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Question Number 213606 by CrispyXYZ last updated on 10/Nov/24

△ABC. 2a+b=2c. Find the minimum of (3/(sin C)) + (1/(tan A)).

$$\bigtriangleup{ABC}.\:\mathrm{2}{a}+{b}=\mathrm{2}{c}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\:{A}}. \\ $$

Answered by Ghisom last updated on 10/Nov/24

wlog b=1 ⇒ c=a+(1/2) a^2 +b^2 −2abcos γ =c^2 b^2 +c^2 −2bccos α =a^2 cos γ =((3−4a)/(8a)) cos α =((4a+5)/(4(2a+1))) (3/(sin γ))+(1/(tan α))=((28a+5)/( (√(3(4a−1)(4a+3))))) ((d[((28a+5)/( (√(3(4a−1)(4a+3)))))])/da)=0 ((8(4a−13))/( (√(3(4a−1)^3 (4a+3)^3 ))))=0 a=((13)/4) min ((3/(sin γ))+(1/(tan α))) =4

$$\mathrm{wlog}\:{b}=\mathrm{1}\:\Rightarrow\:{c}={a}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\gamma\:={c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:\alpha\:={a}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\gamma\:=\frac{\mathrm{3}−\mathrm{4}{a}}{\mathrm{8}{a}} \\ $$$$\mathrm{cos}\:\alpha\:=\frac{\mathrm{4}{a}+\mathrm{5}}{\mathrm{4}\left(\mathrm{2}{a}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:\gamma}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{28}{a}+\mathrm{5}}{\:\sqrt{\mathrm{3}\left(\mathrm{4}{a}−\mathrm{1}\right)\left(\mathrm{4}{a}+\mathrm{3}\right)}} \\ $$$$\frac{{d}\left[\frac{\mathrm{28}{a}+\mathrm{5}}{\:\sqrt{\mathrm{3}\left(\mathrm{4}{a}−\mathrm{1}\right)\left(\mathrm{4}{a}+\mathrm{3}\right)}}\right]}{{da}}=\mathrm{0} \\ $$$$\frac{\mathrm{8}\left(\mathrm{4}{a}−\mathrm{13}\right)}{\:\sqrt{\mathrm{3}\left(\mathrm{4}{a}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{4}{a}+\mathrm{3}\right)^{\mathrm{3}} }}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\mathrm{min}\:\left(\frac{\mathrm{3}}{\mathrm{sin}\:\gamma}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}\right)\:=\mathrm{4} \\ $$

Answered by CrispyXYZ last updated on 11/Nov/24

sin B = 2(sin C − sin A) sin(A + C) = 2(sin C − sin A) 2sin((A + C)/2) cos((A + C)/2) = 4cos((A+C)/2) sin((C−A)/2) sin((A + C)/2) = 2sin((C−A)/2) sin(A/2) cos(C/2) + cos(A/2) sin(C/2) = 2sin(C/2) cos(A/2) − 2cos(C/2) sin(A/2) 3tan(A/2) = tan(C/2) Let x = tan(A/2). (3/(sin C)) + (1/(tan A)) = ((3(1+tan^2 (C/2)))/(2tan(C/2))) + ((1−tan^2 (A/2))/(2tan(A/2))) = ((1+9x^2 )/(2x)) + ((1−x^2 )/(2x)) = (1/x)+4x ≥ 4

$$\mathrm{sin}\:{B}\:=\:\mathrm{2}\left(\mathrm{sin}\:{C}\:−\:\mathrm{sin}\:{A}\right) \\ $$$$\mathrm{sin}\left({A}\:+\:{C}\right)\:=\:\mathrm{2}\left(\mathrm{sin}\:{C}\:−\:\mathrm{sin}\:{A}\right) \\ $$$$\mathrm{2sin}\frac{{A}\:+\:{C}}{\mathrm{2}}\:\mathrm{cos}\frac{{A}\:+\:{C}}{\mathrm{2}}\:=\:\mathrm{4cos}\frac{{A}+{C}}{\mathrm{2}}\:\mathrm{sin}\frac{{C}−{A}}{\mathrm{2}} \\ $$$$\mathrm{sin}\frac{{A}\:+\:{C}}{\mathrm{2}}\:=\:\mathrm{2sin}\frac{{C}−{A}}{\mathrm{2}} \\ $$$$\mathrm{sin}\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\frac{{C}}{\mathrm{2}}\:+\:\mathrm{cos}\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\frac{{C}}{\mathrm{2}}\:=\:\mathrm{2sin}\frac{{C}}{\mathrm{2}}\:\mathrm{cos}\frac{{A}}{\mathrm{2}}\:−\:\mathrm{2cos}\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\frac{{A}}{\mathrm{2}} \\ $$$$\mathrm{3tan}\frac{{A}}{\mathrm{2}}\:=\:\mathrm{tan}\frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{Let}\:{x}\:=\:\mathrm{tan}\frac{{A}}{\mathrm{2}}. \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\:{A}} \\ $$$$=\:\frac{\mathrm{3}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{{C}}{\mathrm{2}}\right)}{\mathrm{2tan}\frac{{C}}{\mathrm{2}}}\:+\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}{\mathrm{2tan}\frac{{A}}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}{x}}\:+\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$=\:\frac{\mathrm{1}}{{x}}+\mathrm{4}{x} \\ $$$$\geqslant\:\mathrm{4} \\ $$

Commented by mnjuly1970 last updated on 10/Nov/24

((1/(x )) +4x)_(min) =4

$$\left(\frac{\mathrm{1}}{{x}\:}\:+\mathrm{4}{x}\right)_{{min}} =\mathrm{4} \\ $$

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