If A,B,C are acute positive angles such that A+B+C=π and cotAcotBcotC=k, then
A
k≤13√3
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B
k>13
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C
k≥13√3
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D
k<19
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Solution
The correct option is Ak≤13√3 A+B+C=π and 0<A,B,C<π2, using A.M. ≥ G.M. ⇒cotA+cotB+cotC3≥k1/3 cotAcotBcotC=k≤13√3 (the minimum value of A.M. is G.M. and in this case A=B=C=π3)