If the sides of an arbitrary triangle $ABC$ are unambiguously extended by their own lengths, then the triangle formed by the endpoints is seven times the original.
Let $A_{1} B=A B, B_{1} C=B C$ and $C_{1} A=C A$. Then

$$
\begin{gathered}
A A_{1} C_{1} \Delta=2 A B C_{1} \Delta=2 A B C \Delta \\
B B_{1} A_{1} \Delta=2 B C A_{1} \Delta=2 A B C \Delta \text { and } \\
C C_{1} B_{1} \Delta=2 C A B_{1} \Delta=2 A B C \Delta
\end{gathered}
$$

Therefore,

$$
A_{1} B_{1} C_{1} \Delta=A A_{1} C_{1} \Delta+B B_{1} A_{1} \Delta+C C_{1} B_{1} \Delta+A B C \Delta=7 A B C \Delta
$$

(Zoltán Jármay, Budapest.)