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 \(\displaystyle DE=DS+SE=AD+BE\).
Statistics:
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 solutionss.
Problems in Mathematics of KöMaL, September 2013