Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem K. 236. (January 2010)

K. 236. The intersection of two interior angle bisectors of a parallelogram lies on a side of the parallelogram. In what ratio does the point of intersection divide the side?

(6 pont)

Deadline expired on February 10, 2010.

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

Megoldás. Az \(\displaystyle ABCD\) paralelogramma két szögfelezőjének metszéspontja legyen \(\displaystyle P\) a \(\displaystyle CD\) oldalon. \(\displaystyle \alpha /2 = BAP\angle = APD\angle\), mert váltószögek, ezért \(\displaystyle APD\triangle\) egyenlőszárú. Ugyanígy \(\displaystyle \beta /2 = ABP\angle = BPC\angle\), ezért \(\displaystyle BCP\triangle\) is egyenlőszárú. Tehát \(\displaystyle PD=DA=BC=CP\): \(\displaystyle P\) felezőpontja \(\displaystyle CD\) oldalnak.


148 students sent a solution.
6 points:86 students.
5 points:39 students.
4 points:8 students.
3 points:3 students.
2 points:2 students.
1 point:7 students.
0 point:1 student.
Unfair, not evaluated:2 solutionss.

Problems in Mathematics of KöMaL, January 2010