Mathematical and Physical Journal
for High Schools
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Problem C. 1161. (March 2013)

C. 1161. The real numbers a, b, c in the equation \frac{x-b-c}{a} +\frac{x-a-c}{b} +\frac{x-a-b}{c}=10 satisfy a+b+c=0, abc=-48, and bc+ac+ab=-28. Solve the equation.

(5 pont)

Deadline expired on April 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Az eredeti egyenlet így írható:

\(\displaystyle \frac{x+a}{a}+\frac{x+b}{b}+\frac{x+c}{c}=10,\)

\(\displaystyle \frac xa+\frac xb+\frac xc=7.\)

Vagyis: \(\displaystyle x=7:\left(\frac1a+\frac1b+\frac1c\right)=7\cdot\frac{abc}{bc+ac+ab}=7\cdot\frac{-48}{-28}=12\).

Statistics:

228 students sent a solution.
5 points:191 students.
4 points:12 students.
3 points:6 students.
2 points:11 students.
1 point:3 students.
0 point:3 students.
Unfair, not evaluated:2 solutionss.

Problems in Mathematics of KöMaL, March 2013