Fie ABCD un patrulater convex si M un punct situat in interiorul patrulaterului. Se noteaza cu \( G_1,G_2,G_3,G_4 \) respectiv centrele de greutate ale triunghiurilor MAB , MBC , MCD , MDA.
a) Sa se arate ca patrulaterul \( G_1G_2G_3G_4 \) are arie constanta oricare ar fi pozitia punctului M.
b) Sa se arate ca \( aria (G_1G_2G_3G_4)\le \frac{1}{9}(AB\cdot CD+AD\cdot BC). \)
In ce conditii are loc egalitatea ?
Concursul Gh. Titeica, 2004
Patrulater cu arie constanta
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Marius Mainea
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a) Se arata usor din asemanari ca:
\( G_1G_2=G_3G_4=\frac{AC}{3}\ si\ G_1G_4=G_2G_3=\frac{BD}{3}. \)
\( \Longrightarrow aria(G_1G_2G_3G_4)=G_1G_2\cdot G_2G_3\cdot sin(\angle G_1G_2,\ G_2G_3)=G_1G_2\cdot G_2G_3\cdot sin(\angle AC,\ BD) \)
\( \Longleftrightarrow aria(G_1G_2G_3G_4)=\frac{AC\cdot BD\cdot sin(\angle AC,\ BD)}{9} \)
\( \Longleftrightarrow aria(G_1G_2G_3G_4)=\frac{2}{9}\cdot aria(ABCD) \)
b) Inegalitatea de demonstrat devine: \( 2\cdot aria(ABCD)\leq AB\cdot CD+AD\cdot BC \)
\( 2\cdot aria(ABCD)=AC\cdot BD\cdot sin(\angle AC,\ BD)\leq AC\cdot BD\leq AB\cdot CD+AD\cdot BC \), ultima inegalitate fiind cea a lui Ptolemeu pentru patrulaterul convex \( ABCD. \) Egalitatea are loc cand \( AC\bot BD \) si patrulaterul \( ABCD \) este inscriptibil.
\( G_1G_2=G_3G_4=\frac{AC}{3}\ si\ G_1G_4=G_2G_3=\frac{BD}{3}. \)
\( \Longrightarrow aria(G_1G_2G_3G_4)=G_1G_2\cdot G_2G_3\cdot sin(\angle G_1G_2,\ G_2G_3)=G_1G_2\cdot G_2G_3\cdot sin(\angle AC,\ BD) \)
\( \Longleftrightarrow aria(G_1G_2G_3G_4)=\frac{AC\cdot BD\cdot sin(\angle AC,\ BD)}{9} \)
\( \Longleftrightarrow aria(G_1G_2G_3G_4)=\frac{2}{9}\cdot aria(ABCD) \)
b) Inegalitatea de demonstrat devine: \( 2\cdot aria(ABCD)\leq AB\cdot CD+AD\cdot BC \)
\( 2\cdot aria(ABCD)=AC\cdot BD\cdot sin(\angle AC,\ BD)\leq AC\cdot BD\leq AB\cdot CD+AD\cdot BC \), ultima inegalitate fiind cea a lui Ptolemeu pentru patrulaterul convex \( ABCD. \) Egalitatea are loc cand \( AC\bot BD \) si patrulaterul \( ABCD \) este inscriptibil.