Mathematical and Physical Journal
for High Schools
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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