Chứng minh nếu trong `ΔABC` có `(a.cosA+b.cosB+c.cosC)/(a.sinB+b.sinC+c.sinA)=(2p)/(9R)` thì `ΔABC` đều (trong đó `p` là nửa chu vi, `R` là bán kính đường tròn ngoại tiếp `ΔABC`)

Áp dụng định lý hàm số sine:

$\begin{array}{l}
a = 2R\sin A,b = 2R\sin B,c = 2R\sin C\\
\sin A = \dfrac{a}{{2R}},\sin B = \dfrac{b}{{2R}},\sin C = \dfrac{c}{{2R}}
\end{array}$

$\begin{array}{l} VT = \dfrac{{a\cos A + b\cos B + c\cos C}}{{a\sin B + b\sin C + c\sin A}}\\  = \dfrac{{2R\sin A\cos A + 2R\sin B\cos B + 2R\sin C\cos C}}{{a.\dfrac{b}{{2R}} + b.\dfrac{c}{{2R}} + c.\dfrac{a}{{2R}}}}\\  = \dfrac{{R\left( {2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C} \right)}}{{\dfrac{1}{{2R}}\left( {ab + bc + ca} \right)}}\\  = \dfrac{{R\left( {\sin 2A + \sin 2B + \sin 2C} \right)}}{{\dfrac{1}{{2R}}\left( {ab + bc + ca} \right)}} = \dfrac{{2{R^2}\left( {\sin 2A + \sin 2B + \sin 2C} \right)}}{{\left( {ab + bc + ca} \right)}} \end{array}$

Cần chứng minh

$\begin{array}{l} \sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\\ C/m:\\ \sin 2A + \sin 2B + \sin 2C\\  = 2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) + \sin 2C\\  = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C\\  = 2\sin \left( {{{180}^o} - C} \right)\cos \left( {A - B} \right) + 2\sin C\cos C\\  = 2\sin C\cos \left( {A - B} \right) + 2\sin C\cos C\\  = 2\sin C\left[ {\cos \left( {A - B} \right) + \cos C} \right]\\  = 2\sin C\left( {2\cos \left( {\dfrac{{A - B + C}}{2}} \right)\cos \left( {\dfrac{{A - B - C}}{2}} \right)} \right)\\  = 4\sin C\left[ {\cos \left( {\dfrac{{A + C + B}}{2} - B} \right)\cos \left( {A - \dfrac{{B + C + A}}{2}} \right)} \right]\\  = 4\sin C\left[ {\cos \left( {{{90}^o} - B} \right)\cos \left( {A - {{90}^o}} \right)} \right]\\  = 4\sin A\sin B\sin C \end{array}$

$\begin{array}{l}
 \Rightarrow VT = \dfrac{{8{R^2}\sin A\sin B\sin C}}{{ab + bc + ca}}\\
 = \dfrac{{8{R^2}.\dfrac{{abc}}{{8{R^3}}}}}{{ab + bc + ca}} = \dfrac{{abc}}{{R\left( {ab + bc + ca} \right)}}\\
VP = \dfrac{{2p}}{{9R}} = \dfrac{{a + b + c}}{{9R}}\\
VT = VP \Leftrightarrow \dfrac{{abc}}{{R\left( {ab + bc + ca} \right)}} = \dfrac{{a + b + c}}{{9R}}\\
 \Leftrightarrow \dfrac{{abc}}{{ab + bc + ca}} = \dfrac{{a + b + c}}{9}\\
 \Leftrightarrow \left( {a + b + c} \right)\left( {ab + bc + ca} \right) = 9abc\left( * \right)\\
AM - GM:a + b + c \ge 3\sqrt[3]{{abc}}\\
ab + bc + ca \ge 3\sqrt[3]{{{{\left( {abc} \right)}^2}}}\\
 \Rightarrow \left( {a + b + c} \right)\left( {ab + bc + ca} \right) \ge 9abc\\
\left( * \right) \Rightarrow ' = ' \Leftrightarrow a = b = c
\end{array}$

Từ đó suy ra tam giác $ABC$ đều.