About the equality $\dfrac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt3$ in a triangle. American Mathematical Problems.

Question

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 $60^{\circ}$.

Answer 1

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 $2\,\arctan(\tfrac{\sqrt3}3)=60^\circ$.

Answer 2

Hint:

$$\sin A-\sqrt3\cos A=2\sin\left(A-60^\circ\right)$$

Let $A-60^\circ=2x$ etc. $\implies x+y+z=0,x+y=-z$

$$\sin2x+\sin2y+\sin2z$$ $$=2\sin(x+y)\cos(x-y)+2\sin z\cos z$$ $$=2\sin(-z)\cos(x-y)+2\sin z\cos(-(x+y))$$ $$=-2\sin z(2\sin x\sin y)$$

Answer 3

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!