Mathematical and Physical Journal
for High Schools
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Problem K. 266. (November 2010)

K. 266. Let a, b, c, d denote positive integers. Given that \frac{a}{b}<\frac{c}{d}, show that \frac{a+c}{b+d} always lies between \frac{a}{b} and \frac{c}{d}.

(6 pont)

Deadline expired on December 10, 2010.

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

Megoldás. Ha \(\displaystyle \frac ab < \frac cd\), akkor \(\displaystyle ad<bc\), továbbá \(\displaystyle \frac{a+c}{b+d}-\frac ab=\frac{ab+cb-ab-ad}{b(b+d)}>0\) és \(\displaystyle \frac{cb-ad}{b^2+bd}<\frac{cb-ad}{bd}=\frac ab - \frac cd\). Az egyenlőtlenség-sorozatból következik a bizonyítandó állítás: \(\displaystyle \frac ab < \frac{a+c}{b+d} < \frac cd\).


170 students sent a solution.
6 points:113 students.
5 points:25 students.
4 points:5 students.
3 points:2 students.
2 points:2 students.
1 point:5 students.
0 point:12 students.
Unfair, not evaluated:6 solutionss.

Problems in Mathematics of KöMaL, November 2010