{bc}+\frac b{ac}\]
Using cosine rule:
\[\frac{2(b^2+c^2-a^2)}{2abc} + \frac{(a^2+c^2-b^2)}{2abc} + \frac{2(a^2+b^2-c^2)}{2abc} = \frac {a^2+b^2}{abc}\]
\[\Rightarrow 2(b^2+c^2-a^2) + (a^2+c^2-b^2) + 2(a^2+b^2-c^2) = 2(a^2+b^2)\]
\[\Rightarrow a^2+3b^2+c^2 = 2a^2+2b^2\]
\[\Rightarrow a^2 = b^2 + c^2\]
This implies the sides with lengths b and c are perpendicular to each other and the side with length a is hypotaneuse.
This also implies A = 90°
