Problem B. 3976. (February 2007)

B. 3976. Through an interior point of a triangle of sides a, b, c a parallel line is drawn to each side. Given that the segments of these lines intercepted by the triangle have the same length, find that length.

(4 pont)

Deadline expired on March 19, 2007.

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

Megoldás: Az ábra jelöléseivel élve BC''=A'P= $\alpha$ c, CB'=A''P= $\alpha$ b, AC'=B''P= $\beta$ c, CA''=B'P= $\beta$ a, BA'=C''P= $\gamma$ a, AB''=C'P= $\gamma$ b. A PC'C'' háromszög hasonló a CAB háromszöghöz, ahol a hasonlóság aránya $\gamma$ . Ezért C'C''= $\gamma$ c, vagyis $\alpha$ + $\beta$ + $\gamma$ =1.

A feltétel szerint

B'C''=(1- $\alpha$ )a=C'A''=(1- $\beta$ )b=A'B''=( $\alpha$ + $\beta$ )c.

Innen 1- $\alpha$ =( $\alpha$ + $\beta$ )c/a és 1- $\beta$ =( $\alpha$ + $\beta$ )c/b. összeadva a

$2-\alpha-\beta=(\alpha+\beta)\Bigl(\frac{c}{a}+\frac{c}{b}\Bigr)$

összefüggésre jutunk, ahonnan

$2=(\alpha+\beta)\frac{ab+ac+bc}{ab}.$

Következésképpen a keresett hosszúság

$A'B''=(\alpha+\beta)c=\frac{2abc}{ab+ac+bc}$

éppen az oldalak harmonikus közepének 2/3 része.

Statistics:

93 students sent a solution.

4 points: 78 students.

3 points: 3 students.

2 points: 1 student.

1 point: 2 students.

0 point: 8 students.

Unfair, not evaluated: 1 solutions.

Problems in Mathematics of KöMaL, February 2007

93 students sent a solution.
4 points:	78 students.
3 points:	3 students.
2 points:	1 student.
1 point:	2 students.
0 point:	8 students.
Unfair, not evaluated:	1 solutions.

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