Mathematical and Physical Journal
for High Schools
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Problem C. 855. (May 2006)

C. 855. We need a 60o 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 60o.

(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=90o-\alpha, CB'F\angle=90o-2\alpha, KB'P\angle=2\alpha, vagyis B'KP\angle=90o-\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 60o-os szög van, így \alpha=30o. Ebből következik, hogy DCP\angle=60o.


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 solutions.

Problems in Mathematics of KöMaL, May 2006