The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ with $S_{ABC} = S_{ADC}$ intersect at $E$. The lines through $E$ parallel to $AD$, $DC$, $CB$, $BA$ meet $AB$, $BC$, $CD$, $DA$ at $K$, $L$, $M$, $N$, respectively. Compute the ratio $\frac{S_{KLMN}}{S_{ABC}}$
Problem
Source: Turkey TST 1996 Problem 4
Tags: ratio, geometry, parallelogram, geometry proposed
TDemir
28.09.2011 20:58
If I understood this problem true, it's too easy I think... $ S_{AKEN}= 2S_{NKE} $ $ S_{KBLE}= 2S_{KLE} $ $ S_{ELCM}= 2S_{ELM} $ $ S_{NEMD}= 2S_{NEM} $(Because AKEN, KBLE, ELCM, NEMD are paralelogram) so $ S_{ABCD}= 2S_{KLMN} $ $ S_{ABC}= S_{ADC}=1/2S_{ABCD} $ ----> $ S_{ABC}= S_{KLMN} $ $ \frac{S_{KLMN}}{S_{ABC}} $=1 I think I translated wrong! Because that's too easy for TST question...
xeroxia
14.07.2012 15:44
TDemir wrote: If I understood this problem true, it's too easy I think... ... (Because AKEN, KBLE, ELCM, NEMD are paralelogram) ... I think I translated wrong! Because that's too easy for TST question... The problem is easy, you're right. These are the first steps of Turkish IMO studies. So there are some very simple questions, and some very hard questions because the proposer sometimes asks questions about his academic works. It needs some time to standard the questions like IMO. Anyway, you actually did mistake while solving. $KBLE$ is not a parallelogram. $[ANE]=[AKE]=[KNE]$ since $AKEN$ is parallelogram. $[EMC]=[ECL]=[LEM]$ since $ELCM$ is parallelogram. We know that $[ACD]=[ABC] \Rightarrow [DNEM]=[KBLE]$. $\frac{[NEM]}{[DNEM]}=\frac{AN}{AD}$ and $\frac{[KBL]}{[KBLE]}=\frac{BK}{AB}$. Since $NE \parallel AB$, $\frac{AN}{AD}=\frac{BE}{BD}$. Since $KE \parallel AD$, $\frac{BK}{AB}=\frac{BE}{BD}$. So $[NEM]=[KBL]$ and $[KEL]=[DEM]$. This implies $[KLMN]=\frac{[ABCD]}2$.