Let's simplify the expression

$$
K=\frac{\frac{1}{a+b}-\frac{2}{b+c}+\frac{1}{c+a}}{\frac{1}{b-a}-\frac{2}{b+c}+\frac{1}{c-a}}+\frac{\frac{1}{b+c}-\frac{2}{c+a}+\frac{1}{a+b}}{\frac{1}{c-b}-\frac{2}{c+a}+\frac{1}{a-b}}+\frac{\frac{1}{c+a}-\frac{2}{a+b}+\frac{1}{b+c}}{\frac{1}{a-c}-\frac{2}{a+b}+\frac{1}{b-c}}
$$

and calculate its value when \( a=5, b=7, c=9 \). How many operations (i.e., how many additions, subtractions, multiplications, and divisions) are performed when calculating \( K \) from the simplified form and from the original form? Also, examine the case when \( a=5, b=7, c=1 \)! What advantages did we see here in algebraic transformations?
To make the expression meaningful at first glance, we must assume that among the numbers $a, b, c$, no two have a sum or difference of 0, meaning $a, b, c$ are numbers with distinct absolute values. Let's denote the consecutive terms of $K$ as $K_{1}, K_{2}, K_{3}$. $K_{2}$ and $K_{3}$ can also be obtained from the preceding terms by replacing $a, b, c$ with $b, c, a$ respectively, so it is sufficient to simplify $K_{1}$ for now. The numerator and denominator of $K_{1}$, after simplifying the fractions, are

\[
\begin{aligned}
& \frac{(b+c)(c+a)-2(a+b)(c+a)+(a+b)+(b+c)}{(a+b)(b+c)(c+a)}=\frac{b^{2}+c^{2}-2 a^{2}}{(a+b)(b+c)(c+a)} \\
& \frac{(b+c)(c-a)-2(b-a)(c-a)+(b-a)(b+c)}{(b-a)(b+c)(c-a)}=\frac{b^{2}+c^{2}-2 a^{2}}{(b-a)(b+c)(c-a)}
\end{aligned}
\]

To ensure that the denominators of the latter, as well as $K_{2}$ and $K_{3}$, which can be derived from the above, are not zero, we must assume that

\[
b^{2}+c^{2}-2 a^{2} \neq 0, \quad c^{2}+a^{2}-2 b^{2} \neq 0, \quad a^{2}+b^{2}-2 c^{2} \neq 0
\]

Thus, simplifying by $b^{2}+c^{2}-2 a^{2}$,

\[
K_{1}=\frac{(b-a)(b+c)(c-a)}{(a+b)(b+c)(c+a)}
\]

(Simplifying by the factor $b+c$ would only be a superficial result, as this factor also appears in the further common denominator.) Now, using the mentioned regularity,

\[
K=\frac{(b-a)(b+c)(c-a)}{(a+c)(b+c)(c+a)}+\frac{(c-b)(c+a)(a-b)}{(b+c)(c+a)(a+b)}+\frac{(a-c)(a+b)(b-c)}{(c+a)(a+b)(b+c)}
\]

The denominators differ only in the order of the factors, so they are equal, and it is sufficient to form the sum of the numerators. Considering the expected simplification possibilities, it seems good to multiply out; in this process, each numerator, being the product of three binomials, gives $2^{3}=8$ terms. The first numerator can be written as:

\[
S_{1}=a^{2} b+a^{2} c+b^{2} c+b c^{2}-a b^{2}-a c^{2}-2 a b c
\]

Based on this,

\[
\begin{aligned}
& S_{2}=b^{2} c+b^{2} a+c^{2} a+c a^{2}-b c^{2}-b a^{2}-2 a b c \\
& S_{3}=c^{2} a+c^{2} b+a^{2} b+a b^{2}-c a^{2}-c b^{2}-2 a b c
\end{aligned}
\]

and their sum is

\[
S=a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b-6 a b c
\]

Without writing out the terms, it is clear that all terms of $S$ appear in the product of the common denominator, except for the $-6 a b c$ term. For example, $a^{2} b$ appears when forming $(a+b)(b+c)(c+a)$ by multiplying the first, first, and second terms of the consecutive parentheses. Multiplying the first terms of two parentheses and the second term of one gives a term of the form $x^{2} y$, and this happens in 6 different ways, just as the first 6 terms of $S$ are distinct. The remaining products of the denominator give $2 a b c$ - the product of the 3 first and the 3 second terms - so the sum $S$ differs from the denominator by $8 a b c$, being "smaller" by this amount:

\[
S=(a+b)(b+c)(c+a)-8 a b c
\]

and thus

\[
K=1-\frac{8 a b c}{(a+b)(b+c)(c+a)}
\]

The numbers $a=5, b=7, c=9$ have distinct absolute values; it is sufficient to calculate the value of $a^{2}+c^{2}-2 b^{2}$ among the expressions (1), which is clearly not 0, so $K$ is meaningful. From (2), $K=1 / 16=0.0625$, and to calculate this, we performed 3 additions, $2+3=5$ multiplications, 1 division, and 1 subtraction, totaling 10 operations.

Calculating $K$ in its original form would involve first forming 3 sums and 3 differences from $a, b, c$, then 6 reciprocals from these, and doubling the reciprocals of the sums; this totals 15 operations. The assembly of the 6 "large" numerators and denominators from the prepared "elements" would require 2-2 simplifications, followed by 3 divisions and 2 additions to complete the calculation. Altogether, this would be $15+12+3+2=32$ operations.

For the data set $a=5, b=7, c=1$, the "similar" value $b^{2}+c^{2}-2 a^{2}$ turns out to be 0, so $K$ is undefined.

The advantage of the algebraic transformations in the first example was that we could calculate the value of $K$ from the simpler form with significantly fewer operations, and in the second example, it saved us a considerable amount of unnecessary calculation. Of course, recognizing the symmetry of $K$ and the fact that $K$ was built from only 3 numbers also provided an advantage. Although we performed significant work with the transformations, the simpler form could be beneficial if applied multiple times.

Pollai Marion (Bp. V., Veres Pálné lg. II. o. t.)

Remarks. 1. The expression can also be written as

\[
K=\frac{a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}}{(a+b)(b+c)(c+a)}
\]

From this form, the value of $K$ can be obtained with 17 operations.

Kéry Gerzson (Sopron, Széchenyi I. g. I. o. t.)

2. Many solutions show 53 operations for calculating from the original form, not taking into account the repeated expressions.
3. Numerous solutions were not accepted due to the claim $0 / 0=0$. The incorrectness of this claim could have been indicated by the fact that the transformed form of $K$ also has a meaning for the second set of values and gives a non-zero value of $37 / 72$.