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: $\displaystyle \frac ab < \frac{a+c}{b+d} < \frac cd$.

Statistics:

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

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.

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