Problem B. 4033. (November 2007)

B. 4033. All the participants in a dance course are married couples. Someone noticed that the difference between the heights of the two members of each couple is at most 10 cm. The dance instructor asks both the men and the women to stand in a line in increasing order of heights. Dancing couples are made by this rule: The tallest woman dances with the tallest man, and so on. Prove that the differences between the members of every dancing couple are also at most 10 cm.

(4 pont)

Deadline expired on December 17, 2007.

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

Megoldás: Legyenek a táncospárok (N_i,F_i), ahol 1 $\le$ i $\le$ $\ell$ , az N_i nő magassága cm-ekben mérve n_i, az F_i férfié f_i, továbbá $n_1\le n_2\le \ldots\le n_\ell$ és $f_1\le f_2\le \ldots\le f_\ell$ .

Tegyük fel, hogy az állítás nem igaz, ekkor van olyan 1 $\le$ i $\le$ $\ell$ , amelyre n_i<f_i-10 vagy f_i<n_i-10. Szimmetria okok miatt elég az első esettel foglalkoznunk. Ekkor minden 1 $\le$ j $\le$ i $\le$ k $\le$ $\ell$ esetén n_j $\le$ n_i<f_i-10 $\le$ f_k-10, vagyis F_k nem lehet az N_j házastársa. Tehát az $N_1,\ldots, N_i$ nők házastársa az $F_1,\ldots,F_{i-1}$ férfiak között keresendő, ami a skatulya-elv miatt nem lehetséges; indirekt feltevésünk ellentmondásra vezetett.

Statistics:

149 students sent a solution.

4 points: 96 students.

3 points: 4 students.

2 points: 11 students.

1 point: 9 students.

0 point: 27 students.

Unfair, not evaluated: 2 solutionss.

Problems in Mathematics of KöMaL, November 2007

149 students sent a solution.
4 points:	96 students.
3 points:	4 students.
2 points:	11 students.
1 point:	9 students.
0 point:	27 students.
Unfair, not evaluated:	2 solutionss.

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