Problem K. 188. (December 2008)

K. 188. The base AB of an isosceles triangle ABC is extended beyond vertex B by the length of the leg. The endpoint obtained is C₁. A perpendicular is erected onto the base AB at vertex A. C₂ is the point on the perpendicular that lies in the same half plane as vertex C, at a distance from A that equals the length of the leg. Given that the points C₁, C and C₂ are collinear, find the angles of triangle ABC.

(6 pont)

Deadline expired on January 10, 2009.

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

Megoldás. A vázlatrajzon bejelöljük a szögeket. CC₁B szög legyen $\alpha$ , CC₂A szög legyen $\beta$ . Mivel CC₁B és CC₂A háromszögek egyenlő szárúak, ezért a C pontnál az ábrán jelölt módon megjelenik $\alpha$ és $\beta$ . Mivel C₁, C és C₂ pontok egy egyenesre illeszkednek, ezért a C-nél 180^o-os szög van. C₁AC₂ háromszög derékszögű, így $\alpha$ + $\beta$ =90^o, tehát az ABC háromszög C-nél levő szöge is 90^o-os. Az ABC háromszög egyenlő szárú, tehát szögei: 45^o, 45^o, 90^o.

Statistics:

166 students sent a solution.

6 points: 110 students.

5 points: 14 students.

4 points: 6 students.

3 points: 5 students.

2 points: 7 students.

1 point: 4 students.

0 point: 13 students.

Unfair, not evaluated: 7 solutionss.

Problems in Mathematics of KöMaL, December 2008

166 students sent a solution.
6 points:	110 students.
5 points:	14 students.
4 points:	6 students.
3 points:	5 students.
2 points:	7 students.
1 point:	4 students.
0 point:	13 students.
Unfair, not evaluated:	7 solutionss.

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