Finding $frac{cotgamma}{cot alpha+cotbeta}$, given $a^2+b^2=2019c^2$

Question

This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:

In a triangle with sides $a,b,c$ opposite angles $alpha,beta,gamma$, it is known that $$a^2+b^2=2019c^2$$ Find $$frac{cotgamma}{cotalpha+cotbeta} $$

Well, by the Sine Law we have $$sin^2alpha+sin^2beta=2019sin^2gamma$$ and by the Cosine Law, $$cosgamma=frac{a^2+b^2-c^2}{2ab} = frac{1009c^2}{ab}$$ I’m stuck here. I tried to convert everything in our target expression to sines and cosines, but that makes the expression more complicated. I guess we can use the fact that $cotgamma=-cot(alpha+beta)$.
How can you tackle this question? (also, apparently there are no worked solutions online)

Michael Rozenberg · Answer

$$frac{cotgamma}{cotalpha+cotbeta}=frac{cosgammasinalphasinbeta}{singammasin(alpha+beta)}=frac{cosgammasinalphasinbeta}{sin^2gamma}=$$
$$=frac{frac{a^2+b^2-c^2}{2ab}cdotfrac{a}{2R}cdotfrac{b}{2R}}{frac{c^2}{4R^2}}=frac{frac{2018c^2}{2ab}cdotfrac{ab}{4R^2}}{frac{ c^2}{4R^2}}=1009.$$

Toby Mak · Answer

Here is the complete solution from AoPS.
Continuing from Aqua's answer, we know that $sin A = frac{a}{2R}, sin B = frac{b}{2R}, sin C = frac{c}{2R}$. Furthermore, as you have found, $cos C = frac{1009c^2}{ab}$. Substituting these values in gives:
$$frac{cotgamma}{cot alpha+cotbeta} = frac{cosgamma sin alpha sin beta}{sin ^2gamma} = frac{frac{1009c^2}{ab} cdot frac{a}{2R} cdot frac{b}{2R}}{frac{c^2}{4R}} = frac{frac{1009c^2}{ab} cdot ab}{c^2} = frac{1009c^2}{c^2} = boxed{1009}.$$

Aqua · Answer

begin{align}
frac{cotgamma}{cot alpha+cotbeta} &= frac{cotgamma sin alpha sin beta}{cos alpha sin beta+sin alpha cosbeta}\
\
&= frac{cotgamma sin alpha sin beta}{sin (alpha +beta)}\
\
&= frac{cosgamma sin alpha sin beta}{sin ^2gamma}\
end{align}
Now it should not be difficult to finish.

Finding $frac{cotgamma}{cot alpha+cotbeta}$, given $a^2+b^2=2019c^2$

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