For real numbers $a, b, c, d$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6$, and $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$ hold. What can be the value of $\frac{a}{b}+\frac{c}{d}$?
Let $p=\frac{a}{b}, q=\frac{b}{c}, r=\frac{c}{d}$ and $s=\frac{d}{a}$. Then the two equations are:

$$
p+q+r+s=6 \quad \text{and} \quad q+qr+rs+sp=8
$$

The second equation can be factored as: $(p+r)(q+s)=8$. By substituting $p+r=x$ into the first equation, we get $q+s=6-x$, so from the second equation, $x(6-x)=8$, which simplifies to $-x^{2}+6x-8=0$, thus

$$
\frac{a}{b}+\frac{c}{d}=x=\frac{-6 \pm \sqrt{6^{2}-4 \cdot(-1) \cdot(-8)}}{2 \cdot(-1)}=\frac{-6 \pm 2}{-2}
$$

Therefore, the value of the desired sum can be 2 or 4.

Of course, our solution is only complete if we can show numbers $a, b, c, d$ for which the desired sum is indeed 2 or 4. Let $a=b=1, c=d=2-\sqrt{3}$ in the first case, and $a=d=2+\sqrt{3}, b=c=1$ in the second case.

David Papp (Budapest, Szent István Gymnasium, 10th grade)