Problem 1193

Solution:

We will prove that the areas of triangles $A_{1} C C_{1}$ and $A_{2} B^{\prime} B_{2}$ are equal. By the Law of Sines, we have

$$\frac{A B}{\sin \angle A_{1} C C_{1}}=\frac{A B}{\sin \angle B C A}=\frac{A C}{\sin \angle A B C}=\frac{A^{\prime} C^{\prime}}{\sin \angle A^{\prime} B^{\prime} C^{\prime}}=\frac{A^{\prime} C^{\prime}}{\sin \angle A_{2} B^{\prime} B_{2}}$$

from which $A B \cdot \sin \angle A_{2} B^{\prime} B_{2}=A^{\prime} C^{\prime} \cdot \sin \angle A_{1} C C_{1}$, which is equivalent to $B_{2} B^{\prime} \cdot \sin \angle A_{2} B^{\prime} B_{2}=C C_{1} \cdot \sin \angle A_{1} C C_{1}$.

Then $S_{A_{1} C C_{1}}=\frac{1}{2} C A_{1} \cdot C C_{1} \cdot \sin \angle A_{1} C C_{1}=\frac{1}{2} A_{2} B^{\prime} \cdot B_{2} B^{\prime} \cdot \sin \angle A_{2} B^{\prime} B_{2}=S_{A_{2} B^{\prime} B_{2}}$.

Similarly, it can be shown that $S_{B_{1} A A_{1}}=S_{B_{2} C^{\prime} C_{2}}$ and $S_{C_{1} B B_{1}}=S_{C_{2} A^{\prime} A_{2}}$.

In the end, we have

$$S_{A_{1} B_{1} C_{1}}=S_{A_{1} C C_{1}}+S_{B_{1} A A_{1}}+S_{C_{1} B B_{1}}+S_{A B C}=S_{A_{2} B^{\prime} B_{2}}+S_{B_{2} C^{\prime} C_{2}}+S_{C_{2} A^{\prime} A_{2}}+S_{A^{\prime} B^{\prime} C^{\prime}}=S_{A_{2} B_{2} C_{2}}$$

which is what we needed to prove.