Ratio of triangles areas

In an equilateral triangle ABC, the point T is its center of gravity, the point R is the image of the point T in axial symmetry along the line AB, and the point N is the image of the point T in axial symmetry along the line BC. Find the ratio of the areas of the triangles ABC and TRN.

Correct answer:

r =  3:1

Step-by-step explanation:

$$\begin{gathered} a=1 \\ v=\sqrt{3}/2\cdot a=\sqrt{3}/2\cdot 1\doteq 0.866 \\ {S}_1=\frac{a\cdot v}{2}=\frac{1\cdot 0.866}{2}\doteq 0.433 \\ {t}_1:t_2=2:1 \\ {t}_1+t_2=t=v \\ {t}_2=\frac{1}{3}\cdot v=\frac{1}{3}\cdot 0.866\doteq 0.2887 \\ TR=2\cdot t_2=2\cdot 0.2887\doteq 0.5774 \\ TN=TR=0.5774\doteq 0.5774 \\ \angle BTN=60{}^{\circ } \\ \cos BTN=v_2:TN \\ {v}_2=TN\cdot \cos BTN=TN\cdot \cos 60°=0.5774\cdot \cos 60°=0.5774\cdot 0.5=0.28868 \\ T{N}^2=(RN/2{)}^2+v_2^2 \\ RN=2\cdot \sqrt{T{N}^2-v_2^2}=2\cdot \sqrt{0.5774^2-0.2887^2}=1 \\ {S}_2=\frac{RN\cdot v_2}{2}=\frac{1\cdot 0.2887}{2}\doteq 0.1443 \\ r=S_1/S_2=0.433/0.1443=3=3:1 \end{gathered}$$

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