Ha az $(A B C D)$ symbolum az $\frac{A C}{B C}: \frac{A D}{B D}$ kifejezést jelenti, mily értékei lesznek e symbolumnak, ha benne az $A, B, C$ és $D$ betúket permutáljuk?

If the symbol $(A B C D)$ represents the expression $\frac{A C}{B C}: \frac{A D}{B D}$, what values will this symbol take if the letters $A, B, C$, and $D$ are permuted?
Let $(A B C D)=k_{1}$, then

$$
(A B C D)=\frac{A C}{B C}: \frac{A D}{B D}=k_{1}
$$

$$
\begin{aligned}
& (A B D C)=\frac{A D}{B D}: \frac{A C}{B C}=\frac{1}{k_{1}} \\
& (A C B D)=\frac{A B}{C B}: \frac{A D}{C D}=k_{2} \\
& (A C D B)=\frac{A D}{C D}: \frac{A B}{C B}=\frac{1}{k_{2}} \\
& (A D B C)=\frac{A B}{D B}: \frac{A C}{D C}=k_{3} \\
& (A D C B)=\frac{A C}{D C}: \frac{A B}{D B}=\frac{1}{k_{3}} \\
& (B A C D)=\frac{B C}{A C}: \frac{B D}{A D}=\frac{1}{k_{1}} \\
& (B A D C)=\frac{B D}{A D}: \frac{B C}{A C}=k_{1} \\
& (B C A D)=\frac{B A}{C A}: \frac{B D}{C D}=k_{3} \\
& (B C D A)=\frac{B D}{C D}: \frac{B A}{C A}=\frac{1}{k_{3}} \\
& (B D A C)=\frac{B A}{D A}: \frac{B C}{D C}=k_{2} \\
& (B D C A)=\frac{B C}{D C}: \frac{B A}{D A}=\frac{1}{k_{2}} \\
& (C A B D)=\frac{C B}{A B}: \frac{C D}{A D}=\frac{1}{k_{2}} \\
& (C A D B)=\frac{C D}{A D}: \frac{C B}{A B}=k_{2} \\
& (C B A D)=\frac{C A}{B A}: \frac{C D}{B D}=\frac{1}{k_{3}} \\
& (C B D A)=\frac{C D}{B D}: \frac{C A}{B A}=k_{3} \\
& (C D A B)=\frac{C A}{D A}: \frac{C B}{D B}=k_{1} \\
& (D D B A)=\frac{C B}{D B}: \frac{C A}{D A}=\frac{1}{k_{1}} \\
& (D A B C)=\frac{D B}{A B}: \frac{D C}{A C}=\frac{1}{k_{3}} \\
& (D A C B)=\frac{D C}{A C}: \frac{D B}{A B}=k_{3} \\
& (D B A C)=\frac{D A}{B A}: \frac{D C}{B C}=\frac{1}{k_{2}} \\
& (D B C A)=\frac{D C}{B C}: \frac{D A}{B A}=k_{2} \\
& (D C A B)=\frac{D A}{C A}: \frac{D B}{C B}=\frac{1}{k_{1}} \\
& (D B A): \frac{D A}{C A}=k_{1}
\end{aligned}
$$

We see that the values of the symbol are equal in fours, so the symbol has only 6 different values; i.e., $k_{1}, k_{2}, k_{3}$ and $\frac{1}{k_{1}}, \frac{1}{k_{2}}, \frac{1}{k_{3}}$.

But $k_{2}$, i.e.,

$$
\begin{gathered}
\frac{A B}{C B}: \frac{A D}{C D}=\frac{C B-C A}{C B} \cdot \frac{C A+A D}{A D}= \\
=\frac{C B \cdot C A-C A^{2}+C B \cdot A D-C A \cdot A D}{C B \cdot A D}= \\
=1-\frac{C A(C A-C B+A D)}{C B \cdot A D}= \\
=1-\frac{C A(B A+A D)}{C B \cdot A D}=1-\frac{C A \cdot B D}{C B \cdot A D} \\
=1-\frac{C A}{D A}: \frac{C B}{D B}=1-k_{1}
\end{gathered}
$$

And $k_{3}$, i.e.,

$$
\begin{gathered}
\frac{A B}{D B}: \frac{A C}{D C}=\frac{D B-D A}{D B} \cdot \frac{D A+A C}{A C}= \\
=\frac{D B \cdot D A-D A^{2}+D B \cdot A C-D A \cdot A C}{D B \cdot A C}= \\
=1-\frac{D A(D A-D B+A C)}{D B \cdot A C}=1-\frac{D A(B A+A C)}{D B \cdot A C}= \\
1-\frac{D A \cdot B C}{D B \cdot A C}=1-\frac{A D}{A C}: \frac{B D}{B C} \\
=1-\frac{1}{k_{1}}=\frac{k_{1}-1}{k_{1}}
\end{gathered}
$$

Therefore, the $(A B C D)$ symbol, when the letters are permuted, takes the following 6 values: $k_{1}, \frac{1}{k_{1}}, \frac{1}{1-k_{1}}, \frac{k_{1}-1}{k_{1}}$ and $\frac{k_{1}}{k_{1}-1}$.

(Kálmán Szartorisz, 8th grade student, Losoncz.) The problem was also solved by: Viktor Hónig, 8th grade student, Kaposvár.