Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem K. 384. (September 2013)

K. 384. Triangle ABC has an obtuse angle at vertex A. Denote the centre of the inscribed circle by S. The line drawn through S parallel to AB intersects side AC at D and side BC at E. Prove that DE=AD+BE.

German competition problem

(6 pont)

Deadline expired on October 10, 2013.

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

Megoldás. \(\displaystyle BAS\angle=DSA\angle\), mert váltószögek. \(\displaystyle DAS\angle=BAS\angle\), mert \(\displaystyle AS\) szögfelező. Így \(\displaystyle DSA\angle=DAS\angle\), vagyis az \(\displaystyle ADS\) háromszög egyenlő szárú. Így \(\displaystyle AD=DS\).

Hasonlóan lehet belátni, hogy \(\displaystyle BE=SE\). Így pedig \(\displaystyle DE=DS+SE=AD+BE\).


202 students sent a solution.
6 points:160 students.
5 points:18 students.
4 points:4 students.
3 points:8 students.
2 points:1 student.
0 point:9 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, September 2013