Problem B. 4534. (April 2013)

B. 4534. M and N are points on the longest side, AB of a triangle ABC, such that BM=BC and AN=AC. The parallel drawn through point M to the side BC intersects side AC at P, and the parallel drawn through point N to the side AC intersects side BC at Q. Prove that CP=CQ.

(Kvant)

(3 pont)

Deadline expired on May 10, 2013.

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

Megoldási ötlet: Párhuzamos szelők tétele.

Megoldás. Legyenek a háromszög oldalai BC=a, CA=b és AB=c. A párhuzamos szelők tételét alkamazva, $\frac{CP}{AC} = \frac{BM}{AB}$ , amiből

$CP = \frac{BM\cdot AC}{AB} = \frac{BC\cdot AC}{AB} = \frac{ab}{c}.$

Hasonlóan kapjuk, hogy $\frac{CQ}{CB} = \frac{AN}{AB}$ , és

$CQ = \frac{AN\cdot CB}{AB} = \frac{AC\cdot CB}{AB} = \frac{ba}{c}.$

Tehát $CP=CQ=\frac{ab}c$ .

Statistics:

100 students sent a solution.

3 points: 94 students.

2 points: 1 student.

1 point: 2 students.

0 point: 3 students.

Problems in Mathematics of KöMaL, April 2013

100 students sent a solution.
3 points:	94 students.
2 points:	1 student.
1 point:	2 students.
0 point:	3 students.

Already signed up?
New to KöMaL?