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of diseases, and the resultant profile library, with b from the BamHI BAC library and Peroxisomal matrix proteins are synthe
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Solutia subpunctului a):
1). Notam cu\( \{P\}=AC\cap MN \) si cu Q mijlocul lui [MN].
2). [OQ] este linie mijlocie a trapezului ACNM \( \Rightarrow |OQ|=\frac{|AM|+||CN|}{2}=\frac{4+1}{2}=\frac{5}{2}. \)
3). \( \triangle {PCN}\tilde \ \triangle{PAM} \Rightarrow \frac{|PC|}{|PA|}=\frac{|CN|}{|AM|}=\frac{1}{4}\Rightarrow \frac{|PC|}{|AC|}=\frac{|PC|}{|PA|-|PC|}=\frac{1}{4-1}=\frac{1}{3}\Rightarrow |PC|=\frac{|AC|}{3}=\frac{6.\sqrt{2}}{3}=2.\sqrt{2} \Rightarrow |OP|=|OC|+|PC|=3.\sqrt{2}+2.\sqrt{2}=5.\sqrt{2}. \)
4). Asa ca, gasirea distantei \( d=d(O; (BMN)) \)revine la a gasi lungimea inaltimii tetraedrului OBPQ, tridreptunghic in O, in care: \( |OB|=3.\sqrt{2}; |OP|=5.\sqrt{2} \) si \( |OQ|=\frac{5}{2}. \)
5). Pentru aceasta se poate folosi formula: \( \frac{1}{d^2}=\frac{1}{|OB|^2}+\frac{1}{|OP|^2}+\frac{1}{|OQ|^2}. \) Dupa efectuarea tuturor calculelor gasim: \( d=\frac{25.\sqrt{26}}{26}. \)
In ceea ce priveste solutia subpunctului b), as face urmatoarele sugestii:
1). \( |PM|+|PN|=minim \Leftrightarrow \angle{MPA}=\angle{NPC} \)
(in acest caz avem: \( P\in (AC); \)deci punctul P considerat aici nu coincide cu punctul P din rationamentele de la rezolvarea subpunctului a) al problemei)
2). \( \triangle {MPA}\tilde \ \triangle{NPC} ;\ \dots \)