Problem C. 855. (May 2006)

C. 855. We need a 60^o angle, but we have no other materials but a rectangular sheet of paper. We first fold the rectangle in half along one of its bimedians. Then fold up one corner along a line through an adjacent corner, so that the vertex be on the bimedian (as shown in the Figure). Show that the acute angle of the trapezium hence obtained is 60^o.

(5 pont)

Deadline expired on June 15, 2006.

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

Megoldás: Ha PCB $\angle$ = $\alpha$ , akkor PCB' $\angle$ = $\alpha$ , CPB $\angle$ =CPB' $\angle$ =90^o- $\alpha$ , CB'F $\angle$ =90^o-2 $\alpha$ , KB'P $\angle$ =2 $\alpha$ , vagyis B'KP $\angle$ =90^o- $\alpha$ , Ezért PKB' háromszög egyenlő szárú, B'P=B'K. Azonban a PCB' derékszögű háromszög átfogójának K a felezőpontja, így KB'=KP. Vagyis B'KP háromszög szabályos, P-nél 60^o-os szög van, így $\alpha$ =30^o. Ebből következik, hogy DCP $\angle$ =60^o.

Statistics:

170 students sent a solution.

5 points: 111 students.

4 points: 36 students.

3 points: 5 students.

1 point: 2 students.

0 point: 2 students.

Unfair, not evaluated: 14 solutionss.

Problems in Mathematics of KöMaL, May 2006

170 students sent a solution.
5 points:	111 students.
4 points:	36 students.
3 points:	5 students.
1 point:	2 students.
0 point:	2 students.
Unfair, not evaluated:	14 solutionss.

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