Mathematical and Physical Journal
for High Schools
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Problem B. 4867. (April 2017)

B. 4867. The sum of the real numbers \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) and \(\displaystyle d\) is \(\displaystyle 0\). Let \(\displaystyle M=ab+bc+cd\) and \(\displaystyle N=ac+ad+bd\). Prove that at least one of the sums \(\displaystyle 20M+17N\) and \(\displaystyle 20N+17M\) is non-positive.

(Bulgarian problem)

(4 pont)

Deadline expired on May 10, 2017.


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

Megoldás. A \(\displaystyle 20M+17N\) és \(\displaystyle 20N+17M\) számok összege

\(\displaystyle 37(M+N)=37(ab+ac+ad+bc+bd+cd)=\)

\(\displaystyle =37\cdot \frac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{2}=-\frac{37}{2}(a^2+b^2+c^2+d^2)\leq 0,\)

így legalább az egyikük nem pozitív.


Statistics:

84 students sent a solution.
4 points:78 students.
3 points:2 students.
2 points:1 student.
1 point:1 student.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, April 2017