MHB Can You Solve This Triangle Cotangent Ratio Problem?

[anemone]

Here is this week's POTW:

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Given $a,\,b$ and $c$ are sides of $\triangle ABC$ such that $9a^{2}+9b^{2}=19c^{2}$.

Evaluate $\dfrac{\cot C}{\cot A+\cot B}$.

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[anemone]

Congratulations to the following members for their correct solution:

1. lfdahl
2. BAdhi
3. kaliprasad
4. Opalg

Solution from BAdhi:

Spoiler

Lets rearrange the required result,

\begin{align*}\displaystyle
\frac{\cot C}{\cot A + \cot B} =& \frac{\cos C}{\sin C(\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B})}\\
=& \frac{\cos C \sin A \sin B}{\sin C ( \cos A \sin B + \sin A \cos B)} \\
=& \frac{\cos C \sin A \sin B}{\sin C [ \sin (A+B) ]} \\
\end{align*}

using the property,

$$\sin(A + B) = \sin(\pi - C)$$

we get

\begin{align*}\displaystyle
\frac{\cot C}{\cot A + \cot B}
=& \frac{\cos C \sin A \sin B}{\sin C \sin (\pi - C) } \\
=& \frac{\cos C \sin A \sin B}{\sin^2C } \\
=& \cos C \frac{\sin A }{\sin C}\frac{\sin B}{\sin C}\\
\end{align*}

Applying the sin and cosine rules,

$$\frac{ \sin A }{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

The expression could be reduced as following,

\begin{align*}\displaystyle
\frac{\cot C}{\cot A + \cot B}
= & \cos C \left( \frac{a}{c} \right) \left( \frac{b}{c} \right) \\
=& \frac{(a^2 + b^2 - c^2)}{2ab}\frac{ab}{c^2}\\
=& \frac{1}{2} \left(\left[\frac ac\right]^2 + \left[\frac bc\right]^2 - 1\right) \\
\end{align*}

Now let's adjust the given property,

\begin{align*}\displaystyle
9a^2 + 9 b^2 =& 19c^2 \\
9\left[ \left(\frac a c\right)^2 + \left( \frac bc\right)^2\right] =& 19\\
\left[ \left(\frac a c\right)^2 + \left( \frac bc\right)^2\right] = & \frac{19}{9}
\end{align*}Using this to above expression we get,\begin{align*}\displaystyle
\frac{\cot C}{\cot A + \cot B}
=& \frac 1 2 \left(\frac{19}{9} - 1\right) \\
=& \frac 5 9
\end{align*}

Alternate solution from Opalg:

Spoiler

\[ \frac{\cot C}{\cot A + \cot B} = \frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}} = \frac{\cos C\sin A\sin B}{\sin C(\cos A\sin B + \cos B\sin A)} = \frac{\sin A\sin B\cos C}{\sin C\sin(A+B)} = \frac{\sin A\sin B\cos C}{\sin^2 C} \] (because $\sin C = \sin(\pi-C) = \sin(A+B)$).

By the sine rule, $\sin A$, $\sin B$, $\sin C$ are proportional to $a$, $b$, $c$. Therefore $\dfrac{\sin A\sin B}{\sin^2 C} = \dfrac {ab}{c^2}$.

By the cosine rule $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{\frac{19}9c^2 - c^2}{2ab} = \dfrac{10c^2}{18ab} = \dfrac{5c^2}{9ab}$.

Therefore $\dfrac{\cot C}{\cot A + \cot B} = \dfrac{ab}{c^2}\dfrac{5c^2}{9ab} = \dfrac59$.