About the equality $\dfrac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt3$ in a triangle. American Mathematical Problems. Among my old notes, I see this problem that seems nice to me. It was proposed by W. J. Blunden in American  Mathematical  Monthly several decades ago. I solved it but I did not write down the date of the journal nor the solution. I post it in MSE now hoping it’s interesting for some students. 
If $\dfrac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt3$  is fulfilled in triangle $\triangle ABC$ then at least one of the angles measures $60^{\circ}$.
 A: Hint:
$$\sin A-\sqrt3\cos A=2\sin\left(A-60^\circ\right)$$
Let $A-60^\circ=2x$  etc.  $\implies x+y+z=0,x+y=-z$
$$\sin2x+\sin2y+\sin2z$$
$$=2\sin(x+y)\cos(x-y)+2\sin z\cos z$$
$$=2\sin(-z)\cos(x-y)+2\sin z\cos(-(x+y))$$
$$=-2\sin z(2\sin x\sin y)$$
A: In the standard notation we obtain:
$$\frac{\sum\limits_{cyc}\frac{2S}{bc}}{\sum\limits_{cyc}\frac{b^2+c^2-a^2}{2bc}}=\sqrt3$$ or
$$4S(a+b+c)=\sqrt3\sum_{cyc}(a^2b+a^2c-a^3)$$ or
$$(a+b+c)^3\prod_{cyc}(a+b-c)=3\left(\sum_{cyc}(a^2b+a^2c-a^3)\right)^2$$ or
$$(a^2+b^2-ab-c^2)(a^2+c^2-ac-b^2)(b^2+c^2-bc-a^2)=0$$ and we are done!
A: 
If \begin{align}\dfrac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt3 \tag{1}\label{1}\end{align}
  is fulfilled in triangle $\triangle ABC$ then at least one of the angles measures $60^\circ$.

Let $\rho,\ r$ and $R$ be the semiperimeter and the radii of inscribed and circumscribed circle, respectively.
Then
\begin{align}
\sin A+\sin B+\sin C&=\frac{\rho}R
,\\
\cos A+\cos B+\cos C&=\frac rR+1
\\
\text{and \eqref{1} becomes}\quad
\frac{\rho}{r+R} &= \sqrt3
\tag{2}\label{2}
,
\end{align}
so we can express $\rho$ in terms of $r$ and $R$ for this case as
\begin{align}
\rho&=\sqrt3(r+R)
\tag{3}\label{3}
.
\end{align}
Without loss of generality, we can scale \eqref{3} by $\tfrac1R$
and proceed with new scaled  $\rho,r$ for the case $R=1$,
\begin{align}
\rho&=\sqrt3(r+1)
\tag{4}\label{4}
.
\end{align}
Using a known equation, which binds 
$\tan\tfrac A2$, $\tan\tfrac B2$, $\tan\tfrac C2$
as roots of the cubic for $R=1$,
\begin{align}
x^3-\frac{4+r}\rho\,x^2+x-\frac r\rho&=0
\tag{5}\label{5}
\end{align}
combined with \eqref{4},
we have
\begin{align}
x^3-\frac{\sqrt3\,(r+4)}{3\,(r+1)}\,x^2
+x-\frac{r\,\sqrt3}{3\,(r+1)}&=0
,\\
\frac{((r+1)\,x^2-\sqrt3\,x+r)(3\,x-\sqrt3)}
{3\,(r+1)}
&=0
\tag{6}\label{6}
,
\end{align}
so one root, say, $\tan\tfrac A2=\tfrac{\sqrt3}3$,
in other words, one of the angles must be equal
$2\,\arctan(\tfrac{\sqrt3}3)=60^\circ$.
