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Problema superba de geometrie in spatiu

Posted: Sat Nov 29, 2008 10:45 pm
by Claudiu Mindrila
In tetraedrul \( ABCD \) punctele \( E \) si \( F \) sunt mijloacele medianelor \( AM \) si \( AN \) ale triunghiurilor \( ABC \) respectiv\( ACD \). Daca \( CE \cap AB={P} \), \( CF \cap AD={Q}, DF \cap AC={R} \), demonstrati ca:
a) \( 9\text{Aria}(PQR)=\text{Aria}(BCD) \);
b) \( 12(PQ+EF+MN)=13BD \).
Virginia si Vasile Tica, lista scurta,2002

Posted: Sat Nov 29, 2008 11:34 pm
by Marius Mainea
a) Din teorema lui Menelaus \( \frac{AP}{BP}=\frac{AR}{RC}=\frac{AQ}{QD}=\frac{1}{2} \)
Deci \( \triangle PQR \sim \triangle BDC \) cu raportul de asemanare \( \frac{1}{3} \)

Rezulta imediat din \( BD=3PQ=2MN=4EF \)