Răspuns :
[tex]AB=AC=l_{3} =R \sqrt{3} \\ \\ AO=R \\ \\ AB+AC-3AO= 2R \sqrt{3} -R=R(2 \sqrt{3} -1)[/tex]
ABC- echilateral =: AB=AC=BC= a sa zicem.
Se stie ca AO= 2/3*lungimea medianei (care aici este si inaltime.
Deci fie AA' _|_BC si aplicam Pitagora in triunghi AA'B =>
=> AA'= √AB²-A'B²=√a²-(a/2)²=√3a²/4=a√3/2
Dar AO=2/3*AA'= 2/3*a√3/2= a√3/3
Deci AB+AC-3AO= a+a- 3*a√3/3= 2a-a√3=a(2-√3)
Se stie ca AO= 2/3*lungimea medianei (care aici este si inaltime.
Deci fie AA' _|_BC si aplicam Pitagora in triunghi AA'B =>
=> AA'= √AB²-A'B²=√a²-(a/2)²=√3a²/4=a√3/2
Dar AO=2/3*AA'= 2/3*a√3/2= a√3/3
Deci AB+AC-3AO= a+a- 3*a√3/3= 2a-a√3=a(2-√3)