Simo_the_Wolf wrote: (a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$; Assume that $AH\perp BC=H$ and $A_{1}$ is the midpoint of $BC$ , $A_{2}$ is the midpoint of $AH$. Similary we'll have $B_{1},B_{2},C_{1},C_{2}$. I think , that locus is $A_{1}A_{2}\cup B_{1}B_{2}\cup C_{1}C_{2}$.

For a generalization of this problem, see at http://www.mathlinks.ro/Forum/viewtopic.php?t=80163 Kostas Vittas.

Really, in part (b) answer is $I$ - incentre of triangle $ABC$

pavel kozlov wrote: Really, in part (b) answer is $I$ - incentre of triangle $ABC$ I think that this is not correct. In my opinion, if we denote as $D,$ $E,$ $F,$ the midpoints of the side segments $BC,$ $AC,$ $AB$ respectively and as $D',$ $E',$ $F',$ the midpoints of the altitudes of $\bigtriangleup ABC,$ through vertices $A,$ $B,$ $C$ respectively, then the answer in part $(b)$ is the concurrency point of the segment lines $DD',$ $EE',$ $FF',$ not necessary coincided with the incenter of $\bigtriangleup ABC.$ Kostas Vittas. PS. Please, see also the N.T.TUAN's message.

pavel kozlov wrote: Really, in part (b) answer is $I$ - incentre of triangle $ABC$ really it is the Lemoine point.

[ƒ(Gabriel)³²¹º]¼ wrote: pavel kozlov wrote: Really, in part (b) answer is $I$ - incentre of triangle $ABC$ really it is the Lemoine point. Thank you dear ƒ(Gabriel)³²¹º]¼ for your remark. It is a new result to me and I see that it is true by the drawing. Have you in mind a synthetic proof about it? Kostas vittas.

Well ... it's enough to show that the Lemoine point is the center of three inscribed rectangles at the same time. In order to achieve this, you only have to notice that the antiparallels of the sides through Lemoine point are bisected by it and, considering two of them, you obtain the diagonals of an inscribed rectangle. Obviously, this is not a proof, but merely a hint.

I'm not convinced it's the Lemoine point. I make it the isotomic conjugate of the orthocentre of the medial triangle (which I don't think is the Lemoine point is it? :S). Method: By inspection and thinking about homothety, the loci are the three lines between the midpoints of BC and AD (altitude from A) and the other two symmetric situations. it is clear that the midpoint P (say) of AD lies on the side MN of the medial triangle (if we take it to be LMN) and furthermore that NP/PM=BD/DC by homothety factor 2. Looking at this from the point of view of the medial triangle, which is oppositely oriented, we see that the reflection of P in the midpoint of MN will be the foot of the altitude from L to MN, whence the point of concurrency is the isotomic conjugate of the orthocentre of LMN.

Ilthigore wrote: (which I don't think is the Lemoine point is it? :S). Yes, the symmedian point of the anticomplementary triangle is the isotomic conjugate of the orthocenter.

Ah, that's cool, and putting our two arguments together we've proved it. Is there a nice reason for this to be true? I don't know much about isotomic conjugation beyond the definition and existence properties.