Atrisinājums

Aprēķināsim \(S_{A_{1}B_{1}C_{1}}\) (skat. 4.zīm.).

\(S_{A_{1}B_{1}C_{1}}=S_{ABC}-S_{A_{1}B_{1}C}-S_{AB_{1}C_{1}}-S_{A_{1}BC_{1}} \cdot \Delta A_{1}B_{1}C\) augstums pret malu \(A_{1}C\) ir \(\frac{1}{7} \frac{1}{7}\) no \(\triangle ABC\) augstuma pret malu \(BC\). \(A_{1}C=\frac{6}{7} BC\). Tātad \(S_{A_{1}B_{1}C}=\frac{6}{49} S_{ABC}\). Līdzīgi arī \(S_{AB_{1}C_{1}}=S_{A_{1}BC_{1}}=\frac{6}{49} S_{ABC}\). Tātad \(S_{A_{1}B_{1}C_{1}}=\frac{31}{49} S_{ABC}\).

Līdzīgi var aprēķināt \(S_{A_{2}B_{2}C_{2}}=S_{ABC}-S_{A_{2}B_{2}C}-S_{AB_{2}C_{2}}-S_{A_{2}BC_{2}}\). \(\Delta A_{2}B_{2}C\) augstums pret malu \(A_{2}C\) ir \(\frac{2}{7}\) no \(\triangle ABC\) augstuma pret malu \(BC\). \(A_{2}C=\frac{5}{7} BC\). Tātad \(S_{A_{2}B_{2}C}=\frac{10}{49} S_{ABC}\). Līdzīgi, arī \(S_{AB_{2}C_{2}}=S_{A_{2}BC_{2}}=\frac{10}{49} S_{ABC}\). Tātad \(S_{A_{2}B_{2}C_{2}}=\frac{19}{49} S_{ABC}\) un \(S_{A_{1}B_{1}C_{1}}+S_{A_{2}B_{2}C_{2}}=\frac{50}{49} S_{ABC}>S_{ABC}\).