Mathematical and Physical Journal
for High Schools
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Problem C. 1021. (February 2010)

C. 1021. P is a point on side AC, and Q is a point on side BC of triangle ABC. The line through P, parallel to BC intersects AB at K, and the line through Q, parallel to AC intersects AB at L. Prove that if PQ is parallel to AB then AK=BL.

(5 pont)

Deadline expired on March 10, 2010.

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

Megoldás. Legyen a \(\displaystyle P'\) a \(\displaystyle BC\) oldalon és \(\displaystyle Q'\) az \(\displaystyle AC\) oldalon az a pont, amire \(\displaystyle PP'\) és \(\displaystyle QQ'\) párhuzamos \(\displaystyle AB\)-vel. Ekkor \(\displaystyle \vec{PP'}=\vec{KB}\) és \(\displaystyle \vec{Q'Q}=\vec{AL}\). Másrészről pedig \(\displaystyle \vec{AK}=\vec{AL}-\vec{KL}=\vec{Q'Q}-\vec{KL}\) és \(\displaystyle \vec{LB}=\vec{KB}-\vec{KL}=\vec{PP'}-\vec{KL}\). Ha \(\displaystyle PQ\) párhuzamos \(\displaystyle AB\)-vel, akkor \(\displaystyle PP'\), \(\displaystyle Q'Q\) és \(\displaystyle PQ\) egybeesnek. Így \(\displaystyle \vec{AK}=\vec{LB}\), ami azt is jelenti, hogy \(\displaystyle AK=LB\).


291 students sent a solution.
5 points:163 students.
4 points:84 students.
3 points:11 students.
2 points:20 students.
1 point:6 students.
0 point:6 students.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, February 2010